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[I've decided to rewrite the question, to make the essential point clearer.]

Let $\,\mathbb{R}^{[0,\infty)}:=\{(x_t)_{t \geq 0} : x_t \in \mathbb{R} \ \, \forall t\}$. We say that a set $Y \subset \mathbb{R}^{[0,\infty)}$ is shift-invariant if for all $(x_t)_{t \geq 0} \in Y$ and $\tau \geq 0$, we have that $(x_{\tau+t})_{t \geq 0} \in Y$.

We say that a shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ is nice if the map

$\begin{align} Y \times [0,\infty) \, &\to \, \mathbb{R} \\ ( (x_t)_{t \geq 0} \, , \, \tau ) \, &\mapsto \, x_\tau \end{align}$

is jointly measurable (where $Y$ is equipped with the induced $\sigma$-algebra from $\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)}$). [This is equivalent to saying that for any measurable space $(\Omega,\mathcal{F})$ and any function $f:[0,\infty) \times \Omega \to \mathbb{R}$, if the map $\omega \mapsto f(t,\omega)$ is measurable for each $t$ and $(f(t,\omega))_{t \geq 0} \in Y$ for each $\omega$, then $f$ is jointly measurable. Thus, for example, it is well-known that the set of right-continuous paths in $\mathbb{R}$ is nice.]

We have the following fact:

Theorem. Let $\,P:[0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]\,$ be a measurable Markov transition function (meaning: $P(t,x,\cdot)$ is a probability measure for all $x$ and $t$, $P(0,x,\cdot)=\delta_x$, $P$ satisfies the Chapman-Kolmogorov equation, and $P$ is jointly measurable in its first two inputs.) Let $\rho$ be a Borel probability measure on $\mathbb{R}$ that is ergodic with respect to the Markov transition function $P$. Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0},\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be an $(\mathcal{F}_t)_{t \geq 0}$-adapted real-valued stochastic process such that

1. $\mathbb{P}(X_{s+t} \in A | \mathcal{F}_s) = P(t,X_s,A)$ for all $s,t,A$;
2. the law of $X_0$ is $\rho$;
3. there exists a nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ such that $(X_t(\omega))_{t \geq 0} \in Y$ for all $\omega$.

Then for any $A \in \mathcal{B}(\mathbb{R})$, for $\mathbb{P}$-almost all $\omega \in \Omega$, $$ \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt \; \to \; \rho(A) \hspace{3mm} \textit{as } T \to \infty. $$

One can obtain this theorem as a consequence of the continuous-time Birkhoff ergodic theorem; the dynamical system on which Birkhoff's theorem is applied is the shift dynamical system on $Y$.

(The reason why this works is that, because of the ergodicity of $\rho$, the law of $(X_t)_{t \geq 0}$ is ergodic with respect to the shift dynamical system on $Y$. See e.g. section 4 of my notes for details.)

My question is:

Suppose that in the above theorem, we replace condition 3 with the weaker condition that the map $(t,\omega) \mapsto X_t(\omega)$ is jointly measurable. Does the theorem still remain correct?

Remark. I already know that the limit $\,\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt\,$ will still exist for almost every $\omega \in \Omega$ [see Q1 of here]. But my problem is that I do not know if this limit will still be (almost everywhere) independent of $\omega$.

---

The difficulty is as follows: Fix $A \in \mathcal{B}(\mathbb{R})$. Let $Z \subset \mathbb{R}^{[0,\infty)}$ be the set of all $(x_t) \in \mathbb{R}^{[0,\infty)}$ with the property that $t \mapsto x_t$ is measurable and the limit $$ L\!\left((x_t)_{t \geq 0}\right):=\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(x_t) \, dt $$ exists. Obviously $L:Z \to [0,1]$ is a shift-invariant function. But moreover, for any nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$, the restriction of $L$ to $Z \cap Y$ is a measurable function; and therefore if condition 3 holds then, due to the ergodicity of $\rho$, $L$ is $\mathbf{X}_\ast\mathbb{P}$-almost everywhere constant (where $\mathbf{X}(\omega):=(X_t(\omega))_{t \geq 0}$). However, $L:Z \to \mathbb{R}$ is not itself a measurable function (I don't think!), and so without condition 3, I cannot see how the ergodicity of $\rho$ helps me.

[I've decided to rewrite the question, to make the essential point clearer.]

Let $\,\mathbb{R}^{[0,\infty)}:=\{(x_t)_{t \geq 0} : x_t \in \mathbb{R} \ \, \forall t\}$. We say that a set $Y \subset \mathbb{R}^{[0,\infty)}$ is shift-invariant if for all $(x_t)_{t \geq 0} \in Y$ and $\tau \geq 0$, we have that $(x_{\tau+t})_{t \geq 0} \in Y$.

We say that a shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ is nice if the map

$\begin{align} Y \times [0,\infty) \, &\to \, \mathbb{R} \\ ( (x_t)_{t \geq 0} \, , \, \tau ) \, &\mapsto \, x_\tau \end{align}$

is jointly measurable (where $Y$ is equipped with the induced $\sigma$-algebra from $\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)}$). [This is equivalent to saying that for any measurable space $(\Omega,\mathcal{F})$ and any function $f:[0,\infty) \times \Omega \to \mathbb{R}$, if the map $\omega \mapsto f(t,\omega)$ is measurable for each $t$ and $(f(t,\omega))_{t \geq 0} \in Y$ for each $\omega$, then $f$ is jointly measurable. Thus, for example, it is well-known that the set of right-continuous paths in $\mathbb{R}$ is nice.]

We have the following fact:

Theorem. Let $\,P:[0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]\,$ be a measurable Markov transition function (meaning: $P(t,x,\cdot)$ is a probability measure for all $x$ and $t$, $P(0,x,\cdot)=\delta_x$, $P$ satisfies the Chapman-Kolmogorov equation, and $P$ is jointly measurable in its first two inputs.) Let $\rho$ be a Borel probability measure on $\mathbb{R}$ that is ergodic with respect to the Markov transition function $P$. Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0},\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be an $(\mathcal{F}_t)_{t \geq 0}$-adapted real-valued stochastic process such that

1. $\mathbb{P}(X_{s+t} \in A | \mathcal{F}_s) = P(t,X_s,A)$ for all $s,t,A$;
2. the law of $X_0$ is $\rho$;
3. there exists a nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ such that $(X_t(\omega))_{t \geq 0} \in Y$ for all $\omega$.

Then for any $A \in \mathcal{B}(\mathbb{R})$, for $\mathbb{P}$-almost all $\omega \in \Omega$, $$ \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt \; \to \; \rho(A) \hspace{3mm} \textit{as } T \to \infty. $$

One can obtain this theorem as a consequence of the continuous-time Birkhoff ergodic theorem; the dynamical system on which Birkhoff's theorem is applied is the shift dynamical system on $Y$.

(The reason why this works is that, because of the ergodicity of $\rho$, the law of $(X_t)_{t \geq 0}$ is ergodic with respect to the shift dynamical system on $Y$. See e.g. section 4 of my notes for details.)

My question is:

Suppose that in the above theorem, we replace condition 3 with the weaker condition that the map $(t,\omega) \mapsto X_t(\omega)$ is jointly measurable. Does the theorem still remain correct?

Remark. I already know that the limit $\,\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt\,$ will still exist for almost every $\omega \in \Omega$ [see Q1 of here]. But my problem is that I do not know if this limit will still be (almost everywhere) independent of $\omega$.

---

The difficulty is as follows: Fix $A \in \mathcal{B}(\mathbb{R})$. Let $Z \subset \mathbb{R}^{[0,\infty)}$ be the set of all $(x_t) \in \mathbb{R}^{[0,\infty)}$ with the property that $t \mapsto x_t$ is measurable and the limit $$ L\!\left((x_t)_{t \geq 0}\right):=\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(x_t) \, dt $$ exists. Obviously $L:Z \to [0,1]$ is a shift-invariant function. But moreover, for any nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$, the restriction of $L$ to $Z \cap Y$ is a measurable function; and therefore if condition 3 holds then, due to the ergodicity of $\rho$, $L$ is $\mathbf{X}_\ast\mathbb{P}$-almost everywhere constant (where $\mathbf{X}(\omega):=(X_t(\omega))_{t \geq 0}$). However, $L:Z \to \mathbb{R}$ is not itself a measurable function (I don't think!), and so without condition 3, I cannot see how the ergodicity of $\rho$ helps me.

[I've decided to rewrite the question, to make the essential point clearer.]

Let $\,\mathbb{R}^{[0,\infty)}:=\{(x_t)_{t \geq 0} : x_t \in \mathbb{R} \ \, \forall t\}$. We say that a set $Y \subset \mathbb{R}^{[0,\infty)}$ is shift-invariant if for all $(x_t)_{t \geq 0} \in Y$ and $\tau \geq 0$, we have that $(x_{\tau+t})_{t \geq 0} \in Y$.

We say that a shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ is nice if the map

$\begin{align} Y \times [0,\infty) \, &\to \, \mathbb{R} \\ ( (x_t)_{t \geq 0} \, , \, \tau ) \, &\mapsto \, x_\tau \end{align}$

is jointly measurable (where $Y$ is equipped with the induced $\sigma$-algebra from $\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)}$). [This is equivalent to saying that for any measurable space $(\Omega,\mathcal{F})$ and any function $f:[0,\infty) \times \Omega \to \mathbb{R}$, if the map $\omega \mapsto f(t,\omega)$ is measurable for each $t$ and $(f(t,\omega))_{t \geq 0} \in Y$ for each $\omega$, then $f$ is jointly measurable. Thus, for example, it is well-known that the set of right-continuous paths in $\mathbb{R}$ is nice.]

We have the following fact:

Theorem. Let $\,P:[0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]\,$ be a measurable Markov transition function (meaning: $P(t,x,\cdot)$ is a probability measure for all $x$ and $t$, $P(0,x,\cdot)=\delta_x$, $P$ satisfies the Chapman-Kolmogorov equation, and $P$ is jointly measurable in its first two inputs.) Let $\rho$ be a Borel probability measure on $\mathbb{R}$ that is ergodic with respect to the Markov transition function $P$. Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0},\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be an $(\mathcal{F}_t)_{t \geq 0}$-adapted real-valued stochastic process such that

1. $\mathbb{P}(X_{s+t} \in A | \mathcal{F}_s) = P(t,X_s,A)$ for all $s,t,A$;
2. the law of $X_0$ is $\rho$;
3. there exists a nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ such that $(X_t(\omega))_{t \geq 0} \in Y$ for all $\omega$.

Then for any $A \in \mathcal{B}(\mathbb{R})$, for $\mathbb{P}$-almost all $\omega \in \Omega$, $$ \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt \; \to \; \rho(A) \hspace{3mm} \textit{as } T \to \infty. $$

One can obtain this theorem as a consequence of the continuous-time Birkhoff ergodic theorem; the dynamical system on which Birkhoff's theorem is applied is the shift dynamical system on $Y$.
[See e.g. section 4 of my notes for details.]

My question is:

Suppose that in the above theorem, we replace condition 3 with the weaker condition that the map $(t,\omega) \mapsto X_t(\omega)$ is jointly measurable. Does the theorem still remain correct?

Remark. I already know that the limit $\,\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt\,$ will still exist for almost every $\omega \in \Omega$ [see Q1 of here]. But my problem is that I do not know if this limit will still be (almost everywhere) independent of $\omega$.

---

The difficulty is as follows: Let $Z \subset \mathbb{R}^{[0,\infty)}$ be the set of all $(x_t) \in \mathbb{R}^{[0,\infty)}$ with the property that $t \mapsto x_t$ is measurable and the limit $$ L\left((x_t)_{t \geq 0}\right):=\lim_{T \to \infty} \frac{1}{T} \int_0^T x_t \, dt $$ exists and is finite. Obviously $L:Z \to \mathbb{R}$ is a shift-invariant function. But moreover, for any nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$, the restriction of $L$ to $Z \cap Y$ is a measurable function; and therefore if condition 3 holds then, due to the ergodicity of $\rho$, $L$ is $\mathbf{X}_\ast\mathbb{P}$-almost everywhere constant (where $\mathbf{X}(\omega):=(X_t(\omega))_{t \geq 0}$). However, $L:Z \to \mathbb{R}$ is not itself a measurable function (I don't think!), and so I cannot see how the ergodicity of $\rho$ helps me.

[I've decided to rewrite the question, to make the essential point clearer.]

Let $\,\mathbb{R}^{[0,\infty)}:=\{(x_t)_{t \geq 0} : x_t \in \mathbb{R} \ \, \forall t\}$. We say that a set $Y \subset \mathbb{R}^{[0,\infty)}$ is shift-invariant if for all $(x_t)_{t \geq 0} \in Y$ and $\tau \geq 0$, we have that $(x_{\tau+t})_{t \geq 0} \in Y$.

We say that a shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ is nice if the map

$\begin{align} Y \times [0,\infty) \, &\to \, \mathbb{R} \\ ( (x_t)_{t \geq 0} \, , \, \tau ) \, &\mapsto \, x_\tau \end{align}$

is jointly measurable (where $Y$ is equipped with the induced $\sigma$-algebra from $\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)}$). [This is equivalent to saying that for any measurable space $(\Omega,\mathcal{F})$ and any function $f:[0,\infty) \times \Omega \to \mathbb{R}$, if the map $\omega \mapsto f(t,\omega)$ is measurable for each $t$ and $(f(t,\omega))_{t \geq 0} \in Y$ for each $\omega$, then $f$ is jointly measurable. Thus, for example, it is well-known that the set of right-continuous paths in $\mathbb{R}$ is nice.]

We have the following fact:

Theorem. Let $\,P:[0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]\,$ be a measurable Markov transition function (meaning: $P(t,x,\cdot)$ is a probability measure for all $x$ and $t$, $P(0,x,\cdot)=\delta_x$, $P$ satisfies the Chapman-Kolmogorov equation, and $P$ is jointly measurable in its first two inputs.) Let $\rho$ be a Borel probability measure on $\mathbb{R}$ that is ergodic with respect to the Markov transition function $P$. Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0},\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be an $(\mathcal{F}_t)_{t \geq 0}$-adapted real-valued stochastic process such that

1. $\mathbb{P}(X_{s+t} \in A | \mathcal{F}_s) = P(t,X_s,A)$ for all $s,t,A$;
2. the law of $X_0$ is $\rho$;
3. there exists a nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ such that $(X_t(\omega))_{t \geq 0} \in Y$ for all $\omega$.

Then for any $A \in \mathcal{B}(\mathbb{R})$, for $\mathbb{P}$-almost all $\omega \in \Omega$, $$ \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt \; \to \; \rho(A) \hspace{3mm} \textit{as } T \to \infty. $$

One can obtain this theorem as a consequence of the continuous-time Birkhoff ergodic theorem; the dynamical system on which Birkhoff's theorem is applied is the shift dynamical system on $Y$.

(The reason why this works is that, because of the ergodicity of $\rho$, the law of $(X_t)_{t \geq 0}$ is ergodic with respect to the shift dynamical system on $Y$. See e.g. section 4 of my notes for details.)

My question is:

Suppose that in the above theorem, we replace condition 3 with the weaker condition that the map $(t,\omega) \mapsto X_t(\omega)$ is jointly measurable. Does the theorem still remain correct?

Remark. I already know that the limit $\,\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt\,$ will still exist for almost every $\omega \in \Omega$ [see Q1 of here]. But my problem is that I do not know if this limit will still be (almost everywhere) independent of $\omega$.

---

The difficulty is as follows: Fix $A \in \mathcal{B}(\mathbb{R})$. Let $Z \subset \mathbb{R}^{[0,\infty)}$ be the set of all $(x_t) \in \mathbb{R}^{[0,\infty)}$ with the property that $t \mapsto x_t$ is measurable and the limit $$ L\!\left((x_t)_{t \geq 0}\right):=\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(x_t) \, dt $$ exists. Obviously $L:Z \to [0,1]$ is a shift-invariant function. But moreover, for any nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$, the restriction of $L$ to $Z \cap Y$ is a measurable function; and therefore if condition 3 holds then, due to the ergodicity of $\rho$, $L$ is $\mathbf{X}_\ast\mathbb{P}$-almost everywhere constant (where $\mathbf{X}(\omega):=(X_t(\omega))_{t \geq 0}$). However, $L:Z \to \mathbb{R}$ is not itself a measurable function (I don't think!), and so without condition 3, I cannot see how the ergodicity of $\rho$ helps me.

Let $\,(P_x^t)_{x \in \mathbb{R} , t \geq 0}\,$ be a measurable Markovian family of transition probabilities - that is, a family of Borel probability measures $P_x^t$ on $\mathbb{R}$ such that

1. for all $A \in \mathcal{B}(\mathbb{R})$, the map $(x,t) \mapsto P_x^t(A)$ is Borel-measurable;

2. for all $x \in \mathbb{R}$, $P_x^0=\delta_x$;

3. for all $A \in \mathcal{B}(\mathbb{R})$ and $s,t \geq 0$, $\ P_x^{s+t}(A)=\int_\mathbb{R} P_y^t(A) \, P_x^s(dy)$.

Working over a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t \geq 0}),\mathbb{P})$: suppose we have a progressively measurable real-valued stochastic process $(X_t)_{t \geq 0}$ such that

1. [Markov property] for all $A \in \mathcal{B}(\mathbb{R})$ and $s,t \geq 0$, $\hspace{2mm} \mathbb{E}[\mathbf{1}_A(X_{s+t})|\mathcal{F}_s] \, \overset{\mathbb{P}\textrm{-a.s.}}{=} \, P_{X_s}^t(A)\,$;
2. [stationary] for all $t \geq 0$, $X_{t\ast}\mathbb{P}=X_{0\ast}\mathbb{P}$.

Letting $\,\rho:=X_{0\ast}\mathbb{P}\,$, we will say that a set $A \in \mathcal{B}(\mathbb{R})$ is invariant if for each $t \geq 0$, for $\rho$-almost all $x \in \mathbb{R}$, $P_x^t(A)=\mathbf{1}_A(x)$. Let

$\hspace{5mm} \mathcal{G} \ := \{ X_0^{-1}(A) \, : \, \textrm{invariant } A \in \mathcal{B}(\mathbb{R}) \} \; \subset \, \mathcal{F}$.

Fix any bounded measurable $f:\mathbb{R} \to \mathbb{R}$. By the answer to Q1 in Birkhoff Ergodic Theorem and Ergodic Decomposition Theorem for Continuous-Time Markov Processes, the limit

$\hspace{5mm} L(\omega) \, := \, \lim_{T \to \infty} \frac{1}{T} \int_0^T f(X_t(\omega)) \, dt$

exists for $\mathbb{P}$-almost all $\omega \in \Omega$.

Is it necessarily the case that $\hspace{2mm}\mathbb{E}[f(X_0)|\mathcal{G}] \, \overset{\mathbb{P}\textrm{-a.s.}}{=} \, L \,$?

We can at least start with the following special case: If $\mathcal{G}$ consists only of null sets and full-measure sets (i.e. $\rho$ is an ergodic measure of $(P_x^t)_{x \in \mathbb{R} , t \geq 0}$), is it necessarily the case that $L$ is almost-everywhere equal to $\int_\mathbb{R} f \, d\rho$?

---

 

Update: The statement is certainly true if $(X_t)$ has right-continuous sample paths. More generally, I believe that the statement is true in the following class of cases (but I don't even know whether or not this class covers all cases):

Suppose there exists a real-valued stochastic process $(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{P}},(\tilde{X}_t)_{t \geq 0})$, sharing the same finite-dimensional distributions as $(X_t)$, and a set $Y \subset \mathbb{R}^{[0,\infty)}$ such that

1. for all $(x_t)_{t \geq 0} \in Y$ and $s \geq 0$, $(x_{s+t})_{t \geq 0} \in Y$;
2. the map $(\tau,(x_t)_{t \geq 0}) \mapsto x_\tau$ from $[0,\infty) \times Y$ to $\mathbb{R}$ is jointly measurable (where $Y$ is equipped with the induced $\sigma$-algebra from $\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)}$);
3. for $\tilde{P}$-almost all $\omega \in \tilde{\Omega}$, $(\tilde{X}_t(\omega))_{t \geq 0} \in Y$.

In such a case, the answer to my question should be yes: The stochastic process $(\tilde{X}_t)$ should have the desired property, by Birkhoff's ergodic theorem applied to the shift dynamical system on $Y$ (see e.g. Theorem 55 of my notes http://wwwf.imperial.ac.uk/~jmn07/Ergodic_Theory_for_Semigroups_of_Markov_Kernels.pdf). Therefore, since $(\tilde{X}_t)$ and $(X_t)$ have the same finite-dimensional distributions, it should follow (by the argument presented in the answer to the question Is it true that all stationary measurable stochastic processes are "measurably stationary"?) that $(X_t)$ also has the desired property.

[I've decided to rewrite the question, to make the essential point clearer.]

Let $\,\mathbb{R}^{[0,\infty)}:=\{(x_t)_{t \geq 0} : x_t \in \mathbb{R} \ \, \forall t\}$. We say that a set $Y \subset \mathbb{R}^{[0,\infty)}$ is shift-invariant if for all $(x_t)_{t \geq 0} \in Y$ and $\tau \geq 0$, we have that $(x_{\tau+t})_{t \geq 0} \in Y$.

We say that a shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ is nice if the map

$\begin{align} Y \times [0,\infty) \, &\to \, \mathbb{R} \\ ( (x_t)_{t \geq 0} \, , \, \tau ) \, &\mapsto \, x_\tau \end{align}$

is jointly measurable(where$Y$ is equipped with the induced$\sigma$-algebra from$\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)}$). [This is equivalent to saying that for any measurable space$(\Omega,\mathcal{F})$ and any function$f:[0,\infty) \times \Omega \to \mathbb{R}$, if the map$\omega \mapsto f(t,\omega)$ is measurable for each$t$ and$(f(t,\omega))_{t \geq 0} \in Y$ for each$\omega$, then$f$ is jointly measurable. Thus, for example, it is well-known that the set of right-continuous paths in$\mathbb{R}$ is nice.]

We have the following fact:

Theorem. Let $\,P:[0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]\,$ be a measurable Markov transition function (meaning: $P(t,x,\cdot)$ is a probability measure for all $x$ and $t$, $P(0,x,\cdot)=\delta_x$, $P$ satisfies the Chapman-Kolmogorov equation, and $P$ is jointly measurable in its first two inputs.) Let $\rho$ be a Borel probability measure on $\mathbb{R}$ that is ergodic with respect to the Markov transition function $P$. Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \geq 0},\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be an $(\mathcal{F}_t)_{t \geq 0}$-adapted real-valued stochastic process such that

1. $\mathbb{P}(X_{s+t} \in A | \mathcal{F}_s) = P(t,X_s,A)$ for all $s,t,A$;
2. the law of $X_0$ is $\rho$;
3. there exists a nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$ such that $(X_t(\omega))_{t \geq 0} \in Y$ for all $\omega$.

Then for any $A \in \mathcal{B}(\mathbb{R})$, for $\mathbb{P}$-almost all $\omega \in \Omega$, $$ \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt \; \to \; \rho(A) \hspace{3mm} \textit{as } T \to \infty. $$

One can obtain this theorem as a consequence of the continuous-time Birkhoff ergodic theorem; the dynamical system on which Birkhoff's theorem is applied is the shift dynamical system on $Y$.
[See e.g. section 4 of my notes for details.]

My question is:

Suppose that in the above theorem, we replace condition 3 with the weaker condition that the map $(t,\omega) \mapsto X_t(\omega)$ is jointly measurable. Does the theorem still remain correct?

Remark. I already know that the limit $\,\lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbf{1}_A(X_t(\omega)) \, dt\,$ will still exist for almost every $\omega \in \Omega$ [see Q1 of here]. But my problem is that I do not know if this limit will still be (almost everywhere) independent of $\omega$.

---

The difficulty is as follows: Let $Z \subset \mathbb{R}^{[0,\infty)}$ be the set of all $(x_t) \in \mathbb{R}^{[0,\infty)}$ with the property that $t \mapsto x_t$ is measurable and the limit $$ L\left((x_t)_{t \geq 0}\right):=\lim_{T \to \infty} \frac{1}{T} \int_0^T x_t \, dt $$ exists and is finite. Obviously $L:Z \to \mathbb{R}$ is a shift-invariant function. But moreover, for any nice shift-invariant set $Y \subset \mathbb{R}^{[0,\infty)}$, the restriction of $L$ to $Z \cap Y$ is a measurable function; and therefore if condition 3 holds then, due to the ergodicity of $\rho$, $L$ is $\mathbf{X}_\ast\mathbb{P}$-almost everywhere constant (where $\mathbf{X}(\omega):=(X_t(\omega))_{t \geq 0}$). However, $L:Z \to \mathbb{R}$ is not itself a measurable function (I don't think!), and so I cannot see how the ergodicity of $\rho$ helps me.