I have a couple of questions regarding ergodicity for Markov processes in continuous time. (In particular, the first question seems like it should be particularly basic, and yet I haven't managed to find a proof [or counterexample!].)


We have a family $(P\_x^t)\_{x \in \mathbb{R},t \geq 0}$ of Borel probability measures 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 $s,t \geq 0$, $P\_x^{s+t}(A)=\int\_\mathbb{R} P\_y^t(A) \, P\_x^s(dy)$.


[We can refer to $(P\_x^t)\_{x \in \mathbb{R},t \geq 0}$ as a "measurable stochastic semigroup". In general, "stochastic semigroups" only need to be measurable in $x$ for each $t$.]


We will say that a probability measure $\rho$ on $\mathbb{R}$ is *stationary* if $\rho(A)=\int\_\mathbb{R} P\_x^t(A) \, \rho(dx)$ for all $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. We will say that a probability measure on $\mathbb{R}$ is *ergodic* if it is an extremal point of the convex set of stationary probability measures.



> 
> Q1. Let $(\Omega,\mathcal{F},(\mathcal{F}\_{t \geq 0}),\mathbb{P})$ be a filtered probability space, and let $(X\_t)\_{t \geq 0}$ be a progressively measurable real-valued homogeneous Markov process with transition probabilities given by $(P\_x^t)\_{x \in \mathbb{R},t \geq 0}$ -- that is to say, $P\_{X\_s(\cdot)}^t(A)$ is a conditional probability of $X\_{s+t}^{-1}(A)$ with respect to $\mathcal{F}\_s$ (for all $s,t,A$). Suppose also that $\rho:=X\_{0\ast}\mathbb{P}$ is stationary. Fix any bounded measurable $f:\mathbb{R} \to \mathbb{R}$; is it the case that
> 
> 
> $\hspace{5mm} \lim\_{T \to \infty} \frac{1}{T} \int\_0^T f(X\_t(\omega)) \, dt$
> 
> 
> exists for $\mathbb{P}$-almost all $\omega \in \Omega$?
> 
> 
> (Please note that we do *not* assume any kind of continuity of $(X\_t)$, but only that it is progressively measurable.)
> 
> 
> 


Now in terms of my motivation, what I am really after is an ergodic decomposition theorem for the setting that I'm currently working with; I think that a positive answer to Q1 will be enough for me to prove this. However, I would ideally like to know if ergodic decompositions exist more generally:



> 
> Q2. Suppose $\rho$ is a stationary probability measure. Does there exist a probability measure $Q$ on the set $\mathcal{M}$ of probability measures on $\mathbb{R}$ (equipped with the usual $\sigma$-algebra, which is known to be standard) such that
> 
> 
> 1. $Q$-almost every $\mu \in \mathcal{M}$ is ergodic;
> 2. for all $A \in \mathcal{B}(\mathbb{R})$, $\rho(A) = \int\_\mathcal{M} \mu(A) \, Q(d\mu)$?
> 
> 
> 


The following might be useful:


**Equivalent definitions of ergodicity**: Given a stationary probability measure $\rho$, we will say that a set $A \in \mathcal{B}(\mathbb{R})$ is $\rho$-*almost stationary* if for all $t \geq 0$, $\rho(x \in A: P\_x^t(A)=1)=\rho(A)$.


(1) In analogy to Proposition 7.2.4 of books.google.co.uk/books?isbn=0521515971 (p378) for deterministic systems, we have that a stationary probability measure $\rho$ is ergodic if and only if every $\rho$-almost stationary set has $\rho$-trivial measure: If $\rho(A) \in (0,1)$ and $A$ is $\rho$-almost stationary, then $\rho$ conditioned on $A$ and $\rho$ conditioned on $\mathbb{R} \setminus A$ are stationary probability measures which can be linearly combined in the obvious way to give $\rho$. In the other direction, it suffices to show that if every $\rho$-almost stationary set has trivial measure and $\tilde{\rho}$ is a stationary probability measure that is absolutely continuous with respect to $\rho$, then $\rho=\tilde{\rho}$. Take a density $g$ of $\tilde{\rho}$ with respect to $\rho$. For each $t$, define the probability measure $\rho\_t$ on $\mathbb{R} \times \mathbb{R}$ by $\rho\_t(A \times B) = \int\_A P\_x^t(B) \, \rho(dx)$. The stationarity of $\tilde{\rho}$ implies that


$\hspace{5mm} \int\_{A \times (X \setminus A)} g(x\_1) \, \rho\_t(d(x\_1,x\_2)) \ = \ \int\_{(X \setminus A) \times A} g(x\_1) \, \rho\_t(d(x\_1,x\_2))$


for any $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. Setting $A:=\{x \in X : g(x) \geq 1\}$, the above equation (combined with the stationarity of $\rho$) implies that $A$ is $\rho$-almost stationary, so $A$ has trivial measure. It follows that $\tilde{\rho}=\rho$.


(2) We will say that a set $A \in \mathcal{B}(\mathbb{R})$ is *invariant* if for all $t \geq 0$ and all $x \in A$, $P\_x^t(A)=1$. Given a set $A$ that is $\rho$-almost stationary, there exists a set $A'$ that is invariant, with $\rho(A \triangle A')=0$. Namely, set


$\hspace{5mm} A' \ := \ \{ x \in X : \textrm{Leb}(t \geq 0 : P\_x^t(A)<1) = 0 \}$


where $\textrm{Leb}$ denotes the Lebesgue measure. So a stationary probability measure $\rho$ is ergodic if and only if every invariant set has $\rho$-trivial measure.


(It is perhaps worth pointing out that (1) does not rely on the stochastic semigroup $(P\_x^t)$ being a "measurable" stochastic semigroup, but the construction in (2) does rely on this.)




---


**Update:** I'm pretty sure the answer to Q2 is *yes*, because I think I can prove it using an ergodic theorem for measurable stochastic semigroups; namely, letting $\rho$ be a stationary probability measure, I think I can first prove that for any bounded measurable $f:\mathbb{R} \to \mathbb{R}$,


$\hspace{5mm} \lim\_{T \to \infty} \frac{1}{T}\int\_0^T \int\_\mathbb{R} \! f(y) P\_x^t(dy) \; dt$


exists for $\rho$-almost all $x \in \mathbb{R}$, with the limit (as a function of $x$) being a conditional expectation of $f$ over the probability space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\rho)$ with respect to the $\sigma$-algebra of $\rho$-almost stationary sets. (As mentioned in Kifer's book "Ergodic Theory of Random Transformations", the discrete-time analogue of the above statement can be obtained as a special case of the Chacon-Ornstein ergodic theorem.) Using this fact, it should be possible to prove the ergodic decomposition theorem (by a similar approach as in the proof for deterministic dynamical systems).


However, I suspect that the answer to Q1 is *no* (although I do not have a counterexample!!). More precisely, I suspect that the answer to Q1 is the same as the answer to my question [Is it true that all stationary measurable stochastic processes are "measurably stationary"?](https://mathoverflow.net/questions/181640/is-it-true-that-all-stationary-measurable-stochastic-processes-are-measurably-s) - and I expect that the answer to that question is *no* (although again, I do not have a counterexample).


If the answer to Q1 is *no*, I wonder whether perhaps it becomes *yes* in the particular case that $(X\_t)\_{t \geq 0}$ is a *strong* Markov process.



Birkhoff Ergodic Theorem and Ergodic Decomposition Theorem for Continuous-Time Markov Processes
I have the answers to my two questions. (I've actually had them for a while; apologies for the delay in posting.) I will give them in reverse order:


The answer to Q2 is *yes*; the structure of the proof is exactly as I outlined in the update. Details can be found in section 5 (in particular, Corollary 100) of my notes <http://wwwf.imperial.ac.uk/~jmn07/Ergodic_Theory_for_Semigroups_of_Markov_Kernels.pdf>.


The answer to Q1 is also *yes*: Since $f$ is bounded, it is sufficient just to consider the limit as $T$ tends to $\infty$ in the integers. By the positive answer to the question [Is it true that all stationary measurable stochastic processes are "measurably stationary"?](https://mathoverflow.net/questions/181640/is-it-true-that-all-stationary-measurable-stochastic-processes-are-measurably-s), the discrete-time stochastic process $\left(\int\_n^{n+1} f(X\_t(\cdot)) \, dt \right)\_{n \geq 0}$ is stationary, and therefore Birkhoff's ergodic theorem (applied to the shift map on $\mathbb{R}^{\mathbb{N}\_0}$) gives the desired convergence.


