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Aug 1 at 22:57 comment added No-one If $\overline{\mathcal{A}}$ denotes the closure of $\mathcal{A}$ with respect to pointwise bounded convergence, then one can see that $\overline{\mathcal{A}}$ is a lattice in the same way as done in the proof of the Stone-Weiestrass theorem, and hence $1_{\{A_tf(\cdot,y)<\epsilon\}}=\lim_j j A_t(\epsilon-f(\cdot,y))\wedge 1\vee 0\in \overline{\mathcal{A}}$, and hence since for $f(x',y'):=d(x',y')/(1+d(x',y'))$ $\{A_tf(\cdot,y)<\epsilon\}$ is a basis of open neighbourhoods $y\in L^0$, by the monotone class theorem $\overline{\mathcal{A}}$ contains all bounded measurable functions on $L^0$.
Aug 1 at 22:45 comment added No-one @AlexanderShamov Hi, sorry for commenting on this very old post, but I think it is worth pointing out that to get that the distribution of $(X_t)_{t\geq 0}$ on $L^0$ can be recovered from the distribution of the finite n-uples $(X_{t_1},...,X_{t_n})$ on $E^n$ one needs to show that all Borel functions on $L^0$ to $\mathbb{R}$ can be recovered as bounded limit of functions in the algebra generated by $\{A_tf\}$, that is $\mathcal{A}:=\{\text{finite sums of the form }\sum b_{k_1,...,k_j} (A_{t_1}f_1)^{k_1}...(A_{t_j}f_j)^{k_j} \}$.
Sep 24, 2014 at 17:49 comment added Alexander Shamov Of course you can. But the idea is pretty much standard anyway...
Sep 24, 2014 at 17:13 comment added Julian Newman Can I please cite this MathOverflow post (or if not, then just cite you) within a document that I am writing? (I do not currently intend to submit the document for publication, but only to post it on my website.)
Sep 24, 2014 at 0:13 comment added Julian Newman Thank you very much. I think this is a very nice argument. (I had noticed that the first moment of the r.v. in question can be determined from the finite-dimensional distributions, but hadn't managed to think beyond that observation.)
Sep 24, 2014 at 0:00 vote accept Julian Newman
Sep 23, 2014 at 19:12 comment added Alexander Shamov There is indeed such a map, under the assumptions that a) $X$ generates the whole σ-algebra on $\Omega$, b) $(\Omega,\mathcal{F},\mathsf{P})$ is standard (see the references in en.wikipedia.org/wiki/…). But we don't actually need it. Please see the updated answer...
Sep 23, 2014 at 19:04 history edited Alexander Shamov CC BY-SA 3.0
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Sep 23, 2014 at 17:56 comment added Julian Newman If you have in mind that one can "without loss of generality" replace $\Omega$ with the product space $E^{[0,\infty)}$ (equipped with the law $\rho$ of the stochastic process $\mathbf{X}$), then this would surely require that the map $(t,f)\mapsto f(t)$ from $[0,\infty)\times E^{[0,\infty)}$ to $E$ is measurable with respect to the $\mathrm{Leb}\otimes\rho$-completion of $\mathcal{B}([0,\infty))\otimes E^{\otimes [0,\infty)}$ (or something like that!). But is that true? Or have I missed the mark altogether?
Sep 23, 2014 at 17:54 comment added Julian Newman Thank you for your reply. I do not understand why stationarity implies the existence of the measure-preserving transformation $T_s$ on $\Omega$ that you described. [If I have such a measure-preserving transformation then the answer to my question trivially becomes yes: my second $\mathbb{R}$-valued map is just the composition of my first one with $T_\tau$---and the two maps are certainly measurable since the measure of a section of a measurable set depends measurably on where the section is taken.]
Sep 23, 2014 at 16:12 history answered Alexander Shamov CC BY-SA 3.0