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Let $S$ be a countable set. Consider $X=S^{\mathbb{N}\cup\{0\}}$ the topological Markov shift equipped with the topology generated by the collection of cylinders.

Denoted $\mathcal{B}$ as the Borel $\sigma$-algebra of $X$ this is also the minimal $\sigma$-algebra with respect to which all coordinate function $x\mapsto x_k$, $k\geq 0$, are measurable. Let $\mathcal{B}^{-n}$ denote the smallest sigma algebra with respect to which all the coordinate maps $x\mapsto x_k$ with $k\geq n$ are measurable.

Let $\phi:X\to\mathbb{R}$ be a continuous function. When $\phi$ depends only on the coordinates $x_k$ for $k\geq n$, this is, $$\phi(x)=\phi(x_n,x_{n+1},x_{n+2},\ldots)\quad\mbox{for }x\in X$$ we have that $\phi$ be $\mathcal{B}^{-n}$-measurable.

My question is, If $$\phi(x)=\phi(x_n,x_{n+1},x_{n+2},\ldots)\quad\nu-a.s\mbox{ for }x\in X,$$ where $\nu$ is a measure on $X$, then $\phi$ be $\mathcal{B}^{-n}$-measurable$\mbox{?}$

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The statement

Let $\phi:X\to\mathbb{R}$. When $\phi$ depends only on the coordinates $x_k$ for $k\geq n$, this is, $$\phi(x)=\phi(x_n,x_{n+1},x_{n+2},\ldots)\quad\mbox{for }x\in X$$ we have that $\phi$ be $\mathcal{B}^{-n}$-measurable.

is already false in general, even when $n=0$ and $S=\{0,1\}$. Indeed, in the latter case one may identify $(X,\mathcal{B})$ with $([0,1],\mathcal{B}([0,1]))$, by using the binary expansion of points $x\in[0,1]$. Then the statement quoted above becomes the following:

any function $f\colon[0,1]\to\mathbb R$ is Borel-measurable.

But of course this is false; take e.g. the indicator of a non-Borel subset of $[0,1]$.

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  • $\begingroup$ I forgot to say that $\phi:X\to\mathbb{R}$ be a continuous function. $\endgroup$ – Rusbert Mar 12 at 16:02

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