(Philosophically, the following question is of a similar flavour to [A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary](https://mathoverflow.net/questions/42141/a-stochastic-process-that-is-1st-and-2nd-order-strictly-stationary-but-not-3r), but more "advanced".)


Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space, and let $X:[0,\infty) \times \Omega \to E$ be a measurable function with the property that for any $n \in \mathbb{N}$, for all $\tau,t\_1,\ldots,t\_n \geq 0$ the image measure of $\mathbb{P}$ under the $E^n$-valued map


$\hspace{5mm} \omega \mapsto \left( X(t\_1,\omega),\ldots,X(t\_n,\omega) \right)$


is equal to the image measure of $\mathbb{P}$ under the $E^n$-valued map


$\hspace{5mm} \omega \mapsto \left( X(t\_1+\tau,\omega),\ldots,X(t\_n+\tau,\omega) \right)$.



> 
> Is it necessarily the case that for any $n \in \mathbb{N}$, for all $A \in \mathcal{E}^{\otimes n}$ and all $s,\tau,t\_1,\ldots,t\_n \geq 0$, the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map
> 
> 
> $\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [0,s]: \left( X(t\_1+t,\omega),\ldots,X(t\_n+t,\omega) \right) \in A )$
> 
> 
> is equal to the image measure of $\mathbb{P}$ under the $\mathbb{R}$-valued map
> 
> 
> $\hspace{2mm} \omega \mapsto \textrm{Leb}(t \in [\tau,\tau+s]: \left( X(t\_1+t,\omega),\ldots,X(t\_n+t,\omega) \right) \in A )$
> 
> 
> (where $\textrm{Leb}$ denotes the Lebesgue measure)?
> 
> 
> 


How about in the special case that $(E,\mathcal{E})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$?


Now I've asked the above question in quite a general form, but what I ultimately want is: **if $(X\_t)\_{t \geq 0}$ is a measurable stationary stochastic process in continuous time, then is the discrete-time stochastic process $\left( \int\_n^{n+1} g(X\_t) \, dt \right)\_{n \geq 0}$ necessarily also stationary** (where $g$ is a bounded measurable function)**??**



Is it true that all stationary measurable stochastic processes are "measurably stationary"?
Yes.


I assume that $E$ is a Polish space, or a standard Borel space that can be turned into a Polish space by some auxillary topology.


Basically, you can view your measurable stochastic process as a random element in the space $L^0(\mathbb{R}\_+, E)$ of equivalence classes of $E$-valued measurable functions. $L^0(\mathbb{R}\_+, E)$ comes equipped with the Polish topology of convergence in measure on $[0,n]$ for all $n \in \mathbb{N}$ and its corresponding standard Borel $\sigma$-algebra. While doing that you lose some information (e.g. the value for some specific $t$), but you will be able to calculate your Lebesgue measures as measurable functions on $L^0$.


Now I claim that



> 
> The distribution of the $L^0$ version of $X$ can be recovered from the joint distribution of $(X\_t, t \ge 0)$.
> 
> 
> 


Note that this will immediately imply a positive answer to your question.


Here is a sketch of a proof of my claim. Consider the functions $L^0(\mathbb{R}\_+, E) \to \mathbb{R}$ of the following form:


$$F\_{f,T}: x \mapsto \intop\_0^T f(x\_t) dt,$$


where $f: E \to [0,1]$ is bounded and Borel, and $T \ge 0$. These functions generate the whole Borel $\sigma$-algebra on $L^0$, because the level sets $\{F\_{f,T} < \mathrm{const}\}$ for continuous $f: E \to [0,1]$ generate the topology.


Now it remains to pin down the joint distribution of $F\_{f,T}(X)$. This is done by observing that the moments can be calculated using the joint distribution of $(X\_t)$ by Fubini's theorem:


$$\mathsf{E} \intop\_0^T f\_1(X\_t) dt \dots \intop\_0^T f\_n(X\_t) dt = \intop\_{[0,T]^n} \mathsf{E} f\_1(X\_{t\_1}) \dots f\_n(X\_{t\_n}) dt\_1 \dots dt\_n$$


