[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
- $\mathbb{P}(X_{s+t} \in A | \mathcal{F}_s) = P(t,X_s,A)$ for all $s,t,A$;
- the law of $X_0$ is $\rho$;
- 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.
