# Measurable function

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$$\mbox{?}$$

## 1 Answer

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 $$[0,1]$$.

• I forgot to say that $\phi:X\to\mathbb{R}$ be a continuous function. – Rusbert Mar 12 '19 at 16:02