11
$\begingroup$

Let $(X,\mathcal{F})$ be some measure space and endow $\mathbb{R}^\mathbb{Z}$ with the product topology and borel $\sigma$-field. Let $F$ be a point to set mapping $X^\mathbb{Z}\rightarrow \mathcal{P}(\mathbb{R}^\mathbb{Z})$. There are results that ensure that there exists a measurable function $f:X^\mathbb{Z}\rightarrow \mathbb{R}^\mathbb{Z}$ such that $f(x) \in F(x)$ for all $x$.

For example, the Kuratowski-Ryll-Nardzewski theorem states the following sufficient conditions:

  1. $F(x)$ is closed in $\mathbb{R}^\mathbb{Z}$.
  2. For all open $O\subset\mathbb{R}^\mathbb{Z}$ we have $\{x\in X^\mathbb{Z} \mid F(x)\cap O \neq\emptyset\} \in \mathcal{F}$.

Question: Let $\tau$ be the backshift operator on $X^\mathbb{Z}$ or $\mathbb{R}^\mathbb{Z}$, that is $$ \tau(\ldots,x_{-1},x_0,x_1,\ldots) = (\ldots,x_{0},x_1,x_2,\ldots). $$ Suppose that $F(\tau (x)) = \tau (F(x))$, where $\tau$ of a set is defined as the set of the shifted elements. Can we construct $f$ in such a way that $f\circ\tau = \tau\circ f$?

I tried to adjust the proof of the above theorem. Unfortunately it depends on $\mathbb{R}^\mathbb{Z}$ being Polish, i.e. it depends on a complete metric that induces the product topology. An example of such a metric is $$ \delta(x,y) = \sum_{n=1}^{\infty}2^{-n}\frac{|x_n-y_n|}{1+ |x_n-y_n|}, $$ which unfortunately is not shift invariant and thus I'm stuck.

$\endgroup$
  • $\begingroup$ I don't think you can do this without further assumptions on $F$. For example, if $F$ is single-valued, i.e., $F(x) = \{g(x)\}$ for all $x$, then necessarily $f = g$. But one can clearly choose a function $g$ such that $g \circ \tau = \tau \circ g$ is not satisfied, e.g. $g(x) = (\dots, -1,0,1,\dots)$ for all $x$. $\endgroup$ – PhoemueX Dec 5 '16 at 17:19
  • $\begingroup$ @PhoemueX You are absolutely right, I forgot to mention that my $F$ function is such that $F(\tau x) = \tau F(x)$, where shifting a set is the set of all the shifted elements. I'll edit my question. $\endgroup$ – Marc Dec 5 '16 at 22:42
  • $\begingroup$ @AnthonyQuas Yes, I missed the $\neq\emptyset $ sign... Here is a link to a proof of the Kuratowski-Ryll-Nardzewski theorem. It is lemma 22 in medvegyev.uni-corvinus.hu/StochasticIntegration/lib/exe/… $\endgroup$ – Marc Dec 5 '16 at 22:46

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.