where we denoted $f_{t}=f(t,\cdot)$ for $t\in T$. Recall the notion of
stationary measure of $f$ in (1.2). Also, recall that the convergence of
measures that we are considering is in the weak* topology. That is,
$\mu_{n}\to\mu$ if and only if

$\int\varphi\,d\mu_{n}\to\int\varphi\,d\mu\quad\text{for all $\varphi\in
C_{b}(X)$}$ (5.1)

where $C_{b}(X)$ denotes the set of bounded real-valued continuous functions
of $X$. However, since $X$ is a locally compact Polish space and the measures
$\mu_{n}$ and $\mu$ are assumed to be probabilities, according to
Portemanteau’s theorem (cf. [54, Theorem 13.16]), (5.1) is equivalent to

$\int\varphi\,d\mu_{n}\to\int\varphi\,d\mu\quad\text{for all $\varphi\in
C_{c}(X)$}.$

Here $C_{c}(X)\subset C_{b}(X)$ denotes the set of compactly supported
continuous functions.

######  Lemma 5.1.

Let $\mu$ be an ergodic stationary measure of $f$. Then,

  1. (i)

$\bar{\mu}(B(\mu))=1$ where $\bar{\mu}=\mathbb{P}\times\mu$;

  2. (ii)

$\mu(B_{\omega}(\mu))=1$ for $\mathbb{P}$-almost every $\omega\in\Omega$;

  3. (iii)

$\mathbb{P}(B_{x}(\mu))=1$ for $\mu$-almost every $x\in X$.

Here,

$B_{\omega}(\mu)=\big{\\{}x\in X:\,(\omega,x)\in B(\mu)\big{\\}}\
\quad\text{and}\quad\ B_{x}(\mu)=\big{\\{}\omega\in\Omega:\,(\omega,x)\in
B(\mu)\big{\\}}$ (5.2)

and

$B(\mu)=\left\\{(\omega,x)\in\Omega\times
X:\,\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\delta_{f_{\omega}^{j}(x)}=\mu\right\\}.$

###### Proof.

Since $\mu$ is an ergodic stationary measure of $f$, the measure
$\bar{\mu}=\mathbb{P}\times\mu$ is an ergodic invariant probability measure of
the corresponding skew-product transformation $F$. We now consider the set
$B(\mu)$ of points $(\omega,x)\in\Omega\times X$ such that for any $\varphi\in
C_{c}(X)$, it holds

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\varphi\circ
f_{\omega}^{j}(x)=\int_{X}\varphi d\mu.$

Similarly we define the set $B(\mu,\varphi)$ of points
$(\omega,x)\in\Omega\times X$ for which the above limit holds for a fixed
continuous function $\varphi:X\to\mathbb{R}$. Taking into account that

$\varphi\circ f_{\omega}^{n}(x)=\bar{\varphi}\circ
F^{n}(\omega,x)\quad\text{and}\quad\int\varphi\,d\mu=\int\bar{\varphi}\,d\bar{\mu}$

where

$\bar{\varphi}(\omega,x)=\varphi(x)\quad\text{for }(\omega,x)\in\Omega\times
X\quad(\bar{\varphi}\in C_{c}(\Omega\times X)),$

it follows from the Birkhoff Ergodic Theorem that $B(\mu,\varphi)$ has full
$\bar{\mu}$-measure. Now, since we assumed that $X$ is locally compact Polish
space, $C_{c}(X)$ is separable (cf. [90, Section 42]). Thus, taking a
countable dense subset $S$ of $C_{c}(X)$, it is not difficult to see that

$B({\mu})=\bigcap_{\varphi\in S}B(\mu,\varphi)$

and therefore $B(\mu)$ has also full $\bar{\mu}$-measure. Finally, we observe
that $B_{\omega}(\mu)$ and $B_{x}(\mu)$ are the $\omega$-section and
$x$-section of $B(\mu)$ respectively, i.e., it holds (5.2). Hence, by Fubini’s
theorem, we have that $\mu(B_{\omega}(\mu))=1$ and $\mathbb{P}(B_{x}(\mu))=1$
for $\mathbb{P}$-almost every $\omega\in\Omega$ and $\mu$-almost every $x\in
X$ respectively. ∎

We highlight the following lemma for future reference. This lemma follows
immediately by the Fubini theorem and the definition of $B_{\omega}(\mu)$ and
$B_{x}(\mu)$ in (5.2).

######  Lemma 5.2.

The following are equivalent:

  1. (1)

$\bar{m}(B(\mu_{1})\cup\dots\cup B(\mu_{r}))=1$ where
$\bar{m}=\mathbb{P}\times m$;

  2. (2)

$m(B_{\omega}(\mu_{1})\cup\dots\cup B_{\omega}(\mu_{r}))=1$ for
$\mathbb{P}$-almost every $\omega\in\Omega$;

  3. (3)

$\mathbb{P}(B_{x}(\mu_{1})\cup\dots\cup B_{x}(\mu_{r}))=1$ for $m$-almost
every $x\in X$.

######  Remark 5.3.

The equivalence in Lemma 5.2 also holds asking "$>0$" instead of "$=1$". Thus,
an ergodic stationary probability $\mu$ of $f$ is said to be a _physical
measure_ if one of the following equivalent conditions holds:

  1. (1)

$\bar{m}(B(\mu))>0$ where $\bar{m}=\mathbb{P}\times m$;

  2. (2)

$m(B_{\omega}(\mu))>0$ for $\mathbb{P}$-almost every $\omega\in\Omega$;

  3. (3)

$\mathbb{P}(B_{x}(\mu))>0$ for $m$-almost every $x\in X$.

This definition agrees with the classical notion of physical measure in the
deterministic case (that is when $\Omega$ is a singleton). However, this new
definition differs from previous notions of physical measure for random maps
introduced in the literature (see for instance [4, Equation (2)] or [6, page
313]). This concept was introduced by asking $m(G(\mu))>0$ where

$G(\mu)=\left\\{x\in
X:\,\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\delta_{f_{\omega}^{j}(x)}=\mu\quad\text{for
$\mathbb{P}$-almost every $\omega\in\Omega$}\right\\}.$

Since $\bar{m}(B(\mu))\geq m(G(\mu))$, one has that $m(G(\mu))>0$ implies
$\bar{m}(B(\mu))>0$. However, a priori, it is not expected the converse. Also,
defining $G(\mu)$ as the statistical basin of attraction does not seem
appropriate, as the following example shows.

######  Example 5.4.

Let $X=T=[-1,1]$ be equipped with the normalized Lebesgue measure and let
$X_{-}=[-1,0)$, $X_{+}=(0,1]$. Consider a continuous map $f_{0}:X\to X$ which
has exactly two sinks $p_{-}=-\frac{1}{2}$, $p_{+}=\frac{1}{2}$ whose basin of
attraction is $X_{-}$, $X_{+}$, respectively, and $0$ is the fixed point of
$f_{0}$. With the notation $B(x,r)$ for the ball of radius $r$ centered at
$x$, we also assume that $f_{0}(X)\subset B(0,\frac{3}{4})$ and
$f_{0}(B(p_{\pm},\frac{1}{4}))\subset B(p_{\pm},\frac{1}{8})$. Then, for the
random map $f:T\times X\to X$ with additive noise given by
$f_{t}(x)=f_{0}(x)+\frac{1}{8}t$, it holds that
$f_{t}(B(p_{\pm},\frac{1}{4}))\subset B(p_{\pm},\frac{1}{4})$ for any $t\in
T$. See Figure 4.

(a) The map $f_{0}$ on $[-1,1]$.

(b) The map $f_{t}$ on $[-1,1]$

Figure 4. Illustrations of the random map $f$ in Example 5.4.

By the argument in Remark 1.4, the annealed Perron–Frobenius operators
associated with the restrictions of $f$ on both $B(p_{-},\frac{1}{4})$ and
$B(p_{+},\frac{1}{4})$ satisfy (FPM). Therefore, $f$ admits at least two
absolutely continuous ergodic stationary measures $\mu_{-}$, $\mu_{+}$ whose
supports are included in $B(p_{-},\frac{1}{4})$, $B(p_{+},\frac{1}{4})$,
respectively. Since the annealed Perron–Frobenius operator associated with $f$
itself also satisfies (FPM), $f$ admits finitely many absolutely continuous
ergodic stationary measures $\mu_{1},\ldots,\mu_{r}$, two of which are
$\mu_{-}$, $\mu_{+}$, such that

$m(B_{\omega}(\mu_{1})\cup\cdots\cup B_{\omega}(\mu_{r}))=1\quad\text{for
$\mathbb{P}$-almost every $\omega$.}$

On the other hand, by the continuity of $f_{0}$, there is a neighborhood $U$
of $0$ such that $f_{0}(U)\subset B(0,\frac{1}{16})$, so that for each $x\in
U$, both $\\{f_{t}(x):t\in T\\}\cap X_{+}$ and $\\{f_{t}(x):t\in T\\}\cap
X_{-}$ have positive Lebesgue measure. Consequently, one can find positive
$\mathbb{P}$-measure sets $\Gamma_{-}$, $\Gamma_{+}$ and $n_{0}\geq 1$ such
that if $\omega\in\Gamma_{\pm}$ then $f_{\omega}^{n}(x)\in
B(p_{\pm},\frac{1}{4})$ for any $n\geq n_{0}$ and $x\in U$. This concludes
that $U\not\subset\bigcup_{j=1}^{r}G(\mu_{j})$. Therefore,

$0<m(G(\mu_{1})\cup\dots\cup G(\mu_{r}))<1.$

In conclusion, if one expects finitely many physical measures where the union
of their basins of attraction covers the whole space almost everywhere, then
the good notion of the basin should be the fiberwise statistical basin.

The following lemma will be essential to prove Theorem C. In this lemma, the
support $\operatorname*{supp}\eta$ of a measure $\eta$ on $X$ (which is not
necessarily absolutely continuous with respect to $m$) is defined as the set
of all points $x$ in $X$ for which every open neighborhood $V$ of $x$ has
positive measure.

######  Lemma 5.5.

Let $\mu$ be a stationary measure of $f$. Then for any probability measure
$\eta$ with $\operatorname*{supp}\eta\subset B_{\omega}(\mu)$,

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}(f^{j}_{\omega})_{*}\eta=\mu.$

In particular, if $x\in B_{\omega}(\mu)$, then

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\varphi(f^{j}_{\omega}(x))=\int\varphi\,d\mu\quad\text{for
all $\varphi\in C_{b}(X)$.}$ (5.3)

###### Proof.

Take any probability measure $\eta$ with support contained in
$B_{\omega}(\mu)$. Given any $\varphi\in C_{c}(X)$, by definition of
$B_{\omega}(\mu)$, we have that for each $x\in\operatorname*{supp}\eta$ it
holds

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\varphi(f^{j}_{\omega}(x))=\int\varphi\,d\mu.$

Taking integrals over $\operatorname*{supp}\eta$ with respect to the
probability measure $\eta$ on both sides of the equality, the dominated
convergence theorem gives

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\int\varphi(f^{j}_{\omega}(x))\,d\eta=\int\varphi\,d\mu.$

Since

$\int\varphi\,d(f^{j}_{\omega})_{*}\eta=\int\varphi\circ
f^{j}_{\omega}\,d\eta\quad\text{for every $j\geq 0$}$

we get the first part of the lemma. To prove (5.3), it suffices to apply the
first part of this lemma to $\eta=\delta_{x}$ for $x\in B_{\omega}(\mu)$. ∎

Let us show that (ii) implies (i) in Theorem C. To do this, according to the
equivalences shown in Theorem D, it suffices to show the following
proposition.

######  Proposition 5.6.

If $f$ satisfies (FPM), then ${\mathcal{L}}_{f}$ satisfies the condition
(FED).

###### Proof.

Since $f$ satisfies (FPM) we have only finitely many absolutely continuous
(with respect to $m$) ergodic stationary probability measures
$\mu_{1},\dots,\mu_{r}$ which have pairwise disjoint supports and
$m(B_{\omega}(\mu_{1})\cup\dots\cup B_{\omega}(\mu_{r}))=1$ for
$\mathbb{P}$-almost every $\omega\in\Omega$. Let $h_{i}$ be the Radon–Nikodým
derivative of the measure $\mu_{i}$ with respect to $m$ for $i=1,\dots,r$.
Observe that $h_{i}$ is an invariant density of ${\mathcal{L}}_{f}$ because
$\mu_{i}$ is a stationary probability measure of $f$. Moreover, since the
supports of such measures are pairwise disjoints, we also get that the
densities $h_{1},\dots,h_{r}$ have mutually disjoint supports. To conclude
(FED) for ${\mathcal{L}}_{f}$ we need to prove that the density
$h=\frac{1}{r}(h_{1}+\dots+h_{r})$ has the maximal support. This means that

$\lim_{n\to\infty}{\mathcal{L}}_{f}^{n*}1_{\operatorname*{supp}h}(x)=1\quad\text{for
$m$-almost every $x\in X$}$ (5.4)

where ${\mathcal{L}}_{f}^{\,*}$ is the adjoint operator of
${\mathcal{L}}_{f}$.

Given $x\in X$ and $\omega\in B_{x}(\mu_{i})$ for some $i\in\\{1,\dots,r\\}$,
we have that $x\in B_{\omega}(\mu_{i})$. Therefore, by Lemma 5.5,

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}1_{\operatorname*{supp}h}\circ
f^{j}_{\omega}(x)=\mu_{i}({\operatorname*{supp}h})=1.$

That is, for every $x\in X$,

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}1_{\operatorname*{supp}h}\circ
f^{j}_{\omega}(x)=1\quad\text{for all $\omega\in B_{x}(\mu_{1})\cup\dots\cup
B_{x}(\mu_{r})$.}$ (5.5)

Taking into account Lemma 5.2, we have that
$\mathbb{P}(B_{x}(\mu_{1})\cup\dots\cup B_{x}(\mu_{r}))=1$ for $m$-almost
every $x\in X$. Then, using the notation introduced in Appendix A, by Lemma
A.4 (4), the dominated convergence theorem, Lemma A.4 (2) and (5.5) imply
that, for $m$-almost every $x\in X$,

$\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}{\mathcal{L}}_{f}^{j*}1_{\operatorname*{supp}h}(x)$
$\displaystyle=\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\int\mathcal{L}_{\omega}^{j*}1_{\operatorname*{supp}h}(x)\,d\mathbb{P}(\omega)$
$\displaystyle=\int_{\cup_{i}B_{x}(\mu_{i})}\,\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}1_{\operatorname*{supp}h}\circ
f^{j}_{\omega}(x)\,d\mathbb{P}(\omega)=1.$

Further, according to Remark 4.5, we have that
$({\mathcal{L}}_{f}^{j*}1_{\operatorname*{supp}h})_{j\geq 1}$ is monotonic
increasing and thus, mean converges imply sequence converges (this follows as
a consequence of the Stolz–Cesàro theorem). That is, we obtain (5.4)
concluding the proof. ∎

Now we will prove that (i) implies (ii). According to Theorem D, if
${\mathcal{L}}_{f}$ is mean constrictive, then ${\mathcal{L}}_{f}$ satisfies
(APM). That is, ${\mathcal{L}}_{f}$ admits finitely many ergodic invariant
densities $h_{1},\dots,h_{r}$ in $D(m)$ with mutually disjoint supports and
$\lambda_{1},\dots,\lambda_{r}$ bounded linear functionals such that for any
$\varphi\in L^{1}(m)$,

$\displaystyle\lim_{n\to\infty}\left\lVert
A_{n}\varphi-\sum_{i=1}^{r}\lambda_{i}(\varphi)h_{i}\right\rVert=0.$ (5.6)

Since $h_{i}$ is an ergodic invariant density $h_{i}$ for ${\mathcal{L}}_{f}$,
the measure $\mu_{i}$ is an absolutely continuous (with respect to $m$)
ergodic stationary measures of $f$ where $d\mu_{i}=h_{i}\,dm$. Moreover, since
the support $\operatorname*{supp}\nu$ of a measure $\nu$ absolutely continuous
with respect to $m$ is defined as the support of the Radon–Nikodým derivative
$\frac{d\nu}{dm}$ of $\nu$ with respect to $m$, we also obtain that
$\operatorname*{supp}\mu_{1},\dots,\operatorname*{supp}\mu_{r}$ are pairwise
disjoints. To prove that $f$ satisfies (FPM) it remains to show that the union
of fiberwise basins of attraction of these measures is $X$ modulus a set of
zero $m$-measure. This will be obtained in the following proposition.

######  Proposition 5.7.

Assume that ${\mathcal{L}}_{f}$ satisfies (APM) as above. Then

$m(X\setminus(B_{\omega}(\mu_{1})\cup\dots\cup B_{\omega}(\mu_{r})))=0$ for
$\mathbb{P}$-almost every $\omega\in\Omega$.

###### Proof.

Let $\Omega_{0}=\\{\omega\in\Omega:\mu_{i}(B_{\omega}(\mu_{i}))=1,\ \text{for
$i=1,\dots,r$}\\}$. From item (i) in Lemma 5.1 we get
$\mathbb{P}(\Omega_{0})=1$. Set

$A_{\omega}=X\setminus\left(B_{\omega}(\mu_{1})\cup\dots\cup
B_{\omega}(\mu_{r})\right).$

Then, for any $\omega\in\Omega_{0}$ it holds that $\mu_{i}(A_{\omega})=0$ for
all $i=1,\dots,r$ because $A_{\omega}\subset X\setminus B_{\omega}(\mu_{i})$.
Consequently, since $d\mu_{i}=h_{i}\,dm$

$\int_{A_{\omega}}h_{i}\,dm=0\quad\text{for all $i=1,\dots,r$ and
$\omega\in\Omega_{0}$}.$ (5.7)

Set

$\Omega_{1}=\bigcap_{n=0}^{\infty}\sigma^{-n}(\Omega_{0}),$

which is also a full $\mathbb{P}$-measure set. Notice that
$\omega\in\Omega_{1}$ implies $\sigma^{n}\omega\in\Omega_{0}$ for any $n\geq
0$. Thus, from (5.7)

$\int_{A_{\sigma^{n}\omega}}h_{i}\,dm=0\quad\text{ for each $i=1,\ldots,r$ and
$n\geq 1$ and $\omega\in\Omega_{1}$.}$ (5.8)

For simplicity of notation, we write $\bar{m}=\mathbb{P}\times m$.

######  Claim 5.8.

For $\bar{m}$-almost every $(\omega,x)\in\Omega\times X$, there is $k\geq 1$
such that $f_{\omega}^{k}(x)\in X\setminus A_{\sigma^{k}\omega}$.

###### Proof.

By contradiction, assume that the claim does not hold. Therefore,

$\bar{m}(B_{0})>0,\quad B_{0}=\\{(\omega,x)\in\Omega\times
X:\,f_{\omega}^{n}(x)\in A_{\sigma^{n}\omega}\ \ \text{for all $n\geq 1$}\\}.$

Set $B=B_{0}\cap(\Omega_{1}\times X)$, which is still a positive
$\bar{m}$-measure set because $\Omega_{1}\times X$ is a full measure set.
Denoting by $B_{\omega}=\\{x\in X:(\omega,x)\in B\\}$ the $\omega$-section of
$B$, Fubini’s theorem implies that for every $n\geq 1$,

$\int_{\Omega}\int_{B_{\omega}}1_{A_{\sigma^{n}\omega}}\circ f^{n}_{\omega}\
dm\,d\mathbb{P}=\int_{\Omega}\int_{B_{\omega}}\,d{m}\,d\mathbb{P}=\bar{m}(B)>0.$

Hence, with the notation from Appendix A, see Lemma A.4 (2),

$\frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}\int_{B_{\omega}}\mathcal{L}_{\omega}^{i*}1_{A_{\sigma^{i}\omega}}\,dm\,d\mathbb{P}=\bar{m}(B)>0.$
(5.9)

Given $\varphi\in L^{1}(m)$ and $\omega\in\Omega$, let us write

$\mathcal{L}^{n}_{\omega}\varphi=\sum_{j=1}^{r}\lambda_{j}(\varphi)h_{j}+\mathcal{Q}^{n}_{\omega}\varphi\quad\text{and}\quad
A_{n}\varphi=\sum_{j=1}^{r}\lambda_{j}(\varphi)h_{j}+\mathcal{Q}_{n}\varphi$

where we recall that (see Lemma A.4 (4)),

$A_{n}\varphi=\frac{1}{n}\sum_{i=0}^{n-1}{\mathcal{L}}^{i}_{f}\varphi=\frac{1}{n}\sum_{i=0}^{n-1}\int\mathcal{L}_{\omega}^{i}\varphi\,d\mathbb{P}.$

Then,

$\displaystyle\frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}\int_{X}$
$\displaystyle\mathcal{Q}_{\omega}^{i}1_{X}\,\,{dm}\,d\mathbb{P}=\int\big{(}A_{n}1_{X}\
-\sum_{j=1}^{r}\lambda_{j}(1_{X})h_{j}\big{)}\,{dm}=\int\mathcal{Q}_{n}1_{X}\,{dm}\leq\|\mathcal{Q}_{n}1_{X}\|.$

On the other hand, by Lemma A.4 (3), Equation (5.8) and the above inequality,

$\displaystyle\frac{1}{n}\sum_{i=0}^{n-1}$
$\displaystyle\int_{\Omega}\int_{B_{\omega}}\mathcal{L}_{\omega}^{i*}1_{A_{\sigma^{i}\omega}}\,dm\,d\mathbb{P}$
$\displaystyle\leq\frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}\int_{X}\mathcal{L}_{\omega}^{i*}1_{A_{\sigma^{i}\omega}}\,{dm}\,d\mathbb{P}=\frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}\int_{A_{\sigma^{i}\omega}}\mathcal{L}_{\omega}^{i}1_{X}\,\,{dm}\,d\mathbb{P}$
$\displaystyle=\frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}\int_{A_{\sigma^{i}\omega}}\mathcal{Q}_{\omega}^{i}1_{X}\,\,{dm}\,d\mathbb{P}\leq\frac{1}{n}\sum_{i=0}^{n-1}\int_{\Omega}\int_{X}\mathcal{Q}_{\omega}^{i}1_{X}\,\,{dm}\,d\mathbb{P}\leq\|\mathcal{Q}_{n}1_{X}\|\to
0.$

The last limit (as $n\to\infty$) follows from the assumption (5.6). But this
limit provides a contradiction with (5.9). ∎

Fix $(\omega,x)\in\Omega\times M$ (up to zero measure sets) and let $k\geq 1$
be the integer of Claim 5.8. Set $y=f^{k}_{\omega}(x)\in X\setminus
A_{\sigma^{k}\omega}$. By definition of $A_{\sigma^{k}\omega}$, there is $i$
such that $y\in G_{\sigma^{k}\omega}(\mu_{i})$. Then, in view of (5.2),
$(\sigma^{k}\omega,y)\in B(\mu_{i})$ and hence

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\delta_{f^{j}_{\sigma^{k}\omega}(y)}=\mu_{i}.$

On the other hand,

$\frac{1}{n}\sum_{j=0}^{n-1}\delta_{f^{j}_{\omega}(x)}=\frac{n-k}{n}\left(\frac{1}{n-k}\sum_{j=0}^{n-k-1}\delta_{f_{\sigma^{k}\omega}^{j}(y)}\right)+\frac{1}{n}\sum_{j=0}^{k-1}\delta_{f^{j}_{\omega}(x)}\to\mu_{i}$

as $n\to\infty$. This show that $x\in B_{\omega}(\mu_{i})$ concluding the
proof. ∎

Finally, we summarize and complete the last details of the proof of Theorem C:

###### Proof of Theorem C.

If $f$ satisfies (FPM), then $\mathcal{L}_{f}$ is (FED) by Proposition 5.6 and
hence Theorem D implies that $\mathcal{L}_{f}$ is (MC). This proves (ii)
implies (i) in Theorem C. On the other hand, if $\mathcal{L}_{f}$ is (MC),
then it is also (APM) by Theorem D and hence, one concludes that $f$ satisfies
(FPM) from Proposition 5.7. This shows (i) implies (ii). Finally, (1) and (2)
follow from Lemma 5.2. ∎

##  6\. Sub-hierarchy in (UC)

In this section, we complete the proof of the implications in the hierarchies
of Figures 2, as well as provide practical sufficient conditions for (FPM)
other than conditions in Section 1.1. In the sequel, $(X,\mathscr{B},m)$
denotes a Polish probability space, $P:L^{1}(m)\to L^{1}(m)$ is a Markov
operator and $P^{n}(x,A)$ is an $n$-transition probability that induces
$P^{n}$. We first give equivalent conditions for (UC).

######  Proposition 6.1.

The following assertions are equivalent:

  1. (1)

$P$ satisfies (UC);

  2. (2)

$P:L^{1}(m)\to L^{1}(m)$ is a quasi-compact operator;

  3. (3)

for any $\varepsilon>0$, there are $\delta>0$ and $n_{0}\geq 1$ such that

$\sup_{\varphi\in D(m)}\int_{A}P^{n_{0}}\varphi<\varepsilon\quad\text{for all
$A\in\mathscr{B}$ with $m(A)<\delta$;}$

  4. (4)

there are $n_{0}\geq 1$, $0<\varepsilon<1$ and $\delta>0$ such that

$\sup_{\varphi\in D(m)}\int_{A}P^{n_{0}}\varphi<\varepsilon\quad\text{for all
$A\in\mathscr{B}$ with $m(A)<\delta$;}$

  5. (5)

for any $\varepsilon>0$, there are $\delta>0$ and $n_{0}\geq 1$ such that
$P^{n_{0}}(x,A)<\varepsilon$ for all $A\in\mathscr{B}$ with $m(A)<\delta$ and
$m$-almost every $x\in X$ (depending on $A$);

  6. (6)

there are $n_{0}\geq 1$, $0<\varepsilon<1$ and $\delta>0$ such that
$P^{n_{0}}(x,A)<\varepsilon$ for all $A\in\mathscr{B}$ with $m(A)<\delta$ and
$m$-almost every $x\in X$ (depending on A);

  7. (7)

for any $\varepsilon>0$, there are $\delta>0$, $n_{0}\geq 1$ and a probability
$\mu$ absolutely continuous with respect to $m$ such that
$P^{n_{0}}(x,A)<\varepsilon$ for all $A\in\mathscr{B}$ with $\mu(A)<\delta$
and $m$-almost every $x\in X$ (depending on $A$);

  8. (8)

there are $n_{0}\geq 1$, $0<\varepsilon<1$, $\delta>0$ and a probability $\mu$
absolutely continuous with respect to $m$ such that
$P^{n_{0}}(x,A)<\varepsilon$ for all $A\in\mathscr{B}$ with $\mu(A)<\delta$
and $m$-almost every $x\in X$ (depending on $A$).

###### Proof.

The equivalence between (1)–(4) follows from [20, Theorem 2]. On the other
hand, observe that

$\int_{A}P^{n}\varphi\,dm=\int\varphi(x)\,P^{n*}1_{A}(x)\,dm=\int\varphi(x)P^{n}(x,A)\,dm$
(6.1)

for all $\varphi\in L^{1}(m)$, $A\in\mathscr{B}$ and $n\geq 1$. Taking into
account that if $\varphi\in D(m)$ then $\int\varphi\,dm=1$, from (6.1) one
immediately gets that (5) implies (3) and (6) implies (4).

We are going to see now that (3) implies (5). By contradiction, assume that
there exists $\varepsilon>0$ satisfying that for any $\delta>0$ and $n\geq 1$
there are $A\in\mathscr{B}$ with $m(A)<\delta$ and set $E\subset X$ with
$m(E)>0$ such that $P^{n}(x,A)\geq\varepsilon$ for all $x\in E$. Take
$\varphi=\frac{1}{m(E)}\,1_{E}\in D(m)$. Hence, (3) implies that
$\int_{A}P^{n}\varphi\,dm<\varepsilon$. However, by (6.1)

$\int_{A}P^{n}\varphi\,dm=\int\varphi(x)P^{n}(x,A)\,dm=\frac{1}{m(E)}\int_{E}P^{n}(x,A)\,dm\geq\varepsilon$

we get a contradiction.

The implication of (6) from (4) follows analogously as the implication of (5)
from (3). This concludes the equivalence from (1) to (6). To complete the
proof, we will show that (5) is equivalent to (7) and (6) is equivalent to
(8).

Observe that clearly, (5) implies (7) and (6) implies (8) by taking $\mu=m$.
As before, the converse implications are analogous and thus we only prove (7)
implies (5). To do this, by (7) we have that $\mu$ is absolutely continuous
with respect to $m$. In particular, according to [43, Lemma 1], $\mu$ is
uniform absolutely continuous with respect to $m$. That is, for any $\eta>0$
there is $\alpha>0$ such that $\mu(A)<\eta$ for all $A\in\mathscr{B}$ with
$m(A)<\alpha$. Then taking $\eta=\delta$ in (7), we get that for any
$\varepsilon>0$, there are $\alpha>0$ and $n\geq 1$ such that
$P^{n}(x,A)<\varepsilon$ for all $A\in\mathscr{B}$ with $m(A)<\alpha$ and
$m$-almost every $x\in X$ (depending on $A$). This concludes (5). ∎

As already mentioned in Section 1.3.4, it follows from ((8) of) Proposition
6.1 that (D) implies (UC). The following proposition shows that the converse
is also true if $P$ is strong Feller.

######  Proposition 6.2.

Let $P:L^{1}(m)\to L^{1}(m)$ be a Markov operator. Assume that $P$ is strong
Feller, that is, there exists $k\in\mathbb{N}$ such that an associated $k$-th
transition probability $P^{k}(x,A)$ is strong Feller continuous. Then $P$
satisfies (UC) if and only if $P$ satisfies (D).

###### Proof.

In view of Proposition 6.1, (D) implies (UC) even if we do not assume that $P$
is strong Feller. We will prove now the converse.

Assume first that $P(x,A)$ is strong Feller continuous and (UC). That is,
$k=1$ in the statement of the proposition. The first observation in this case
is that $P^{n}(x,A)$ is strong Feller continuous for all $n\geq 1$. To see
this, recall that $P^{n+1}(x,A)=\int P(y,A)\,P^{n}(x,dy)$ for $n\geq 1$ where
$P^{1}(x,A)=P(x,A)$ is strong Feller continuous by assumption. Arguing by
induction, if $P^{n}(x,A)$ is strong Feller continuous, it is Feller
continuous and thus $P^{n}(x,\cdot)$ varies continuously in the weak*
topology. Moreover, according to Proposition 2.2, since $P(x,A)$ is strong
Feller continuous, $y\mapsto P(y,A)$ is a bounded continuous function for all
$A\in\mathscr{B}$. Thus,

$P^{n+1}(x^{\prime},A)=\int P(y,A)\,P^{n}(x,dy)\to\int
P(y,A)\,P^{n}(x,dy)=P^{n+1}(x,A)\quad\text{if $x^{\prime}\to x$}.$

Therefore, $P^{n+1}(x,A)$ is strong Feller continuous. Now, having the
continuity of $x\mapsto P^{n_{0}}(x,A)$ in mind, we get that (8) in
Proposition 6.1 holds for all $x\in X$. This concludes (D).

Let us prove the proposition for $k>1$, i.e. assume now that $P$ satisfies
(UC) and $P^{k}(x,A)$ is strong Feller continuous for $k>1$. To see the
conclusion, consider the operator $Q=P^{k}$. By definition, $Q$ satisfies
(UC). Moreover, the associated transition probability of $Q$ is
$Q(x,A)=P^{k}(x,A)$. Thus, we can apply the case with $k=1$ for $Q$,
concluding that $Q$ satisfies (D). In particular, $P$ satisfies (D). ∎

######  Remark 6.3.

If a random map $f$ satisfies the Araújo or Brin–Kifer conditions, Proposition
2.6, Theorem 2.7 and Proposition 6.2 implies that the Perron–Frobenius
operator $\mathcal{L}_{f}$ satisfies (D).

Another important consequence of Proposition 6.1 is the following practical
sufficient conditions for (FPM).

######  Theorem 6.4.

Assume that $P^{n_{0}}(x,dy)=p(x,y)\,dm(y)$ and one of the following holds:

  1. (1)

there are $\gamma>0$ and $\alpha>0$ such that

    1. (a)

$m(\operatorname*{supp}p(x,\cdot))>\gamma$ for $m$-almost every $x\in X$, and

    2. (b)

$\alpha\leq p(x,y)$ for $m$-almost every $x\in X$ and $m$-almost every
$y\in\operatorname*{supp}p(x,\cdot)$;

  2. (2)

there is $\beta>0$ such that $p(x,y)\leq\beta$ for $m$-almost every $x\in X$
and $m$-almost every $y\in\operatorname*{supp}p(x,\cdot)$.

Then, $\mathcal{L}_{f}$ satisfies (UC). In particular, if $X$ is locally
compact, $f$ satisfies (FPM).

###### Proof.

By Proposition 6.1 (6), the condition (2) implies that $\mathcal{L}_{f}$
satisfies (UC). Therefore, it follows from the implications in Figure 1 and
Theorem C that $f$ satisfies (FPM).

Assume the condition (1). Then, for any $A\in\mathscr{B}$ with
$m(A)>1-\frac{\gamma}{2}$, it holds

$\displaystyle P^{n_{0}}(x,A)$
$\displaystyle=\int_{A}p(x,y)\,dm(y)=\int_{A\cap\mathrm{supp}\,p(x,\cdot)}p(x,y)\,dm(y)$
$\displaystyle\geq\alpha
m(A\cap\mathrm{supp}\,p(x,\cdot))>\alpha\,\frac{\gamma}{2}.$

In the last inequality, we used that
$m(A\cap\mathrm{supp}\,p(x,\cdot))=m(A)+m(\mathrm{supp}\,p(x,\cdot))-m(A\cup\mathrm{supp}\,p(x,\cdot))>(1-\frac{\gamma}{2})+\gamma-1=\frac{\gamma}{2}$.
Hence, by considering complements and using Proposition 6.1 (6) again, we get
(UC) for $\mathcal{L}_{f}$, and $f$ satisfies (FPM). ∎

######  Remark 6.5.

As previously mentioned, Theorem 6.4 (2) is another weakening of Brin–Kifer’s
condition (see (2) in page 1.1). Moreover, we find Araújo–Aytaç [8] for
Theorem 6.4 (1). Indeed, they considered the case when $X$ is a compact
manifold equipped with the normalized Lebesgue measure $m$, and assumed that
there exist $\alpha,\beta,\gamma>0$ and $t_{*}\in T$ such that for every $x\in
X$,

  * (i)

$\operatorname*{supp}p(x,\cdot)$ includes the ball of radius $\gamma$ centered
at $f_{t_{*}}(x)$, and

  * (ii)

$\alpha\leq p(x,y)\leq\beta$ for $m$-almost every
$y\in\operatorname*{supp}p(x,\cdot)$.

Obviously, the conditions in Theorem 6.4 relaxed Araújo–Aytaç’s condition to
get that (UC).101010They also assumed an aperiodicity condition to ensure
uniform ergodicity, meaning that $X$ cannot be decomposed into $\ell$ subsets
$X=X_{1}\cup\cdots\cup X_{\ell}$ ($\ell\geq 2$) such that
$p(x,X_{i+1\pmod{\ell}})$ for all $x\in X_{i}$ and $1\leq i\leq\ell$. We do
not assume it, as indicated in the second example in Section 1.4.1, which
satisfies the conditions in Theorem 6.4 but violates the aperiodicity
condition. Moreover, in view of the following section, under the assumptions
in [8, Thm. A], one gets (D*).

###  6.1. (D*) and uniform ergodicity

Let us also give equivalent conditions for (D*).

######  Proposition 6.6.

The following assertions are equivalent:

  1. (1)

$P$ satisfies (D*);

  2. (2)

there is a probability measure $\pi$ absolutely continuous with respect to $m$
such that

$\lim_{n\to\infty}\sup_{x\in X}\|P^{n}(x,\cdot)-\pi\|_{TV}=0;$

  3. (3)

there are a probability measure $\pi$ absolutely continuous with respect to
$m$ and constants $C>0$, $0<\lambda<1$ such that

$\|P^{n}(x,\cdot)-\pi\|_{TV}\leq C\lambda^{n}\quad\text{for all $x\in X$}.$

###### Proof.

The equivalence between (2) and (3) follows from [69, Theorem 16.0.2]. It is
also clear that (2) implies (1) by taking the measure $\mu$ in (1) as the
measure $\pi$ in (2). Finally, the implication of (2) from (1) follows
basically from [33, Theorem 2]. Namely, Dorea and Pereira proved that (1)
implies the convergence in (2) but they do not conclude that $\pi$ is
absolutely continuous with respect to $m$. To prove this, observe that the
convergence in (2) implies that $\pi$ is the unique measure such that

$\pi(A)=\int P(x,A)\,d\pi(x)\quad\text{for all $A\in\mathscr{B}$}.$

See Remark 6.7 for more details. On the other hand, (1) implies condition (D)
adapted to Markov operators in $L^{1}(m)$ (see Section 1.3.4). In view of
Theorem E (together with the observation that $P(x,A)=\int_{A}P^{n}1_{X}dm$),
one has that (D) implies (S). That is, there is $g\in D(m)$ such that $Pg=g$.
Hence,

$\displaystyle m_{g}(A)$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\rm
def}}{{=}}\int_{A}g\,dm=\int_{A}Pg\,dm=\int g\,P^{*}1_{A}\,dm=\int
P(x,A)\,dm_{g}\quad\text{for all $A\in\mathscr{B}$}.$ (6.2)

Therefore, by the uniqueness, $m_{g}=\pi$, proving that $\pi$ is absolutely
continuous. ∎

######  Remark 6.7.

In Markov processes theory, condition (2) in Proposition 6.6 without the
absolute continuity of $\pi$ with respect to $m$ is called uniform ergodicity
(cf. [69]). Apart from the background, it would still make sense to call any
of the equivalent conditions in Proposition 6.6 uniform ergodicity (for Markov
operators on $L^{1}(m)$) because the conditions imply that there exists a
unique invariant density of $P$. To see this, observe first that (2) in
Proposition 6.6 implies that $\pi$ is the unique invariant probability in the
sense that

$\pi(A)=\int P(x,A)\,d\pi(x)\quad\text{for all $A\in\mathscr{B}$}.$

Indeed, if $\nu$ is a probability satisfying also the above relation, then

$\nu(A)=\int P(x,A)\,d\nu(x)=\int P(x,A)P(y,dx)\,d\nu(y)=\dots=\int
P^{n}(x,A)\,d\nu.$

Since $P^{n}(x,A)$ converges uniformly to $\pi(A)$ on $x$ as $n\to\infty$, the
right-hand side of the above expression converges to $\pi(A)$, and therefore,
$\nu=\pi$. Since $\pi$ is absolutely continuous with respect to $m$, let us
write $d\pi=h\,dm$ with $h\in D(m)$. Then,

$\displaystyle\int_{A}h(x)\,dm$ $\displaystyle=\pi(A)=\int P(x,A)\,d\pi=\int
h(x)P(x,A)\,dm$ $\displaystyle=\int h(x)P^{*}1_{A}(x)\,dm=\int_{A}Ph(x)\,dm.$

This implies that $Ph=h$. Moreover, by the uniqueness shown above, arguing as
in (6.2), we conclude that $h$ is the unique invariant density of $P$.

##  7\. Examples and counterexamples: proof of propositions

In this section, we prove all propositions given in Section 1.4.

###  7.1. Proof of Propositions 1.11 and 1.12: Additive type noise

We first prove Proposition 1.11. Recall that our random map is given by
(1.10).

To see (UC), notice that if $x\neq 0$ then

$\\{t\in T:f_{t}(x)\in A\\}=\\{t\in T:f_{0}(x)+t\in A\pmod{1}\\}$

is the translated set of $A$ by $f_{0}(x)$. Hence, by the invariance of the
Lebesgue measure for translations,

$P(x,A)=p\left(\\{t\in T:f_{t}(x)\in A\\}\right)=p(A)=m(A).$ (7.1)

Now, (UC) immediately follows from the interpretation of (UC) in Section 1.3.4
(e.g. for $n_{0}=1$, $\delta=\varepsilon=\frac{1}{2}$ and $\mu=m$).

On the other hand, since $\\{t\in T:f_{t}(0)=0\\}=T$, we have
$P(0,\\{0\\})=1$, namely $P(0,\cdot)=\delta_{0}$. Thus it follows from (1.4)
that $P^{n}(0,\\{0\\})=1$ for any $n\geq 1$. This means that (D) is violated
with $x=0$ and $A=\\{0\\}$.

(a) The maps $\iota$ and $f_{0}$ on $[-1,1]$.

(b) The map $f_{t}$ on $[-1,1]$

Figure 5. Illustrations of the random map $f$ in Proposition 1.11.

We next prove Proposition 1.12. Recall the random map displayed in Figure 5.
Notice that $p(A)=2m(A)$ for any Borel set $A\subset X_{+}$. Furthermore, by
the argument obtaining (7.1),

$p\left(\left\\{t:\widetilde{f}_{t}(y)\in
B\right\\}\right)=2m(B)\quad\text{for any $y\in X_{+}$ and Borel set $B\subset
X_{+}$}$

and thus

$P(x,A)=\begin{cases}p\left(\left\\{t:\widetilde{f}_{t}(x)\in\iota(A)\right\\}\right)=2m(\iota(A))=2m(A)&\text{for
$x\in X_{+}$}\\\ p\left(\left\\{t:\widetilde{f}_{t}(\iota(x))\in
A\right\\}\right)=2m(A)&\text{for $x\in X_{-}$}\end{cases}$

for each $A\in\mathscr{B}$. By (1.4), $P^{n}(x,A)=2m(A)$ for each $n\geq 1$,
$x\in X$ and $A\in\mathscr{B}$. Therefore, obviously (D) holds (e.g. for
$n_{0}=1$, $\delta=\varepsilon=\frac{1}{2}$ and $\mu=m$). On the other hand,
given $n_{0}\geq 1$, $0<\varepsilon<1$, $\delta>\frac{1}{2}$ and a probability
measure $\mu$, it holds that $\mu(X_{+})\leq\frac{1}{2}$ or
$\mu(X_{-})\leq\frac{1}{2}$. For simplicity assume that
$\mu(X_{+})\leq\frac{1}{2}$, and set $A=X_{+}$. Then, $\mu(A)<\delta$ but
$P^{n_{0}}(x,A)=2m(A)=1>\varepsilon$. This means that (D*) is violated.

###  7.2. Proof of Propositions 1.13 and 1.14: Multiplicative noise

Recall condition (b) in Remark 1.2. Also recall that condition almost-(b) is
the natural relaxation of the condition (b) from “for all $x\in X$” to “for
$m$-almost every $x\in X$”. Let $f$ be the random map given in (1.11), i.e.
random perturbation of a measurable map $f_{0}:[0,1]\to[0,1]$ by
multiplicative noise. To prove Propositions 1.13, we need the following lemma.

######  Lemma 7.1.

Denote the set of fixed points and zeros of $f_{0}$ by $F$ and $Z$,
respectively, that is, $F=\left\\{x:f_{0}(x)=x\right\\}$ and
$Z=\left\\{x:f_{0}(x)=0\right\\}$. Then, the following holds:

  1. (1)

If $0\in F\cap Z$, then $f$ does not satisfy (b).

  2. (2)

If $Z=\emptyset$, then $f$ satisfies (b).

  3. (3)

If $m(Z)=0$, then $f$ satisfies almost-(b).

###### Proof.

Assume that $0\in F\cap Z$. Then,
$\left\\{f_{\omega}^{n}(0):\omega\in\Omega\right\\}=\\{0\\}$ for all $n\geq
1$, implying that $P_{n}(0,\cdot)=\delta_{0}$ for all $n\geq 1$. Since
$\delta_{0}$ is not absolutely continuous with respect to the Lebesgue measure
$m$ of $[0,1]$, (b) does not hold.

Take a point $x$ such that $x\not\in Z$. Then $f_{0}(x)>0$, so
$\left\\{f_{\omega}(x):\omega\in\Omega\right\\}$ is a closed interval which is
not a point set, and $P(x,\cdot)$ is the normalized Lebesgue measure on the
closed interval, implying that $P(x,\cdot)$ is absolutely continuous for all
$x\not\in Z$. Hence, $Z=\emptyset$ implies (b), and $m(Z)=0$ implies
almost-(b). ∎

###### Proof of Proposition 1.13.

Let $f_{0}$ be the measurable map in (1) or (2) of Proposition 1.13, that is,
$f_{0}(x)=\frac{x}{2}$ for $x\in(0,1]$ and $f_{0}(0)=0$ or $\frac{1}{2}$. To
see $\mathcal{L}_{f}$ does not satisfy (S) for this $f$, recall that for a
given Markov operator $P:L^{1}(m)\to L^{1}(m)$, according to Theorem E, the
existence of a $P$-invariant density is equivalent to the following condition:
there is $\delta>0$ such that

$\sup_{n\geq 1}\int_{A}P^{n}1_{X}\,dm<1\quad\text{for any}\ \ A\in\mathscr{B}\
\ \text{with}\ \ m(A)<\delta.$

By duality, this implies that

$\int P^{n*}1_{A}\,dm<1\quad\text{for all $n\geq 1$ and $A\in\mathscr{B}$ with
$m(A)<\delta$.}$

Since $P^{n*}1_{A}(x)=P^{n}(x,A)\leq 1$ for any $x\in X$, we conclude the
following necessary condition for the existence of a $P$-invariant density:

  1. ($\star$)

there is $\delta>0$ such that for all $n\geq 1$ and $A\in\mathscr{B}$ with
$m(A)<\delta$, there exists $E\in\mathscr{B}$ with $m(E)>0$ satisfying that
$P^{n}(x,A)<1$ for all $x\in E$.

We will prove that $\mathcal{L}_{f}$ does not satisfy ($\star$). Notice that,
for that purpose, it suffices to show that for any $\delta>0$ there are
$n_{0}\geq 1$ and $A_{0}\in\mathscr{B}$ such that $m(A_{0})<\delta$ but
$P^{n_{0}}(x,A_{0})=1$ for $m$-almost every $x$. Observe also that since
$0<f_{t}(x)\leq f_{0}(x)$ for all $t\in[0,1]$ and $x\neq 0$, we have that, if
$x\neq 0$ then

$0<f^{n}_{\omega}(x)\leq f_{0}^{n}(x)=\frac{1}{2^{n}}\quad\text{ for any
$n\geq 1$ and $\omega\in\Omega$.}$

Hence, the support of
$P^{n}(x,\cdot)=\mathbb{P}(\\{\omega:f^{n}_{\omega}(x)\in\,\cdot\,\\})$ is
contained in the interval $(0,\frac{1}{2^{n}}]$ of length $\frac{1}{2^{n}}$ if
$x\neq 0$. Therefore, for any $\delta>0$, by taking
$n_{0}\geq-\frac{\log\delta}{\log 2}$ and $A_{0}=(0,\frac{1}{2^{n_{0}}}]$, we
have that $m(A_{0})<\delta$ and $P^{n_{0}}(x,A_{0})=1$ whenever $x\neq 0$.
From these observations, ($\star$) does not hold, and consequently, neither
does (S) for $\mathcal{L}_{f}$. In particular, due to the implications in
Figure 1, (FPM) does not hold for $\mathcal{L}_{f}$.

Let us complete the proof of items (1) and (2) of Proposition 1.13. When
$f_{0}$ is the continuous map in (1), obviously (a) in Remark 1.2 holds and
$Z=\\{0\\}$. So, by Lemma 7.1, $f$ does not satisfy (b) but satisfies
almost-(b). When $f_{0}$ is the map in (2) having discontinuity at $0$, (a)
does not hold and $Z=\emptyset$. Therefore, by using Lemma 7.1 again, we
conclude that $f$ satisfies (b).

We next prove item (3) of Proposition 1.13. Let $f_{0}(x)=2x\pmod{1}$ for
$x\in[0,1]$. Then, since $0\in F\cap Z$, it follows from Lemma 7.1 that $f$
does not satisfy (b). On the other hand, by [48, Theorem 3.1], we find that
$\mathcal{L}_{f}$ satisfies (C). Therefore, $f$ satisfies (FPM) by Theorem C
together with the implications in Figure 1. ∎

Now, we will prove Proposition 1.14. Hence, let $f$ be the random map given in
(1.12), i.e. random perturbation of a measurable map $f_{0}:[0,1]\to[0,1]$ by
multiplicative type noise. First, we need the following lemma.

######  Lemma 7.2.

Denote the set of fixed points of $f_{0}$ by $F$. The following hold:

  1. (1)

$F=\emptyset$ if and only if then $f$ satisfies (b);

  2. (2)

If $m(F)=0$, then $f$ satisfies almost-(b).

###### Proof.

Take a point $x\not\in F$. Then $f_{0}(x)-x\neq 0$ and thus
$I_{x}=\left\\{f_{t}(x):\,t\in T\right\\}$ is a closed interval which is not a
point set. Hence, $P(x,\cdot)$ is the normalized Lebesgue measure on the
closed interval $I_{x}$. This implies that $P(x,\cdot)$ is absolutely
continuous for all $x\not\in F$. On the other hand, if $x\in F$, then
$\left\\{f_{\omega}^{n}(x):\,\omega\in\Omega\right\\}=\\{x\\}$ for all $n\geq
1$. Consequently $P^{n}(x,\cdot)=\delta_{x}$ for all $n\geq 1$. Since
$\delta_{x}$ is not absolutely continuous with respect to $m$, (b) does not
hold. The claim immediately follows from these observations. ∎

###### Proof of Proposition 1.14.

Now let $f_{0}$ be the $C^{1}$ map given by the gradient flow with potential
(1.13). As mentioned, $f_{0}$ has infinitely many sinks $(s_{k})_{k\geq 1}$
and sources $(r_{k})_{k\geq 1}$ such that the union of the basins of $s_{k}$
for $f_{0}$ coincides with $X\setminus\\{r_{1},r_{2},\ldots\\}$. In
particular, $F=\\{r_{k},s_{k}:\,k\geq 1\\}$ is the set of fixed point of
$f_{0}$. Hence, it follows from Lemma 7.2 that $f$ satisfies almost-(b), but
not (b).

Furthermore, take $x$ in the basin of $s_{k}$ for $f_{0}$. In view of (1.12),
$f_{t}(x)$ is a convex combination of $x$ and $f_{0}(x)$. Thus, $f_{t}(x)$ is
in the closed interval between these points. In particular (from the
orientation preserving) in the closed interval whose endpoints are $x$ and
$s_{k}$. Arguing recursively, $f_{\omega}^{n}(x)$ is in the closed interval
whose endpoints are $s_{k}$ and $f_{0}^{n-1}(x)$ for any $\omega\in\Omega$ and
$n\geq 1$. Since $f_{0}^{n}(x)\to s_{k}$, this implies that
$f_{\omega}^{n}(x)\to s_{k}$ as $n\to\infty$ for any $\omega\in\Omega$. This
completes the proof of Proposition 1.14. ∎

###  7.3. Proof of Proposition 1.15: Random expanding maps

The constrictivity of $\mathcal{L}_{f}$ under the condition (1.14) is just the
consequence of the work by Boyarsky and Levesque [29, Theorem 2]. Furthermore,
it is not difficult to see that $\mathcal{L}_{f}$ is not uniformly
constrictive as follows. Notice that, for any $n_{0}\geq 1$ and $x\in X$,

$P^{n_{0}}(x,\cdot)=\sum_{\omega\in\\{1,\ldots,k\\}^{n_{0}}}p_{\omega}\delta_{f_{\omega}^{n_{0}}(x)}$

where $p_{\omega}=p^{n_{0}}(\\{\omega\\})$ and
$f_{\omega}^{n_{0}}=f_{\omega_{{n_{0}}-1}}\circ\cdots\circ f_{\omega_{0}}$ for
$\omega=(\omega_{0},\ldots,\omega_{{n_{0}}-1})$. Thus, if we consider the
finite set $A:=\\{f_{\omega}^{n_{0}}(x):\omega\in\\{1,\ldots,k\\}^{n_{0}}\\}$,
then $P^{n_{0}}(x,A)=1$ but $\mu(A)=0$ for any $m$-absolutely continuous
probability measure $\mu$. By virtue of the interpretation of (UC) in Section
1.3.4, this concludes the violation of (UC).

###  7.4. Proof of Proposition 1.16: Random contracting maps

For the proof of Proposition 1.16, the next proposition is important. Recall
the mixing property and exactness of an invariant density for a Markov
operator given in Section 1.5.

######  Proposition 7.3.

For $f$ in Proposition 1.16, $1_{X}$ is an $\mathcal{L}_{f}$-invariant
density. Moreover, $1_{X}$ is mixing, but is not exact.

###### Proof.

It is obvious that $1_{X}$ is an invariant density for $\mathcal{L}_{f}$.
Furthermore, we immediately find that

$\displaystyle\mathcal{L}_{f}^{*}\varphi(x)=\frac{1}{2}\mathcal{L}_{1}^{*}\varphi(x)+\frac{1}{2}\mathcal{L}_{2}^{*}\varphi(x)=\frac{1}{2}\varphi\left(\frac{x}{2}\right)+\frac{1}{2}\varphi\left(\frac{x+1}{2}\right)$

which is equivalent to the Perron–Frobenius operator $\mathcal{L}_{g}$ of the
(deterministic) dyadic map $g(x)=2x\pmod{1}$. Then, for any
$A,B\in\mathscr{B}$, we have

$\displaystyle\int_{X}\mathcal{L}_{f}^{n}1_{A}\cdot
1_{B}\,dm=\int_{X}1_{A}\cdot\mathcal{L}_{g}^{n}1_{B}\,dm\to m(A)m(B)$

as $n\to\infty$ since $m$ is mixing for $g$. Therefore, (using a simple
function approximation) we conclude that $1_{X}$ is mixing.

On other other hand, for $\varphi=1_{A}-1_{A^{c}}$ with $A=[0,\frac{1}{2}]$
and any $n\in\mathbb{N}$, we have

$\int_{X}|\mathcal{L}_{f}^{n}\varphi|dm=\sum_{k=0}^{2^{n}-1}\int_{X}\left|1_{\left[\frac{2k}{2^{n+1}},\frac{2k+1}{2^{n+1}}\right]}-1_{\left[\frac{2k+1}{2^{n+1}},\frac{2k+2}{2^{n+1}}\right]}\right|\,dm=\int_{X}1\,\,dm=1.$

If $1_{X}$ will be exact then,
$\mathcal{L}_{f}^{n}\varphi\to\int\varphi\,dm=0$ in $L^{1}$-norm. Thus, the
above computation implies that $1_{X}$ is not exact. ∎

Let us move to the proof of Proposition 1.16. It is well-known that, in the
class of constrictive Markov operators preserving $1_{X}$, the mixing property
implies the exactness (cf. [63, Remark 5.5.1]). Thus, the above proposition
already concludes that $P$ is not constrictive. As another proof, we can also
check it directly as follows.

Recall the Dunford–Pettis interpretation of (C) (given in Section 1.3.3):
$\mathcal{L}_{f}$ is constrictive if and only if for any $\varepsilon>0$ there
exists $\delta>0$ such that for any $h\in D(m)$, there is $n_{0}\geq 1$ for
which

$\int_{A}\mathcal{L}_{f}^{n}h\,dm<\varepsilon\quad\text{for any $n\geq n_{0}$
and $A\in\mathscr{B}$ with $m(A)\leq\delta$}.$

The key point is that the set $A$ can depend on $n$, compare it with the
definition of (AC). Fix $\varepsilon>0$ and $\delta>0$. Let $A=[0,\delta]$ and
$h=\frac{1}{\delta}1_{A}\in D(m)$. Then

$\mathcal{L}_{f}^{n}h=\frac{1}{\delta}\sum_{k=0}^{2^{n}-1}1_{\left[\frac{k}{2^{n}},\frac{k}{2^{n}}+\frac{\delta}{2^{n}}\right]}.$

Thus, by letting $B_{n}=\operatorname*{supp}\mathcal{L}_{f}^{n}h$, we have

$\int_{B_{n}}\mathcal{L}_{f}^{n}h\,dm=1\quad\text{and }\quad
m(B_{n})=\delta\,\text{ for any $n\geq 1$},$

which implies that $\mathcal{L}_{f}$ is not constrictive.

On the other hand, since $\mathcal{L}_{f}$ is mixing with respect to $m$, we
can conclude $\mathcal{L}_{f}$ is asymptotically constrictive as follows. For
any $\varepsilon>0$, set $\delta=\varepsilon$. Then for any $h\in D(m)$ and
$B\in\mathscr{B}$ with $m(B)<\delta$, by the mixing property, we have

$\lim_{n\to\infty}\int_{B}\mathcal{L}_{f}^{n}h\,dm=\int_{X}h\,dm\cdot
m(B)<\delta=\varepsilon.$

This completes the proof of Proposition 1.16.

###  7.5. Proof of Proposition 1.19: Random rotations

To refrain notations, in this subsection we identify any closed interval in
${S}^{1}$ with its corresponding set in $[0,1)$ (a closed interval or a set
$[0,a]\cup[b,1)$ with some $a<b$).

####  7.5.1. Case (1): Irrational $\alpha-\beta$

It is sufficient to show that $\mathcal{L}_{f}$ is asymptotically stable, that
is, $\|\mathcal{L}_{f}^{n}\varphi-1_{X}\|\to 0$ as $n\to\infty$ for any
$\varphi\in D(m)$, since an asymptotically stable Markov operator is
constrictive (see [63]). For $\varphi\in D(m)$, we can calculate
$\mathcal{L}_{f}^{n}\varphi$ as

$\displaystyle\mathcal{L}_{f}^{n}\varphi(x)$
$\displaystyle=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}\varphi(x-(n-k)\alpha-k\beta)=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}\varphi(x-n\alpha+k(\alpha-\beta))$
$\displaystyle=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}\mathcal{L}_{\beta-\alpha}^{k}\varphi(x-n\alpha)$

where $\mathcal{L}_{\gamma}$ denotes the Perron–Frobenius operator for the
irrational rotation with angle $\gamma$. Since $\mathcal{L}_{f}1_{X}=1_{X}$
and $\alpha-\beta$ is irrational, using the mean ergodic theorem weighted with
binomial coefficients (see [35, Theorem 4.1 and Corollary 4.3]),

$\mathcal{L}_{f}^{n}\mathcal{L}_{-\alpha}^{n}\varphi=\mathcal{L}_{f}^{n}\varphi(x+n\alpha)=\frac{1}{2^{n}}\sum_{k=0}^{n}\binom{n}{k}\mathcal{L}_{\beta-\alpha}^{k}\varphi(x)\to
1_{X}\text{ in $L^{1}(m)$ as $n\to\infty$.}$

It is clear that
$\mathcal{L}_{f}\mathcal{L}_{-\alpha}=\mathcal{L}_{-\alpha}\mathcal{L}_{f}$,
$\mathcal{L}_{-\alpha}1_{X}=1_{X}$ and
$\|\mathcal{L}_{-\alpha}\psi\|=\|\psi\|$ for any $\psi\in L^{1}(m)$.
Therefore, we have

$\|\mathcal{L}_{f}^{n}\varphi-1_{X}\|=\|\mathcal{L}_{-\alpha}^{n}(\mathcal{L}_{f}^{n}\varphi-1_{X})\|=\|\mathcal{L}_{-\alpha}^{n}\mathcal{L}_{f}^{n}\varphi-\mathcal{L}_{-\alpha}^{n}1_{X}\|=\|\mathcal{L}_{f}^{n}\mathcal{L}_{-\alpha}^{n}\varphi-1_{X}\|\to
0$

as $n\to\infty$, which completes the proof.

####  7.5.2. Case (2): Irrational $\alpha$ and $\beta$ with rational
$\alpha-\beta$

Let $\alpha-\beta=\frac{\ell}{N}$ for some integer $\ell$ and $N$. Fix
$0<\delta<1$. Set $B_{i}=[\frac{i}{N},\frac{i}{N}+\frac{\delta}{N}]$ for
$i=0,\cdots,N-1$ and $B=B_{0}\cup\dots\cup B_{N-1}$. Notice that
$m(B)=\delta$.

We first show that $\mathcal{L}_{f}1_{B}=1_{B+\alpha}$. Indeed, observe that
for any $i=0,\dots,N-1$ there is a unique $j=0,\dots,N-1$ such that
$B_{i}+\alpha=B_{j}+\beta$, where $A+t=[a+t,b+t]$ for $A=[a,b]$ and
$t\in[0,1)$. This follows for $j=i-\ell\pmod{N}$, since

$\left(\frac{i}{N}+\alpha\right)-\left(\frac{j}{N}+\beta\right)=\frac{i}{N}-\frac{j}{N}+\alpha-\beta=\frac{i}{N}-\frac{j}{N}+\frac{\ell}{N}=0\pmod{1},$

and such $j$ is unique. Then we have

$\mathcal{L}_{f}1_{B}=\frac{1}{2}\sum_{i=0}^{N-1}(1_{B_{i}+\alpha}+1_{B_{i}+\beta})=\frac{1}{2}\sum_{i=0}^{N-1}(1_{B_{i}+\alpha}+1_{B_{i}+\alpha})=\sum_{i=0}^{N-1}1_{B_{i}+\alpha}=1_{B+\alpha}.$

Moreover, we have that $\mathcal{L}_{f}^{n}1_{B}=1_{B+\alpha n}$ for any
$n\geq 1$.

Take $\varphi\coloneqq\frac{1}{\delta}1_{B}\in D(m)$ since $m(B)=\delta$.
Since $\alpha\not\in\mathbb{Q}$, we can consider a sequence
$\\{n_{j}\\}_{j\geq 1}$ such that $\alpha n_{j}\to 0\pmod{1}$ as $j\to\infty$.
Hence,

$\int_{B}\mathcal{L}_{f}^{n_{j}}\varphi\,dm\to 1\quad\text{as $j\to\infty$}.$

Therefore, $\mathcal{L}_{f}$ does not satisfies (AC).

We next prove that $\mathcal{L}_{f}$ is mean constrictive. It is clear that
the function $1_{X}$ is invariant for $\mathcal{L}_{f}$. Then, from Remark
1.8, it is sufficient to show that $1_{X}$ is an ergodic density. Let
$\mathcal{L}_{f}^{*}1_{A}=1_{A}$, where $\mathcal{L}_{f}^{*}$ is the adjoint
operator for $\mathcal{L}_{f}$. We will prove $m(A)\in\\{0,1\\}$. Since
obviously $1_{A}\in L^{2}(m)$, we have the Fourier series of $1_{A}$ as
follows, $1_{A}(x)=\sum_{n=-\infty}^{\infty}a_{n}e^{2n\pi ix}$. Then we can
calculate

$\mathcal{L}_{f}^{*}\varphi(x)=\frac{1}{2}\sum_{n=-\infty}^{\infty}a_{n}e^{2n\pi
i(x+\alpha)}+\frac{1}{2}\sum_{n=-\infty}^{\infty}a_{n}e^{2n\pi i(x+\beta)}.$

Then $\mathcal{L}_{f}^{*}\varphi=\varphi$ and

$\mathcal{L}_{f}^{*}\varphi(x)=\frac{1}{2}\mathcal{L}_{1}^{*}\varphi(x)+\frac{1}{2}\mathcal{L}_{2}^{*}\varphi(x)=\frac{1}{2}\varphi(x+\alpha)+\frac{1}{2}\varphi(x+\beta)$

imply

$\frac{1}{2}\sum_{n=-\infty}^{\infty}a_{n}e^{2n\pi
i(x+\alpha)}+\frac{1}{2}\sum_{n=-\infty}^{\infty}a_{n}e^{2n\pi
i(x+\beta)}=\sum_{n=-\infty}^{\infty}a_{n}e^{2n\pi ix}.$

Due to the uniqueness of Fourier series, it must be satisfied that, for any
$n\in\mathbb{Z}$,

$\frac{1}{2}e^{2n\pi i\alpha}+\frac{1}{2}e^{2n\pi i\beta}=1.$

It can be rewritten as

$e^{2n\pi i\gamma}+1=2e^{-2n\pi i\beta}\quad\text{with
$\gamma\coloneqq\alpha-\beta$}.$

That is,

$\cos(2n\pi\gamma)+1=2\cos(-2n\pi\beta)\quad{\rm
and}\quad\sin(2n\pi\gamma)=2\sin(-2n\pi\beta).$

Hence

$\left(2\cos(2n\pi\beta)-1\right)^{2}+\left(-2\sin(2n\pi\beta)\right)^{2}=1,$

which leads to $\cos(2n\pi\beta)=1$. This equality holds only for $n=0$ since
$\beta$ is irrational. Thus, all coefficient $a_{n}$ must be 0 for any $n\neq
0$, which shows $1_{A}$ is constant and hence $m(A)=0$ or $1$.

####  7.5.3. Case (3): Rational $\alpha$ and $\beta$

By the same arguments, we have $\mathcal{L}_{f}1_{B}=1_{B+\alpha}$. Unlike the
case $\alpha,\beta\in[0,1]\backslash\mathbb{Q}$, we further have
$\mathcal{L}_{f}1_{B}=1_{B}$ by taking $N$ as the least common multiple of the
denominator for $\alpha$ and $\alpha-\beta$. Then, for any $\delta>0$, taking
$\varphi=\frac{1}{\delta}1_{B}$, we get $m(B)=\delta$ but

$\int_{B}A_{n}\varphi\,dm=\frac{1}{n}\sum_{j=0}^{n-1}\int_{B}\mathcal{L}_{f}^{j}\varphi\,dm=1,$

since the support of $\mathcal{L}_{f}^{n}\varphi$ and $A_{n}\varphi$ is always
on $B$. This concludes that $\mathcal{L}_{f}$ does not satisfy (MC).

Finally, since $1_{X}$ is clearly invariant for $\mathcal{L}_{f}$, by the fact
that (WAP) is equivalent to the existence of an invariant density with the
maximal support [88, Theorem 3.1], $\mathcal{L}_{f}$ satisfies (WAP). This
completes the proof.

###  7.6. Proof of Proposition 1.20: Direct sums of random maps

Notice that the random dynamics $f_{+}$ on $X_{+}$ generated by
$\tau_{1}^{+}$, $\tau_{2}^{+}$ satisfies (S) because $f$ satisfies (S) and
preserves $X_{+}$. That is, $\mathcal{L}_{f_{+}}$ admits an invariant density,
denoted by $h_{+}$. Let $h$ be the density function on $X$ given by $h(x)=0$
on $X_{-}$ and $h(x)=h_{+}(x)$ on $X_{+}$. Then, $h$ is obviously an invariant
density of $\mathcal{L}_{f}$, namely, (S) holds for $f$.

On the other hand, (WAP) does not hold. If $\mathcal{L}_{f}$ has an invariant
density $h$ with the maximal support, then an integral function $h_{-}$ on
$X_{-}$, given by $h_{-}(x)=h(x)$, should be invariant for
$\mathcal{L}_{f_{-}}$, where $f_{-}$ is the random dynamics generated by
$\tau_{1}^{-}$, $\tau_{2}^{-}$, due to the invariance of $X_{-}$, $X_{+}$ for
$f$. This contradicts the assumption that $f_{-}$ is not in (S).

###  7.7. Proof of Proposition 1.21: Deterministic systems

All statements except one for the baker’s transformation can be proven by
literally repeating the argument in the previous subsections. On the other
hand, for the statement for the baker’s transformation, it immediately follows
from the fact that the baker’s transformation is mixing but not exact with
respect to the Lebesgue measure $m$ (cf. [3]).

##  Appendix A Annealed Perron–Frobenius operators

Let $(X,\mathscr{B},m)$ be a Polish probability space and consider
$(\Omega,\mathscr{F})=(T^{\mathbb{N}},\mathscr{A}^{\mathbb{N}})$ an infinite
product space of a measure space. Now, we introduce a probability measure
$\mathbb{P}$ on $(\Omega,\mathscr{F})$ which is shift invariant but not
necessarily a Bernoulli probability. Let $f:T\times X\to X$ be a measurable
map. We consider iterations $f^{n}_{\omega}=f_{\omega_{n}}\circ\dots\circ
f_{\omega_{1}}$ for $n\geq 1$ where the sequence $\omega=(\omega_{i})_{i\geq
1}$ is chosen from $\Omega$ according to the probability $\mathbb{P}$. To
emphasize this random choice, we call $f$ as a $\mathbb{P}$-random map. We can
introduce the annealed Perron–Frobenius operator ${\mathcal{L}}_{f}$ given by

${\mathcal{L}}_{f}\varphi=\int\mathcal{L}_{\omega}\varphi\,d\mathbb{P}(\omega)\quad\text{for
$\varphi\in L^{1}(m)$}$ (A.1)

where $\mathcal{L}_{\omega}$ is the Perron–Frobenius operator of $f_{\omega}$.
Recall that we introduced $\mathcal{L}_{\omega}\varphi$ as the Radon–Nikodým
derivative of the signed measure $(f_{\omega})_{*}m_{\varphi}$ with respect to
$m$ where

$m_{\varphi}(B)=\int_{B}\varphi\,dm\quad\text{for all $B\in\mathscr{B}$}.$

That is, for each $\varphi\in L^{1}(m)$,

$\mathcal{L}_{\omega}\varphi\in L^{1}(m)\ \ \text{such that}\ \
(f_{\omega})_{*}m_{\varphi}(B)=\int_{B}\mathcal{L}_{\omega}\varphi\,dm\ \
\text{for all $B\in\mathscr{B}$}.$ (A.2)

In other words, $(f_{\omega})_{*}m_{\varphi}=m_{\mathcal{L}_{\omega}\varphi}$,
i.e.,

$\int_{(f_{\omega})^{-1}(B)}\varphi\,dm=\int_{B}\mathcal{L}_{\omega}\varphi\,dm\quad\text{for
all $B\in\mathscr{B}$}.$ (A.3)

The following lemma characterizes $\mathcal{L}_{f}$ in similar terms.

######  Lemma A.1.

For each $\varphi\in L^{1}(m)$, ${\mathcal{L}}_{f}\varphi$ is the
Radon–Nikodým derivative of the measure $\bar{m}_{\varphi}$ with respect to
$m$ where

$\bar{m}_{\varphi}(B)=\int(f_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}\quad\text{for
all $B\in\mathscr{B}$}.$

That is,

$\mathcal{L}_{f}\varphi\in L^{1}(m)\ \ \text{such that}\ \
\bar{m}_{\varphi}(B)=\int_{B}{\mathcal{L}}_{f}\varphi\,dm\ \ \text{for all
$B\in\mathscr{B}$}.$

In particular,

${\mathcal{L}}_{f}\varphi=\varphi\quad\iff\quad
m_{\varphi}(B)=\int(f_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}\quad\text{for
all $B\in\mathscr{B}$}.$ (A.4)

###### Proof.

According to the definition of the annealed operator $\mathcal{L}_{f}$ in
(A.1) and the Perron–Frobenius operator of $f_{\omega}$ in (A.2), it holds

$\int_{B}\mathcal{L}_{f}\varphi\,dm=\int_{B}\int\mathcal{L}_{\omega}\varphi\,d\mathbb{P}\,dm=\int\int_{B}\mathcal{L}_{\omega}\varphi\,dm\,d\mathbb{P}=\int(f_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}.$

This proves the first part of the lemma. The second part follows immediately
by observing that $\mathcal{L}_{f}\varphi=\varphi$ if and only if
$\int_{B}\mathcal{L}_{f}\varphi\,dm=m_{\varphi}(B)$. Hence, the above equation
reads as ${m}_{\varphi}=\bar{m}_{\varphi}$ and thus (A.4) follows. ∎

######  Remark A.2.

Since the Perron–Frobenius operator also acts on the space of finite signed
measures as

$\mathcal{L}_{f}\mu(B)=\int 1_{B}\circ
f_{\omega}(x)\,d\mathbb{P}(\omega)d\mu(x)\quad\text{for all
$B\in\mathscr{B}$,}$

the equivalence (A.4) can be generalized as follows:

${\mathcal{L}}_{f}\mu=\mu\quad\iff\quad\mu(B)=\int(f_{\omega})_{*}\mu(B)\,d\mathbb{P}\quad\text{for
all $B\in\mathscr{B}$}.$

######  Remark A.3.

Note that when $\mathbb{P}=p^{\mathbb{N}}$ is a Bernoulli measure on $\Omega$,

$\int(f_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}=\int(f_{\omega_{1}})_{*}m_{\varphi}(B)\,dp(\omega_{1}).$

Thus, from (A.4), $\varphi$ is an $\mathcal{L}_{f}$-invariant density if and
only if $m_{\varphi}$ is an $f$-stationary measure in the sense introduced in
(1.2) that is absolutely continuous with respect to $m$.

Let us write

$\mathcal{L}^{n}_{\omega}=\mathcal{L}_{\omega_{n}}\circ\dots\circ\mathcal{L}_{\omega_{1}}\quad\text{for
$\psi\in L^{\infty}(m)$ and $\omega\in\Omega$}$

where $\sigma$ is the left shift on $\Omega$ preserving $\mathbb{P}$. As
usual, we introduce the adjoint operator of ${\mathcal{L}}_{f}:L^{1}(m)\to
L^{1}(m)$ as the operator ${\mathcal{L}}^{\,*}_{f}:L^{\infty}(m)\to
L^{\infty}(m)$ satisfying

$\langle{\mathcal{L}}^{\,*}_{f}\psi,\varphi\rangle=\langle\psi,{\mathcal{L}}_{f}\varphi\rangle\quad\text{for
$\psi\in L^{\infty}(m)$ and $\varphi\in L^{1}(m)$ \ \ where \ \ $\langle
h,g\rangle=\int hg\,dm$.}$

Recall that since $(\mathcal{L}^{n}_{f})^{*}=(\mathcal{L}^{*}_{f})^{n}$ where
$\mathcal{L}^{n}_{f}$ and $(\mathcal{L}^{*}_{f})^{n}$ are the $n$-th iteration
of $\mathcal{L}_{f}$ and $\mathcal{L}^{*}_{f}$ respectively, we simply write
it as $\mathcal{L}^{n*}_{f}$. Similarly, we write
$(\mathcal{L}^{n}_{\omega})^{*}=\mathcal{L}^{n*}_{\omega}$.

######  Lemma A.4.

For any $n\geq 1$, $\varphi\in L^{1}(m)$ and $\psi\in L^{\infty}(m)$, it holds

  1. (1)

$\mathcal{L}_{\omega}^{n}=\mathcal{L}_{f^{n}_{\omega}}$;

  2. (2)

$\mathcal{L}^{n*}_{\omega}\psi=\psi\circ f^{n}_{\omega}$;

  3. (3)

$\langle{\mathcal{L}}^{n*}_{\omega}\psi,\varphi\rangle=\langle\psi,{\mathcal{L}}^{\,n}_{\omega}\varphi\rangle$.

Moreover, if $\mathbb{P}$ is a Bernoulli measure on $\Omega$ (i.e., an
infinite product $p^{\mathbb{N}}$ of a probability measure ${p}$ on $T$), it
holds

  1. (4)

${\mathcal{L}}_{f}^{n}\varphi=\int\mathcal{L}^{n}_{\omega}\varphi\,d\mathbb{P}$;

  2. (5)

${\mathcal{L}}_{f}^{n*}\psi=\int\mathcal{L}_{\omega}^{n*}\psi\,d\mathbb{P}$.

###### Proof.

Let us prove (1). To do this we proceed by induction. It is clear that (1)
holds for $n=1$. Assume that
$(f^{n-1}_{\omega})_{*}m_{\varphi}=m_{\mathcal{L}^{n-1}_{\omega}\varphi}$.
Then

$(f^{n}_{\omega})_{*}m_{\varphi}=(f_{\sigma^{n-1}\omega})_{*}((f^{n-1}_{\omega})_{*}m_{\varphi})=(f_{\sigma^{n-1}\omega})_{*}m_{\mathcal{L}^{n-1}_{\omega}\varphi}.$

From this, having in mind (A.3), we get

$\displaystyle(f^{n}_{\omega})_{*}m_{\varphi}(B)=\int_{(f_{\sigma^{n-1}\omega})^{-1}(B)}\mathcal{L}_{\omega}^{n-1}\varphi\,dm=\int_{B}\mathcal{L}_{\sigma^{n-1}\omega}\circ\mathcal{L}_{\omega}^{n-1}\varphi\,dm=\int_{B}\mathcal{L}_{\omega}^{n}\varphi\,dm.$

This implies that $\mathcal{L}_{f^{n}_{\omega}}=\mathcal{L}_{\omega}^{n}$.

Now, (2) follows immediately from (1) since
$\mathcal{L}_{\omega}^{n*}=(\mathcal{L}_{\omega}^{n})^{*}=(\mathcal{L}_{f^{n}_{\omega}})^{*}$.
Hence, as is well-known, $(\mathcal{L}_{f^{n}_{\omega}})^{*}\psi=\psi\circ
f^{n}_{\omega}$. Finally, observe that (3) is just the duality relation. To
conclude the proof, from now on we assume that the probability $\mathbb{P}$ is
Bernoulli.

######  Claim A.5.

For any $n\geq 1$,

$m_{\mathcal{L}^{n}_{f}\varphi}(B)=\int(f^{n}_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}\quad\text{for
all $B\in\mathscr{B}$ and $\varphi\in L^{1}(m)$}.$

###### Proof.

We prove the claim by induction. For $n=1$, we need to prove that
$m_{\mathcal{L}_{f}\varphi}(B)=\int(f^{n}_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}$.
Observe that this is just Lemma A.1. Thus, we can assume that the claim holds
for $n-1$. Then, using again Lemma A.1,

$\displaystyle m_{\mathcal{L}^{n}_{f}\varphi}(B)$
$\displaystyle=\int_{B}\mathcal{L}^{n}_{f}\varphi\,dm=\int_{B}\mathcal{L}_{f}(\mathcal{L}^{n-1}_{f}\varphi)\,dm=\int_{B}(f_{\omega})_{*}m_{\mathcal{L}^{n-1}_{f}\varphi}(B)\,d\mathbb{P}.$

By induction we have that

$\displaystyle\int_{B}(f_{\omega})_{*}m_{\mathcal{L}^{n-1}_{f}\varphi}(B)\,d\mathbb{P}$
$\displaystyle=\iint(f^{n-1}_{\bar{\omega}})_{*}m_{\varphi}(f^{-1}_{\omega}(B))\,d\mathbb{P}(\bar{\omega})\,d\mathbb{P}(\omega)$
$\displaystyle=\iint(f_{\omega}\circ
f^{n-1}_{\bar{\omega}})_{*}m_{\varphi}(B)\,d\mathbb{P}(\bar{\omega})\,d\mathbb{P}(\omega).$

Taking into account that $\mathbb{P}$ is a Bernoulli measure and that
$f_{\omega}=f_{\omega_{1}}$ and
$f^{n-1}_{\bar{\omega}}=f_{\bar{\omega}_{n-1}}\circ\dots\circ
f_{\bar{\omega}_{1}}$ where $\omega=(\omega_{i})_{i\geq 1}$ and
$\bar{\omega}=(\bar{\omega}_{i})_{i\geq 1}$, we have

$\iint(f_{\omega}\circ
f^{n-1}_{\bar{\omega}})_{*}m_{\varphi}(B)\,d\mathbb{P}(\bar{\omega})\,d\mathbb{P}(\omega)=\int(f^{n}_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}.$

Putting together all the above equations, we get the claim. ∎

Now we conclude (4). To do this, we use the first item in this lemma, the fact
that the Perron–Frobenius operator of $f^{n}_{\omega}$ satisfies that
$m_{\mathcal{L}_{f^{n}_{\omega}}\varphi}=(f^{n}_{\omega})_{*}m_{\varphi}$, and
finally Claim A.5. Then we get that

$\displaystyle\int_{B}\int\mathcal{L}_{\omega}^{n}\varphi\,d\mathbb{P}\,dm$
$\displaystyle=\int\int_{B}\mathcal{L}_{f^{n}_{\omega}}\varphi\,dm\,d\mathbb{P}=\int(f^{n}_{\omega})_{*}m_{\varphi}(B)\,d\mathbb{P}=\int_{B}P^{n}\varphi\,dm.$

From this, since $B\in\mathscr{B}$ is arbitrary, we obtain that
$\mathcal{L}_{f}^{n}\varphi=\int\mathcal{L}_{\omega}^{n}\varphi\,d\mathbb{P}\,dm$
as desired.

Finally, we will prove (5). By (3), (4) and the duality, we have

$\displaystyle\langle\mathcal{L}^{n*}_{f}\psi,\varphi\rangle$
$\displaystyle=\langle\psi,\mathcal{L}^{n}_{f}\varphi\rangle=\langle\psi,\int\mathcal{L}^{n}_{\omega}\varphi\,d\mathbb{P}\rangle=\int\langle\psi,\mathcal{L}^{n}_{\omega}\varphi\rangle\,d\mathbb{P}$
$\displaystyle=\int\langle\mathcal{L}^{n*}_{\omega}\psi,\varphi\rangle\,d\mathbb{P}=\langle\int\mathcal{L}^{n*}_{\omega}\psi\,d\mathbb{P},\varphi\rangle.$

This implies
$\mathcal{L}^{n*}_{f}\psi=\int\mathcal{L}^{n*}_{\omega}\psi\,d\mathbb{P}$ as
required. ∎

##  Appendix B Generalized restrictions of Markov operators

We will extend the theory of restrictions of Markov operators developed in
Section 3.2. Roughly speaking, we want to replace the reference measure $m$
with an absolutely continuous invariant measure and restrict the Markov
operator to the support of such measure recovering the Markov property among
other ergodic properties.

Let $P:L^{1}(m)\to L^{1}(m)$ be a Markov operator. Take $h\in D(m)$ and define
the probability measure $m_{h}$ as usual by $dm_{h}=h\,dm$. Denote
$S=\operatorname*{supp}h$ and note that $S\in\mathscr{B}$ with $m(S)>0$. Let
us consider $L^{1}(m_{h})=L^{1}(S,\mathscr{B}_{S},m_{h})$ where
$\mathscr{B}_{S}$ denotes the trace $\sigma$-algebra of $S$ in $\mathscr{B}$.
As in Section 3.2, $L^{1}(m_{h})\hookrightarrow L^{1}(m)$. Abusing notation,
we write this inclusion by $1_{S}$ and identify $L^{1}(m_{S})$ with

$1_{S}(L^{1}(m_{S}))=\\{\phi\in L^{1}(m):\operatorname*{supp}\phi\subset
S\\}\subset L^{1}(m).$

We define the operator

$\displaystyle P_{h}:L^{1}(m_{h})\to L^{1}(m_{h}),\qquad
P_{h}\phi=\frac{1_{S}P(1_{S}\phi h)}{h}\quad\text{for $\phi\in
L^{1}(m_{h})$}.$

Actually, $P_{h}$ acts on $L^{1}(m)$ by the same formula. Moreover, if
$h=\frac{1_{S}}{m(S)}$, then $P_{h}\phi=1_{S}P(1_{S}\phi)$. That is, $m_{h}$
and $P_{h}$ coincides with the measure $m_{S}$ and the operator $P_{S}$
introduced in (3.3) and (3.4) respectively.

It is clear that $P_{h}$ is a bounded linear positive operator on
$L^{1}(m_{h})$. Moreover, $P_{h}$ is also a contraction on $L^{1}(m_{h})$
since $\|P_{h}\|_{\rm op}\leq 1$. On the other hand,
$L^{\infty}(m_{h})\hookrightarrow L^{\infty}(m)$ by the same canonical
inclusion $1_{S}$. As before, we identify $L^{1}(m_{h})$ with
$1_{S}(L^{\infty}(m_{h}))$. We also have that $P^{*}_{h}$, the adjoint
operator of $P_{h}$, acting on $L^{\infty}(m_{h})$ coincides with
$1_{S}P^{*}1_{S}$. Indeed, for each $\varphi\in L^{1}(m_{h})$ and $\psi\in
L^{\infty}(m_{h})$,

$\displaystyle\int_{S}\varphi\cdot P_{h}^{*}\psi\,dm_{h}$
$\displaystyle=\int_{S}P_{h}\varphi\cdot\psi\,dm_{h}=\int_{X}1_{S}\frac{P(\varphi
h)}{h}\cdot\psi h\,dm$ $\displaystyle=\int_{X}\varphi h\cdot
P^{*}(1_{S}\psi)\,dm=\int_{S}\varphi\cdot P^{*}(1_{S}\psi)\,dm_{h}$

as desired.

######  Proposition B.1.

Let $P:L^{1}(m)\to L^{1}(m)$ be a Markov operator and consider
$S\in\mathscr{B}$ with $m(S)>0$. The following conditions are equivalent:

  1. (1)

$P^{*}1_{X\setminus S}\leq 1_{X\setminus S}$. Equivalently, $P^{*}1_{S}\geq
1_{S}$;

  2. (2)

$\operatorname*{supp}P1_{S}\subset S$. Equivalently,
$\operatorname*{supp}P^{*}1_{X\setminus S}\subset X\setminus S$;

  3. (3)

$1_{S}P(1_{S}\phi)=P(1_{S}\phi)$ for all $\phi\in L^{1}(m)$. Equivalently,

$P_{S}\phi=P(1_{S}\phi)\ \ \text{for all $\phi\in
L^{1}(m)$}\quad\text{or}\quad P(1_{S}\phi)\in L^{1}(m_{S})\ \ \text{for all
$\phi\in L^{1}(m)$};$

  4. (4)

for every $\phi\in L^{1}(m)$ it holds that

$\int_{S}P(1_{S}\phi)\,dm=\int_{S}\phi\,dm;$

  5. (5)

$P_{S}:L^{1}(m_{S})\to L^{1}(m_{S})$ is a Markov operator;

  6. (6)

$P_{h}:L^{1}(m_{h})\to L^{1}(m_{h})$ is a Markov operator for any $h\in D(m)$
with $S=\operatorname*{supp}h$.

###### Proof.

Let us prove the equivalence between the above conditions.

First, we will prove (1) $\Rightarrow$ (2). Suppose that (2) does not hold.
Then, there is $B\subset\operatorname*{supp}P1_{S}$ with $m(B\setminus S)>0$.
From this and the condition (1), it follows

$0<\int_{B\setminus S}P1_{S}\,dm=\int 1_{S}P^{*}1_{B\setminus S}\,dm\leq\int
1_{S}P^{*}1_{X\setminus S}\,dm\leq\int 1_{S}1_{X\setminus S}\,dm=0$

which is a contradiction. Note that the equivalence indicated in (1)
immediately follows by using $1_{X}=P^{*}1_{S}+P^{*}1_{X\setminus S}$.
Furthermore, the equivalence indicated in (2) follows immediately by the
duality

$\int_{X\setminus S}P1_{S}\,dm=\int_{S}P^{*}1_{X\setminus S}\,dm.$

Now, we will show that (2) $\Rightarrow$ (3). Let $\phi\in L^{1}(m)$. Note
that $\operatorname*{supp}1_{S}\phi\subset S=\operatorname*{supp}1_{S}$. Thus,
by Lemma 3.1 and condition (2),
$\operatorname*{supp}P(1_{S}\phi)\subset\operatorname*{supp}P1_{S}\subset S$
up to $m$-null set. In particular, $1_{S}P(1_{S}\phi)=P(1_{S}\phi)$. That is,
condition (3) holds.

We will see that (3) $\Rightarrow$ (4). Assuming (3) and using that $P$ is a
Markov operator, for any $\phi\in L^{1}(m)$,

$\int_{S}P(1_{S}\phi)\,dm=\int 1_{S}P(1_{S}\phi)\,dm=\int
P(1_{S}\phi)\,dm=\int 1_{S}\phi\,dm=\int_{S}\phi\,dm$

and thus (4) holds.

Next, we will prove that (4) and (5) are equivalent. Clearly, $P_{S}$ is a
bounded linear positive operator. Note that, by definition, for any $\phi\in
L^{1}(m_{S})$,

$\int
P_{S}\phi\,dm_{S}=\frac{1}{m(S)}\int_{S}P(1_{S}\phi)\,dm\quad\text{and}\quad\int\phi\,dm_{S}=\frac{1}{m(S)}\int_{S}\phi\,dm.$

Thus, clearly we get that (4) implies that $\int
P_{s}\phi\,dm_{S}=\int\phi\,dm_{S}$ and viceversa.

We will obtain the implication (4) to (6). Similar as above, for any
$\varphi\in L^{1}(m)$ and $h\in D(m)$ with $S=\operatorname*{supp}h$, $P_{h}$
is clearly a bounded linear positive operator. Moreover, taking into account
that $\varphi h=1_{S}\varphi h$,

$\int P_{h}\varphi\,dm_{h}=\int_{S}P(1_{S}\varphi
h)\,dm\quad\text{and}\quad\int\varphi\,dm_{h}=\int_{S}\varphi h\,dm.$ (B.1)

Hence, applying (4) to $\phi=\varphi h$, we immediately get

$\int P_{h}\varphi\,dm=\int\varphi\,dm_{h}.$

Thus, $P_{h}$ is Markov and we conclude (6).

Since (6) $\Rightarrow$ (5) is clear, the rest to show is (4) $\Rightarrow$
(1). Suppose contrarily that there is some $B\subset S$ such that $m(B)>0$ and
$1_{B}P^{*}1_{S}<1_{B}$. This implies that

$\int_{S}1_{B}\,dm>\int_{S}1_{B}P^{*}1_{S}\,dm=\int_{S}P(1_{B}1_{S})\,dm$

but clearly contradicts to the assumption (4) with $\phi=1_{B}$. ∎

##  Appendix C Ergodicity of invariant densities

Let $P:L^{1}(m)\to L^{1}(m)$ be a Markov operator as introduced in Section
1.2. We associate $P$ with a new operator, which we will still denote by $P$,
acting on the set of probability measures $\mu$ on $(X,\mathscr{B})$ which are
absolutely continuous with respect to $m$ by

$P\mu(A)=\int P^{*}1_{A}(x)\,d\mu(x)\quad\text{for any $A\in\mathscr{B}$}.$

Here $P^{*}:L^{\infty}(m)\to L^{\infty}(m)$ denotes the adjoint operator of
$P:L^{1}(m)\to L^{1}(m)$. Recall that $D(m)=\\{\phi\in L^{1}(m):\phi\geq 0,\
\|\phi\|=1\\}$ and for a given $\phi\in D(m)$, we denote $m_{\phi}$ the
probability measure given by $dm_{\phi}=\phi\,dm$. We say that $\phi\in D(m)$
is a $P$-invariant density if $P\phi=\phi$. Similarly, a probability measure
$\mu$ on $(X,\mathscr{B})$ is said to be $P$-invariant if $P\mu=\mu$. Denote
by $D_{P}(m)$ and $I_{P}(m)$, respectively, the convex sets of $P$-invariant
densities and $P$-invariant probability measures that are absolutely
continuous with respect to $m$.

######  Lemma C.1.

Let $\phi\in D(m)$. Then, $P\phi=\phi$ if and only if ${P}m_{\phi}=m_{\phi}$.
In particular, $D_{P}(m)$ is identified with $I_{m}({P})$.

###### Proof.

If $P\phi=\phi$, then for any $A\in\mathscr{B}$,

$Pm_{\phi}(A)=\int P^{*}1_{A}\,dm_{\phi}=\int
P^{*}1_{A}\phi\,dm=\int_{A}P\phi\,dm=\int_{A}\phi\,dm=m_{\phi}(A).$

Conversely, if $Pm_{\phi}=m_{\phi}$, then for any $A\in\mathscr{B}$,

$m_{\phi}(A)=\int P^{*}1_{A}\phi\,dm=\int_{A}P\phi\,dm.$

Thus $P\phi$ is the Radon–Nikodým derivative of $m_{\phi}$ with respect to
$m$. Since this derivative is exactly $\phi$, we have $P\phi=\phi$. ∎

Recall that an invariant probability measure $\nu$ of a transformation $g$ of
a measurable space $(Y,\mathscr{A})$, i.e., $\nu=g_{*}\nu$, is said to be
_ergodic_ if $\nu(A)\in\\{0,1\\}$ for all set $A\in\mathscr{A}$ such that
$A\subset g^{-1}(A)$. Equivalently, for all $A\in\mathscr{A}$ such that
$A=g^{-1}(A)$. Moreover, it is also well known that one can weaken both
previous conditions of invariance of the set $A$ by simply asking that the
relation holds up to a $\mu$-null set. That is, if
$\mathcal{L}^{*}_{g}1_{A}\geq 1_{A}$ or $\mathcal{L}^{*}_{g}1_{A}=1_{A}$,
respectively, for $\mu$-almost everywhere where
$\mathcal{L}^{*}_{g}\psi=\psi\circ g$ for $\psi\in L^{\infty}(m)$ is the
adjoint Perron–Frobenius operator of $g$. The following proposition shows some
of these equivalences in a more general setting.

######  Proposition C.2.

Let $\mu\in I_{P}(m)$. Then the following conditions are equivalent:

  1. (1)

$\mu(A)\in\\{0,1\\}$ for any $A\in\mathscr{B}$ such that $P^{*}1_{A}\geq
1_{A}$ for $m$-almost everywhere;

  2. (2)

$\mu(A)\in\\{0,1\\}$ for any $A\in\mathscr{B}$ such that $P^{*}1_{A}\geq
1_{A}$ for $\mu$-almost everywhere;

  3. (3)

$\mu(A)\in\\{0,1\\}$ for any $A\in\mathscr{B}$ such that $P^{*}1_{A}=1_{A}$
for $\mu$-almost everywhere.

###### Proof.

(2) $\Rightarrow$ (3): Obvious.

(3) $\Rightarrow$ (2): Let us consider $A\in\mathscr{B}$ such that
$P^{*}1_{A}\geq 1_{A}$ for $\mu$-almost everywhere. Since $\mu\in I_{P}(m)$,
then $\mu(A)=\int P^{*}1_{A}\,d\mu$. On the other hand, $\mu(A)=\int
1_{A}\,d\mu$. Thus, since $P^{*}1_{A}\geq 1_{A}$ for $\mu$-almost everywhere,
we get $0\leq\int P^{*}1_{A}-1_{A}\,d\mu=0$. From this follows that
$P^{*}1_{A}=1_{A}$ for $\mu$-almost everywhere. Consequently, it immediately
follows that (2) implies (3).

(2) $\Rightarrow$ (1): Let us consider $A\in\mathscr{B}$ such that
$P^{*}1_{A}\geq 1_{A}$ for $m$-almost everywhere. This means that there is
$B\in\mathscr{B}$ such that $m(B)=0$ and $P^{*}1_{A}(x)\geq 1_{A}(x)$ for all
$x\in X\setminus B$. Since $\mu$ is absolutely continuous with respect to $m$,
then $\mu(B)=0$. Thus, $P^{*}1_{A}\geq 1_{A}$ for $\mu$-almost everywhere.
From this, it immediately follows that (3) implies (2).

(1) $\Rightarrow$ (2): Let us consider $A\in\mathscr{B}$ such that
$P^{*}1_{A}\geq 1_{A}$ for $\mu$-almost everywhere. Since $\mu\in I_{P}(m)$,
by Lemma C.1, we have $\phi\in D_{P}(m)$ such that $\mu=m_{\phi}$ (i.e.,
$d\mu=\phi\,dm$). Then, by Proposition 4.5, we have that $P^{*}1_{X\setminus
S}\leq 1_{X\setminus S}$ for $m$-almost everywhere where
$S=\operatorname*{supp}\phi$. Then,

$1_{S}P^{*}1_{A}=1_{S}P^{*}1_{S\cap A}+1_{S}P^{*}1_{(X\setminus S)\cap
A}=1_{S}P^{*}1_{S\cap A}\leq P^{*}1_{S\cap A}$ (C.1)

for $m$-almost everywhere. But observe that
$\operatorname*{supp}1_{S}P^{*}1_{A}\subset S$ and $m$ and $\mu$ are
equivalent on $S$. Thus, since $P^{*}1_{A}\geq 1_{A}$ for $\mu$-almost
everywhere we have $1_{S\cap A}=1_{S}1_{A}\leq 1_{S}P^{*}1_{A}$ $m$-almost
everywhere. Putting this together with (C.1), we get that $1_{S\cap A}\leq
P^{*}1_{S\cap A}$ $\mu$-almost everywhere. By assumption, $m(S\cap
A)m(X\setminus(S\cap A))=0$. Consequently, since $\mu(A)=\mu(S\cap A)$ and
$\mu(X\setminus A)=\mu(X\setminus(S\cap A))$, one has that
$\mu(A)\mu(X\setminus A)=0$. ∎

Let $h\in D_{P}(m)$. From, Lemma C.1, we can apply Proposition C.2 to
$\mu=m_{h}$.

######  Definition C.3.

Let $h\in D_{P}(m)$ and set $\mu=m_{h}$. If any of the equivalent items in
Proposition C.2 holds, we say that _$\mu$ is an ergodic $P$-invariant
measure_, _$h$ is an ergodic $P$-invariant density_, _$(P,\mu)$ is ergodic_ or
_$(P,h)$ is ergodic_.111111The reference measure $m$ is always implicit when
we mention $P$.

Observe that if $P1_{X}=1_{X}$, then according to Lemma C.1, it holds that
$Pm=m$, i.e., $m\in I_{P}(m)$. Then, Proposition C.2 implies that $1_{X}$ is
an ergodic $P$-invariant density if and only if $m(A)\in\\{0,1\\}$ for all
$A\in\mathscr{B}$ with $P^{*}1_{A}=1_{A}$. Motivated by this last condition
and following Krengel [59, page 126], we introduce the following definition:

######  Definition C.4.

A Markov operator $P:L^{1}(m)\to L^{1}(m)$ is called _ergodic in the sense of
Krengel_ if the set of $P^{*}$-invariant sets
$\mathscr{B}_{i}=\\{A\in\mathscr{B}:P^{*}1_{A}=1_{A}\ \text{$m$-almost
everywhere}\\}$ is trivial, i.e., $\mathscr{B}_{i}=\\{\emptyset,X\\}$ up to an
$m$-null set.

In view of this definition, the above observation can be written as follows:

######  Remark C.5.

If $P1_{X}=1_{X}$, then the following are equivalent:

  1. (1)

$(P,1_{X})$ is ergodic (or in other words, $(P,m)$ is ergodic);

  2. (2)

$P$ is ergodic in the sense of Krengel.

In general, $P$ could be ergodic in the sense of Krengel and $1_{X}\not\in
D_{P}(m)$. See, for instance, Remark C.9. Note that Definition C.3 requires
that $1_{X}$ is a $P$-invariant density to be ergodic.

Now, recalling the theory of restriction of Markov operators introduced in
Section 3.2 and generalized in Appendix B we have the following:

######  Theorem C.6.

Let $h\in D_{P}(m)$ and $S=\operatorname*{supp}h$. Then the following
conditions are equivalent:

  1. (1)

$(P,h)$ is ergodic (or in other words, $(P,m_{h})$ is ergodic);

  2. (2)

$(P_{h},1_{S})$ is ergodic (or in other words, $(P_{h},m_{h})$ is ergodic);

  3. (3)

$P_{h}$ is ergodic in the sense of Krengel;

  4. (4)

$P_{S}$ is ergodic in the sense of Krengel.

###### Proof.

Recall that $P^{*}_{h}\psi=1_{S}P^{*}(1_{S}\psi)$ for all $\psi\in
L^{\infty}(m_{h})$.

(1) $\Rightarrow$ (3): Let us consider $A\in\mathscr{B}_{S}$ such that
$P^{*}_{h}1_{A}=1_{A}$ for $\mu$-almost everywhere. Then
$1_{S}P^{*}1_{A}=1_{A}$ $\mu$-almost everywhere and hence also $m$-almost
everywhere. In particular, $P^{*}1_{A}\geq 1_{A}$ $m$-almost everywhere and
thus, by assumption, $m_{h}(A)\in\\{0,1\\}$.

(3) $\Leftrightarrow$ (2): Notice that $P_{h}1_{S}=1_{S}$. Thus, the
equivalence follows from Remark C.5.

(2) $\Rightarrow$ (1): Let us consider $A\in\mathscr{B}$ such that
$P^{*}1_{A}\geq 1_{A}$ for $\mu$-almost everywhere. Since $h\in D_{P}(m)$, we
have $P^{*}1_{X\setminus S}\leq 1_{X\setminus S}$ for $m$-almost everywhere.
Hence, as argued in the previous theorem (see (C.1)),

$1_{S\cap A}\leq 1_{S}P^{*}1_{A}=1_{S}P^{*}1_{S\cap
A}+1_{S}P^{*}1_{(X\setminus S)\cap A}=1_{S}P^{*}1_{S\cap A}=P_{h}^{*}1_{S\cap
A}$

$m$-almost everywhere. Then, by assumption we have that $m_{h}(S\cap
A)\in\\{0,1\\}$. But, since $m_{h}(A)=m_{h}(S\cap A)$ we conclude the
implication.

(3) $\Leftrightarrow$ (4): Notice that $P^{*}_{S}$ and $P^{*}_{h}$ are both of
the form $1_{S}P^{*}1_{S}$ acting, respectively, on $L^{1}(m_{S})$ and
$L^{1}(m_{h})$. Since both measures $m_{S}$ and $m_{h}$ are equivalent, we
have that $P^{*}_{S}1_{A}=1_{A}$ $m_{S}$-almost everywhere if and only if
$P^{*}_{h}1_{A}=1_{A}$ $m_{h}$-almost everywhere. From this, it immediately
follows the equivalence. ∎

######  Definition C.7.

A Markov operator $P:L^{1}(m)\to L^{1}(m)$ is called _conservative_ if there
is $\varphi\in L^{1}(m)$ such that $\varphi>0$ on $X$ and

$X=\left\\{x\in
X:\sum_{i=0}^{\infty}P^{i}\varphi(x)=\infty\right\\}\quad\text{up to an
$m$-null set}.$

######  Proposition C.8.

Let $h\in D(m)$ be a $P$-invariant density and set $S=\operatorname*{supp}h$.
Then both, $P_{h}$ and $P_{S}$ are conservative Markov operators. In
particular, if $h\in D_{P}(m)$ is ergodic then,

  1. (1)

$P_{S}:L^{1}(m_{S})\to L^{1}(m_{S})$ is a conservative and ergodic Markov
operator and $D_{P_{S}}(m_{S})=\\{h\\}$;

  2. (2)

$P_{h}:L^{1}(m_{h})\to L^{1}(m_{h})$ is a conservative and ergodic Markov
operator and $D_{P_{h}}(m_{h})=\\{1_{S}\\}$.

###### Proof.

Since $h$ is $P$-invariant, according to Proposition 3.3 and Proposition B.1,
$P_{S}$ and $P_{h}$ are Markov operators and $P_{S}h=h$. Also, it is not
difficult to check that $P_{h}1_{S}=1_{S}$. On the other hand, $P_{S}$ and
$P_{h}$ are conservative since $h>0$ and $1_{S}>0$ and $\sum_{i\geq
0}P_{S}^{i}h=\infty$ and $\sum_{i\geq 0}P_{h}^{i}1_{S}=\infty$ on $S$. If in
addition $h$ is ergodic, then from Theorem C.6 we get that $P_{S}$ and $P_{h}$
are conservative and ergodic Markov operators. Finally, from [37, Theorem A in
Chapter VI] it follows that $D_{P_{S}}(m_{S})=\\{h\\}$ and
$D_{P_{h}}=\\{1_{S}\\}$. ∎

######  Remark C.9.

In view of the above proposition, if $h$ is an ergodic $P$-invariant density
with $h>0$ on $X$ and $1_{X}\not=h$, then $P$ is ergodic in the sense of
Krengel but $1_{X}\not\in D_{P}(m)=\\{h\\}$.

The next proposition is well-known for the case of transformations. One can
generalize this to the case for Markov operators in terms of invariant
densities.

######  Proposition C.10.

Let $h\in D_{P}(m)$ and set $S=\operatorname*{supp}h$. Then, $h$ is ergodic if
and only if $h$ is an extremal point of $D_{P}(m)$. That is, it cannot be
decomposed as

$h=th_{1}+(1-t)h_{2}$ with $t\in(0,1)$ and $h_{1},h_{2}\in D_{P}(m)$.

Moreover, if $h_{1}\in D_{P}(m)$ is ergodic and $h_{2}\in D_{P}(m)$, then
either

$h_{1}=h_{2}\quad\text{or}\quad
m(\operatorname*{supp}h_{1}\cap\operatorname*{supp}h_{2})=0.$

###### Proof.

Suppose that $h$ is ergodic. By Proposition C.8, $P_{S}$ is a Markov operator
and $D_{P_{S}}(m_{S})=\\{h\\}$. In particular, we have $P_{S}g=P(1_{S}g)$ for
all $g\in L^{1}(m_{S})$. See Proposition 3.3 or Proposition B.1. Now, if
$h=ah_{1}+(1-a)h_{2}$ for some $0<a<1$ and $h_{1},h_{2}\in D_{P}(m)$, then
$h=h_{1}=h_{2}$. Indeed, observe that since $h_{i}$ is a $P$-invariant
density, according again to Proposition C.8, $P_{S_{i}}$ is a Markov operator
and $P_{S_{i}}h_{i}=h_{i}$ where $S_{i}=\mathrm{supp}\,h_{i}$ for $i=1,2$.
Then, since $S_{i}\subset S$, we have
$P_{S}h_{i}=P(1_{S}h_{i})=P(1_{S_{i}}h_{i})=P_{S_{i}}h_{i}=h_{i}$ for $i=1,2$.
Consequently, it follows $h=h_{1}=h_{2}$ since $D_{P_{S}}(m_{S})=\\{h\\}$.
This proves that $h$ is an extremal point of $D_{P}(m)$.

Now, we will prove the converse. Suppose $h$ is not ergodic and denote by
$\mu:=m_{h}$. Take a set $A\in\mathscr{B}$ such that $P^{*}1_{A}\geq 1_{A}$
and $0<\mu(A)<1$. Hence, since $\mu(S\setminus A)=1-\mu(A)$, we can write $h$
as a convex combination as follows

$\displaystyle h=\mu(A)\cdot\frac{1_{A}h}{\mu(A)}+\mu({S\setminus
A})\cdot\frac{1_{S\setminus A}h}{\mu({S\setminus A})}.$

Moreover, since $(1_{A}h)/\mu(A)$ and $(1_{S\setminus A}h)/\mu(S\setminus A)$
have both $L^{1}$-norm equals one, to show that $h$ is not an extremal point
in $D_{P}(m)$ it suffices to prove that $h1_{A}$ and $h1_{S\setminus A}$ are
$P$-invariant. To prove this, observe first that since $h=1_{A}h+1_{S\setminus
A}h$ and $h=Ph=P(1_{A}h)+P(1_{S\setminus A}h)$, one can write

$1_{A}h-P(1_{A}h)=P(1_{S\setminus A}h)-1_{S\setminus A}h.$ (C.2)

From Lemma 3.2, it follows that $\operatorname*{supp}P1_{A}\subseteq A$ up to
an $m$-null set. Morevover, since
$\operatorname*{supp}h=S=\operatorname*{supp}1_{S}$, by the $P$-invariance of
$h$ and Lemma 3.1, it follows that
$S=\operatorname*{supp}h=\operatorname*{supp}Ph=\operatorname*{supp}P1_{S}$ up
to an $m$-null set. Hence, $\operatorname*{supp}P1_{S\setminus A}\subseteq
S\setminus A$. Additionally, since
$\operatorname*{supp}(1_{A}h)\subseteq\operatorname*{supp}1_{A}$ and
$\operatorname*{supp}(1_{S\setminus
A}h)\subseteq\operatorname*{supp}1_{S\setminus A}$, Lemma 3.1 implies that
$\operatorname*{supp}P(1_{A}h)\subseteq\operatorname*{supp}P1_{A}\subseteq A$
and $\operatorname*{supp}P(1_{S\setminus
A}h)\subseteq\operatorname*{supp}P1_{S\setminus A}\subseteq S\setminus A$.
Consequentely, $\operatorname*{supp}(1_{A}h-P(1_{A}h))\subseteq A$ and
$\operatorname*{supp}(P(1_{S\setminus A}h)-1_{S\setminus A}h)\subseteq
X\setminus A$ and thus it follows from (C.2) that

$\displaystyle P(1_{A}h)=1_{A}h\quad\text{ and }\quad P(1_{S\setminus
A}h)=1_{S\setminus A}h$

as desired.

Finally, we will prove the second part of the proposition. Let us assume
$h_{1}$ and $h_{2}$ as in the statement. Clearly if $h_{1}=h_{2}$ then
$m(S)>0$ where $S=\operatorname*{supp}h_{1}\cap\operatorname*{supp}h_{2}$.
Conversely, suppose that $m(S)>0$. Since $P1_{S}\leq
P1_{\operatorname*{supp}h_{i}}$, by Lemma 3.1,
$\operatorname*{supp}P1_{S}\subset\operatorname*{supp}P1_{\operatorname*{supp}h_{i}}=\operatorname*{supp}Ph_{i}=\operatorname*{supp}h_{i}$
for $i=1,2$ and thus $\operatorname*{supp}P1_{S}\subset S$. According to
Proposition B.1, $P_{S}$ is a Markov operator, $P_{S}\phi=P(1_{S}\phi)$ and

$\int_{S}P(1_{S}\phi)\,dm=\int_{S}1_{S}\phi\,dm$ (C.3)

for all $\phi\in L^{1}(m)$. This implies that $1_{S}h_{i}$ is $P$-invariant.
Indeed, since $Ph_{i}=h_{i}$, it follows that
$1_{S}P(1_{S}h_{i})=P_{S}h_{i}=P(1_{S}h_{i})\leq h_{i}$. Hence
$P(1_{S}h_{i})\leq 1_{S}h_{i}$. On the other hand, if
$P(1_{S}h_{i})<1_{S}h_{i}$ on a set $A\in\mathscr{B}$ of positive $m$-measure,
then

$\int_{S}P(1_{S}h_{i})\,dm<\int_{A}1_{S}h_{i}\,dm+\int_{S\setminus
A}1_{S}h_{i}\,dm=\int_{S}1_{S}h_{i}\,dm$

contradicting (C.3). Therefore, we have $m(A)=0$ and
$P(1_{S}h_{i})=1_{S}h_{i}$. Then, $g_{i}=\frac{1_{S}h_{i}}{H_{i}}\in D_{P}(m)$
where $H_{i}=\int_{S}h_{i}\,dm$. Moreover, $m_{g_{1}}$ is absolutely
continuous with respect to $m_{h_{1}}$ since
$dm_{g_{1}}=\frac{1_{S}}{H_{1}}\,dm_{h_{1}}$. In particular, since $m_{h_{1}}$
is ergodic, we also have that $m_{g_{1}}$ is ergodic. Since
$S=\operatorname*{supp}g_{1}$, we obtain from Proposition C.8 that
$D_{P_{S}}(m_{S})=\\{g_{1}\\}$. But, since $S=\operatorname*{supp}g_{2}$ and
$g_{2}$ is $P$-invariant, we also have that $g_{2}\in D_{P_{S}}(m_{S})$. Thus,
$g_{2}=g_{1}$. From this, it follows that $h_{1}=h_{2}$ as desired. ∎

######  Remark C.11.

Recall that the support of $h\in L^{1}(m)$ is only well-defined up to a set of
measure zero. Thus, without loss of generality, from the above corollary, we
can assume that the supports of any two ergodic $P$-invariant densities
$h_{1}$ and $h_{2}$ are identical or disjoint.

To conclude, we will relate the previous definition of ergodicity for
invariant measures of Markov operators with the classical approach in random
dynamical systems and probability theory. First of all, let us consider an
annealed Perron–Frobenius operator $\mathcal{L}_{f}:X\to X$ associated with a
random map $f:T\times X\to X$ (with respect to a Bernoulli probability measure
$\mathbb{P}$). Recall that, as we showed in Remark 2.5, when $X$ is a Polish
space, any Markov operator can be represented as an annealed Perron–Frobenius
operator. As explained in Section 1.3.4, associated with this map we have a
transition probability $P(x,A)$ which coincides, for each $A\in\mathscr{B}$,
with $\mathcal{L}_{f}^{*}1_{A}$ on $m$-almost everywhere. Let us consider an
absolutely continuous probability measure $\mu$ on $X$ which is
$\mathcal{L}_{f}$-invariant. A classical approach to introducing the
ergodicity of $\mu$ is first to leave this measure to an invariant measure of
a deterministic dynamical system that represents $f$ in some sense. Then,
ergodicity is defined throughout this leaf measure. A deterministic dynamical
system that represents $f$ is its associated skew-product given by

$F:\Omega\times X\to\Omega\times X,\quad
F(\omega,x)=(\sigma\omega,f_{\omega}(x))$

with $f_{\omega}=f(t,\cdot)$ if $\omega_{0}=t$,
$\omega=(\omega_{i})_{i\in\mathbb{Z}}\in\Omega=T^{\mathbb{N}}$ and
$\sigma:\Omega\to\Omega$ the shift operator. The measure $\mu$ is lifted to
the $F$-invariant measure $\mathbb{P}\times\mu$. But we can also consider the
shift operator acting on $(X^{\mathbb{Z}},\mathscr{B}^{\mathbb{Z}})$. By
Kolmogorov extension theorem, one can construct a shift-invariant probability
measure $\mathbb{P}_{\mu}$ on $X^{\mathbb{Z}}$ from the consistent sequence of
measures $\\{\mathbb{P}_{\mu}^{n}\\}_{n\geq 0}$ where $\mathbb{P}_{\mu}^{n}$
is defined on $X^{2n+1}$ by

$\int\varphi\,d\mathbb{P}^{n}_{\mu}=\int\varphi(x_{-n},\dots,x_{n})\,P(x_{n-1},dx_{n})\dots
P(x_{-n},dx_{-n+1})\mu(dx_{-n}).$

The following result follows basically from classical results of Markov chain
and random dynamical systems. We refer to [41, Section 5.2] and [66,
Proposition 1.3] (see also [51, 74]) for more details.

######  Theorem C.12.

Let $f:T\times X\to X$ be a random map (with respect to a Bernoulli
probability measure $\mathbb{P}$) and denote by $\mathcal{L}_{f}:L^{1}(m)\to
L^{1}(m)$ the associated annealed Perron–Frobenius operator. Let $\mu\in
I_{m}(\mathcal{L}_{f})$. Then the following conditions are equivalent:

  1. (1)

$(\mathcal{L}_{f},\mu)$ is ergodic;

  2. (2)

$\mathbb{P}_{\mu}$ is an ergodic shift invariant measure on $X^{\mathbb{Z}}$;

  3. (3)

$\mathbb{P}\times\mu$ is an ergodic $F$-invariant probability measure on
$\Omega\times X$.

##  Appendix D Mixing and exactness of invariant densities

In this appendix we recall definition of mixing and exactness for
(deterministic) dynamics, and relate them with mixing and exactness for Markov
operators defined in Section 1.5. Although the content of this appendix should
be a folklore among experts, we include it for convenience of the reader.

Let $(X,\mathscr{B},m)$ be a probability space. Let $g:X\to X$ be an
$m$-nonsingular measurable map, and consider a $g$-invariant probability
measure $\mu$ on $X$. Then, $\mu$ is called _m_ ixing if $\mu(g^{-n}A\cap
B)\to\mu(A)\mu(B)$ as $n\to\infty$ for any $A,B\in\mathscr{B}$ (cf. [89]).
Recall also that $\mu$ is said to be _exact_ if

$\bigcap_{n\geq 0}g^{-n}\mathscr{B}=\\{\emptyset,X\\}\pmod{\mu}.$

From Lin’s theorem ([65]), $\mu$ is exact if and only if

$\lim_{n\to\infty}\left\lVert\mathcal{L}_{g}^{n}\varphi\right\rVert_{L^{1}(\mu)}=0\quad\text{for
any $\varphi\in L^{1}_{0}(\mu)\coloneqq\left\\{\psi\in
L^{1}(\mu):\int_{X}\psi\,d\mu=0\right\\}$.}$

Below as in Appendix B, we identify $\varphi\in L^{1}(m_{S})$ with
$1_{S}\varphi\in L^{1}(m)$ for an $m$-positive measure set $S$.

######  Lemma D.1.

Assume that $\mu$ is a $g$-invariant probability measure absolutely continuous
with respect to $m$ with the density map $h$. Let $S:=\operatorname*{supp}h$.
Then, $\mu$ is mixing if and only if

$\lim_{n\to\infty}\mathcal{L}_{g}^{n}\varphi=h\int_{X}\varphi
dm\quad\text{weakly in $L^{1}(m)$ for any $\varphi\in L^{1}(m_{S})$.}$

Furthermore, $\mu$ is exact if and only if

$\lim_{n\to\infty}\mathcal{L}_{g}^{n}\varphi=h\int_{X}\varphi\,dm\quad\text{strongly
in $L^{1}(m)$ for any $\varphi\in L^{1}(m_{S})$.}$ (D.1)

###### Proof.

Note first that the mixing property of $\mu$ is equivalent to require that

$\int\psi\circ
g^{n}\cdot\varphi\,d\mu\to\int\psi\,d\mu\int\varphi\,d\mu\quad\text{as
$n\to\infty$}$

for any $\psi\in L^{\infty}(\mu)$ and $\varphi\in L^{1}(\mu)$. Due to the
duality between $\mathcal{L}_{g}$ and $\psi\mapsto\psi\circ g$, this can be
written as

$\int\psi\cdot\mathcal{L}_{g}^{n}(\varphi h)\,dm\to\int\psi\left(h\int(\varphi
h)\,dm\right)dm\quad\text{as $n\to\infty$}.$

Hence, we immediately get the claim for mixing.

We next show the claim for exactness. Since exactness is hereditary property
between absolutely continuous measures, it follows from Lin’s theorem
mentioned above that $\mu$ is exact if and only if

$\lim_{n\to\infty}\left\lVert\mathcal{L}_{g}^{n}\varphi\right\rVert_{L^{1}(m_{S})}=0\text{
for any $\varphi\in L^{1}_{0}(m_{S})$.}$ (D.2)

Thus, it suffices to show the equivalence between (D.2) and (D.1).

(D.2)$\Rightarrow$(D.1): Note that
$\mathcal{L}_{g}^{n}\varphi-h\int_{X}\varphi
dm=\mathcal{L}_{g}^{n}(\varphi-h\int_{X}\varphi dm)$ and

$\int_{X}\left(\varphi-h\int_{X}\varphi
dm\right)dm=\frac{1}{m(S)}\int_{X}\varphi
dm_{S}-\int_{X}hdm\cdot\frac{1}{m(S)}\int_{X}\varphi dm_{S}=0.$

(D.1)$\Rightarrow$(D.2): It is straightforward if we take $\varphi\in
L^{1}_{0}(m_{S})$. ∎

We finally remark that, if a map $g:X\to X$ admits an ergodic invariant
probability measure $\mu$, then it holds by Birkhoff’s pointwise ergodic
theorem

$\lim_{n\to\infty}\int\frac{1}{n}\sum_{j=0}^{n-1}\varphi\circ
g^{j}\cdot\psi\,d(\mathbb{P}\times\mu)=\int\psi d\mu\int\varphi d\mu$

for any $\varphi\in L^{\infty}(\mu)$, $\psi\in L^{1}(\mu)$. Thus, in the case
when $\mu=h\,dm$ with some $h\in D(m)$,

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\mathcal{L}_{g}^{j}\psi=h\int\psi
d\mu\quad\text{weakly in $L^{1}(m)$ for each $\psi\in L^{1}(m_{S})$.}$

On the other hand, since $h$ is an ergodic invariant density and the
restriction of $\mathcal{L}_{g}$ on $L^{1}(m_{S})$ satisfies (WAP), it follows
from Yoshida–Kakutani’s mean ergodic theorem ([92, Theorem 1]) that

$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\mathcal{L}_{g}^{j}\psi=h\int\psi
d\mu\quad\text{strongly in $L^{1}(m)$ for each $\psi\in L^{1}(m_{S})$.}$

## Acknowledgement

We are sincerely grateful to Víctor Araújo, Hiroki Sumi, Mitsuhiro Shishikura
and Masayuki Asaoka for their valuable comments. P. G. Barrientos was
supported by grant PID2020-113052GB-I00 funded by MCIN, PQ 305352/2020-2
(CNPq) and JCNE E-26/201.305/2022 (FAPERJ). F. Nakamura was supported by JSPS
KAKENHI Grant Number 19K21834. Y. Nakano was supported by JSPS KAKENHI Grant
Numbers 19K14575, 19K21834, 21K03332 and 22K03342. H. Toyokawa was supported
by JSPS KAKENHI Grant Numbers 19K21834 and 21K20330.

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