Class of cylinder sets and sigma algebra I have the following definition for the class of cylinder sets $\mathcal{Z_J}$, with base $J\subset I $, where $I$ is an index set:

$\mathcal{Z_J}=\{X_J^{-1}(A)\subset\Omega: A \in \mathcal{A_J}\}$ where $\Omega$ is the product space, $\mathcal{A}_J$ is the product $\sigma-$algebra and $X_j$, $j \in J$, is the coordinate map.

Then I have the following collection of such $\mathcal{Z}_J$:

$\mathcal{Z}=\bigcup_{J \subset I finite} \mathcal{Z_J}$.

How can I show that:

*

*$\mathcal{Z}_J$ is a $\sigma-$algebra

*$\mathcal{Z}$ is an algebra

In particular, why isn't $\mathcal{Z}$ a $\sigma-$algebra ?
My book of reference is Probability Theory by A. Klenke (3rd version) and I am totally new to this topic. Thank you.
 A: First, suppose $J=j_{1},\cdots,j_{k},\ A\in\mathcal{A}_{J}$, then$$\mathcal{Z}_{J}=\bigcup_{A\in\mathcal{A}_{J}}\{x\in\Omega:(X_{j_{1}}(x),\cdots,X_{j_{k}}(x))\in A\}.$$Since $\mathcal{A}_{J}$ is a $\sigma$- algebra, $\mathcal{Z}_{J}$ is a $\sigma$- algebra.
Next, suppose$$E=\{x\in\Omega:(X_{t_{1}}(x),\cdots,X_{t_{k}}(x))\in A\}\in\mathcal{Z},$$
$$F=\{x\in\Omega:(X_{s_{1}}(x),\cdots,X_{s_{l}}(x))\in B\}\in\mathcal{Z},$$where $\{t_{1},\cdots,t_{k}\}=J_{1}\subset I,\ \{s_{1},\cdots,s_{l}\}=J_{2}\subset I,\ A\in\mathcal{A}_{J_{1}},\ B\in\mathcal{A}_{J_{2}}$. Then$$E^{c}=\{x\in\Omega:(X_{t_{1}}(x),\cdots,X_{t_{k}}(x))\in\Omega_{J_{1}}\backslash A\}\in\mathcal{Z},$$so $\mathcal{Z}$ is closed under complementation. Now let $(u_{1},\cdots,u_{m})$ be an $m$- tuple containing all the $t_{\alpha}$ and all the $s_{\beta}$. Then $(t_{1},\cdots,t_{k})$ must be the initial segment of some permutation of $(u_{1},\cdots,u_{m})$. Thus there is a permutation $\pi$ of $(1,2,\cdots,m)$ such that$$(u_{\pi^{-1}(1)},\cdots,u_{\pi^{-1}(m)})=(t_{1},\cdots,t_{k},t_{k+1},\cdots,t_{m}),$$where $t_{k+1},\cdots,t_{m}$ are elements of $s_{1},\cdots,s_{l}$. Define $\psi:\Omega_{J_{1}\cup J_{2}}\rightarrow\Omega_{J_{1}}$ by$$\psi(X_{1}(x),\cdots,X_{m}(x))=(X_{\pi^{-1}(1)}(x),\cdots,X_{\pi^{-1}(k)}(x)),$$and set $A'=\psi^{-1}A$, then $A'\in\Omega_{J_{1}\cup J_{2}}$ and$$E=\{x\in\Omega:(X_{u_{1}}(x),\cdots,X_{u_{m}}(x))\in A'\}.$$Similarly for $F$ we have
$$F=\{x\in\Omega:(X_{u_{1}}(x),\cdots,X_{u_{m}}(x))\in B'\},$$where $B'\in\Omega_{J_{1}\cup J_{2}}$. But then$$E\cup F=\{x\in\Omega:(X_{u_{1}}(x),\cdots,X_{u_{m}}(x))\in A'\cup B'\}\in\mathcal{Z}.$$This proves that $\mathcal{Z}$ is an algebra.
