Let $(\Omega _0,\mathcal F_0,\mathbb P_0)$ a probability space. Let $\Omega =\Omega _0^{\mathbb Z}$, $\mathcal F=\mathcal F_0^{\otimes \mathbb Z}$ the $\sigma -$algebra generated by cylinder and $\mathbb P=\mathbb P_0^{\otimes \mathbb Z}$.
Let $\theta :\mathbb Z\times \Omega \to \Omega $ defined by $$\theta (n,\omega )=\{\omega _{i+n}\}_{i\in\mathbb Z}.$$ I want to prove that $\theta $ is measurable. Since $\mathcal F$ is generated by the cylinder of the form $$C_{i_1,...,i_m}(B_1,...,B_m)=\{\omega \mid \omega _{i_1}\in B_1,...,\omega _{i_{m}}\in B_m\},$$ where $B_i\in \mathcal F_0$. I just have to prove that $$\theta ^{-1}C_{i_1,...,i_m}(B_1,...,B_n)\in \mathcal B(\mathbb Z)\otimes \mathcal F.$$
Is it correct to say that $$\theta ^{-1}C_{i_1,...,i_n}(B_1,...,B_n)=C_{i_1-n,...,i_m-n}(B_1,...,B_n)\in \mathcal B(\mathbb Z)\otimes \mathcal F \ \ ? \tag{E}$$
The thing is cylinder of $\mathcal B(\mathbb Z)\otimes \mathcal F$ should be of the form $$C_{n,i_1,...,i_m}(U,B_1,...,B_n)=\{(t,\omega )\mid t\in U,\omega _{i_1}\in B_1,...,\omega _{i_m}\in B_m\},$$ $B_i\in \mathcal F_0$, $U\in \mathcal B(\mathbb Z)$, and it's not really what I have in $(E)$. How can I adapt ?