Transformation shift measurable How to prove that this transformation is measurable?
$\sigma:B(n)\rightarrow{B(n)} $ 
$\sigma(x)(k)=x(k+1)$
$\sigma(...,x_{-1},x_{0},x_{1},...)=(...,x_{0},x_{1},x_{2},...)$
where $B(n)$ with product topology of $Y=\left\{{0,1,2,...,n-1}\right\}$, i.e $B(n)=Y^{\mathbb{Z}}$ and $B(n)$ measurable space with $\sigma$-algebra generated by the base of all cylinders.

Progress. First, this transformation shift $\sigma$ prove is continuous, because all function in topological spaces is measurable. But the form of cylinders is very cumbersome.  
Second,I think in use a Sub basis of $Y^{Z}$ ,for example if $\left\{B_{j}\right\}$ basis for $Y$, a sub basis of that topology  $Y^{Z}$ are $U_{i,j}=\pi_i^{-1}(B_j)$ $i\in I,j\in J$, but no more. Where $I =Y$.
 A: Hints: Denote by $\pi_j: B(n) \to Y, x \mapsto \pi_j(x) := x(j)$ the projection onto the $j$-th coordinate for each $j \in \mathbb{Z}$.


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*Recall the following statement: Let $(X,\mathcal{A})$ be a measurable space such that $\mathcal{A}$ is generated by a family of sets $\mathcal{G}$, i.e. $\sigma(\mathcal{G}) = \mathcal{A}$. Then a mapping $\sigma: (X,\mathcal{A}) \to (X,\mathcal{A})$ is measurable if, and only if, $\sigma^{-1}(G) \in \mathcal{A}$ for all $G \in \mathcal{G}$.

*By definition, the cylindrical $\sigma$-algebra is generated by sets of the form $$G = \bigcap_{j \in F} \{\pi_j = i_j\} \tag{1}$$ for $F \subseteq \mathbb{Z}$ finite and $i_j \in Y=\{0,\ldots,n-1\}$ for all $j \in F$.

*By step 1 and 2 it suffices to show that $$\sigma^{-1} \left( \bigcap_{j \in F} \{\pi_j = i_j\} \right) \in \mathcal{A}.$$ To this end, show that $$\sigma^{-1}\left( \bigcap_{j \in F} \{\pi_j = i_j\} \right) = \bigcap_{j \in F} \{\pi_{j-1} = i_j\}$$  and conclude this set is contained in $\mathcal{A}$. (Hint: It is of the form $(1)$.)

