Approximation of a measurable set in a product space

Question

Let $X$ be a finite set and $\mu$ a probability measure on $X$. We consider the probability space $(X^{\mathbb{Z}},\mathcal{A}, \mu^{\otimes\mathbb{Z}})$, where $\mathcal{A}$ is the product $\sigma$-algebra, that is, the direct product of the $\sigma$-algebras of the finite set $X$.

Let $A\in \mathcal{A}$ be a measurable set. For $n\in\mathbb{N}$, we define

$$ A_n = \Bigl\{ (x_h)_{h\in\mathbb{Z}} \in X^{\mathbb{Z}} \,\Big|\, \exists (y_h)_{h\in\mathbb{Z}}\in A, (x_h)_{h\in [\![-n,n]\!]} = (y_h)_{h\in [\![-n,n]\!]} \Bigr\}. $$

Question: Is it true that $\mu^{\otimes\mathbb{Z}}\left( \bigcap_{n\in\mathbb{N}}A_n \right)=\mu^{\otimes\mathbb{Z}}(A)$?

The inequality $\mu^{\otimes\mathbb{Z}}(A)\leq \mu^{\otimes\mathbb{Z}}\left( \bigcap_{n\in\mathbb{N}}A_n \right)$ is easy since $A\subset \bigcap_{n\in\mathbb{N}}A_n$. But I don't know if the other inequality is true.

Answer 1

I think a proposition of measures whose proof may be useful is called Continuity From Above which states that

If $\{E_j\}_1 ^\infty \subset \mathcal{A}, E_1 \supset E_2 \supset ...,$ and $\mu(E_1) < \infty$ then $\mu(\bigcap_1^\infty E_j) = \lim_{j \to \infty} \mu(E_j) $

A proof can be found in Folland's Real Analysis Book.