# Shift invariant measurable selection theorem

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.

• 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 $x$. – PhoemueX Dec 5 '16 at 17:19
• @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. – Marc Dec 5 '16 at 22:42
• @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/… – Marc Dec 5 '16 at 22:46