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:

- $F(x)$ is closed in $\mathbb{R}^\mathbb{Z}$.
- 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.

notsatisfied, e.g. $g(x) = (\dots, -1,0,1,\dots)$ for all $x$. $\endgroup$ – PhoemueX Dec 5 '16 at 17:19