Show that a function is $\sigma$-additive Let $\Omega_0=\{1,2,3\}$, $\Omega=\Omega_0^\mathbb{N},\mathcal{h}=\{E_1\times...\times E_n \times \Omega_0^\mathbb{N}:E_i\subset \Omega_0\}$. So $\mathcal{h}$ is the infinite cartesian product of sets where only finitely many are inequal to $\Omega_0$.  I have already shown that $\mathcal{h}$ is a semi-ring. Now I need to show that if $p$ is a probability measure on $P(\Omega_0)$, and if $E=E_1\times ... \times E_n \times \Omega_0^\mathbb{N}$ and we define $\mathbb{P}(E)=\prod_{i=1}^n p(E_i)$, then $\mathbb{P}$ induces a unique probability measure on $\mathcal{A}_\sigma(\mathcal{h})$, which is the $\sigma$-algebra generated by $\mathcal{h}$. For this I only need to show that $\mathbb{P}$ is $\sigma$-additive on $\mathbb{h}$, as the rest follows from Carathéodory's extension theorem.
So I need to show that for disjoint sets $A_i\in\mathcal{h}$ where $\cup_{i\in\mathbb{N}}A_i$ is also in $\mathcal{h}$ the following holds true: 
$\mathbb{P}(\cup_{i\in\mathbb{N}}A_i)=\sum_{i=1}^\infty\mathbb{P}(A_i)$
However I really have problems doing so. First of all, I am not sure how the union of disjoint sets of $\Omega$ looks like, maybe they have a certain attribute which I cannot see. Or maybe I don't even need the Carathéodory extension theorem. But from now, I don't have a clue how to show this.  So at least a hint would be very nice. 
 A: Hint: Let $S_n = \bigcup_{i = 1}^n A_i$ and $S = \bigcup_{i = 1}^\infty A_i$. Then to show that $\mathbf{P}(S_n) \to \mathbf{P}(S)$ it is equivalent to show that $\mathbf{P}(S \setminus S_n) \to 0$. That is, if you have a decreasing sequence $B_1 \supseteq B_2 \supseteq \cdots$ with $\bigcap_{i \ge 1} B_i = \varnothing$ then $\mathbf{P}(B_i) \to 0$. There are several ways to show this. The simplest way is perhaps to argue that if $\bigcap_{i \ge 1} B_i = \varnothing$ then $B_n = \varnothing$ for $n \gg 0$.
A: I hope I managed to prove it, though a little bit different than you suggested: Let $\cup_{i=1}^\infty A_i = F_1 \times ... \times F_n \times \Omega_0^{\mathbb{N}}$ and $A_i = E_{i_1}\times ... \times E_{i_{n_i}}\times \Omega_0^\mathbb{N}$. Then: $$\chi_{\cup_{i=1}^\infty A_i}(x)=\chi_{F_1\times ... \times F_n \times \Omega_0^n}(x_1,x_2,...)=\sum_{i=1}^\infty \chi_{A_i}(x_1,x_2,...) = \sum_{i=1}^\infty \chi_{E_{i_1}\times ... \times E_{i_{n_i}}\times \Omega_0^\mathbb{N}}(x_1,x_2,...)$$ $$\Rightarrow \chi_{F_1}(x_1)...\chi_{F_n}(x_n)\chi_{\Omega_0^\mathbb{N}}(x_{n+1},...)=\sum_{i=1}^\infty\chi_{E_{i_1}}(x_1)...\chi_{E_{i_{n_i}}}(x_{n_{i}})\chi_{\Omega_0^\mathbb{N}}(x_{n_i+1})$$
Since $\chi_{\Omega_0}(x)$ is $1$ for all $x$ and with integrating over each variable plus $\sigma$-additivity of the Lebesgue-Integral I get: $$\Rightarrow \int_{\Omega_0} \chi_{F_1}(x_1)...\chi_{F_n}(x_n) d(p(x_1))=\sum_{i=1}^\infty\int_{\Omega_0}\chi_{E_{i_1}}d(p(x_1(x_1)...\chi_{E_{i_{n_i}}}(x_{n_{i}})d(p(x_1))$$
$$\Rightarrow p(F_1)\chi_{F_2}(x_2)...\chi_{F_n}(x_n)=\sum_{i=1}^\infty p(E_{i_1})\chi_{E_{i_2}}(x_2)...\chi_{E_{i_{n_i}}}(x_{n_i})$$$$\Rightarrow\text{ }...\text{ }\Rightarrow$$
$$\Rightarrow\prod_{i=1}^np(F_i)=\sum_{i=1}^\infty\prod_{j=1}^{n_i}p(E_{i_j})\Rightarrow \mathbb{P}(\cup_{i=1}^\infty A_i)=\sum_{i=1}^\infty \mathbb{P}(A_i)$$
Is that proof correct?
