We construct a measure for an infinitely often repeated random experiment with finitely many possible outcomes. Let $E$ be the set of possible outcomes. For $e\in E$, let $p_e\geq0$ be the probability that $e$ occurs. Hence $\sum_{e\in E}p_e=1$. For a fixed realization of the repeated experiment, let $\omega_1,\omega_2,\cdots\in E$ be the observed outcomes. Hence the space of all possible outcomes of the repeated experiment is $\Omega=E^\mathbb{N}$. We define the set of all sequences whose first $n$ values are $\omega_1,\cdots\omega_n$:
\begin{equation}
[\omega_1,\cdots\omega_n] := \{\omega'\in\Omega:\omega'_i=\omega_i\;\mathrm{for\;any\;}i=1,\cdots,n\}.
\end{equation}
Let $\mathcal{A}_0=\{\emptyset\}$. For $n\in\mathbb{N}$, define the class of cylinder sets that depend only on the first $n$ coordinates
\begin{equation}
\mathcal{A}_n := \{[\omega_1,\cdots\omega_n] :\omega_1,\cdots\omega_n\in E\},
\end{equation}
and let $\mathcal{A} := \bigcup^\infty_{n=0}\mathcal{A}_n$.
We interpret $[\omega_1,\cdots\omega_n]$ as the event where the outcome of the first experiment is $\omega_1$, the outcome of the second experiment is $\omega_2$ and finally the outcome of the $n$th experiment is $\omega_n$. The outcomes of the other experiments do not play a role for the occurrence of this event. As the individual experiments ought to be independent, we should have for any choice $\omega_1,\cdots\omega_n\in E$ that the probability of the event $[\omega_1,\cdots\omega_n]$ is the product of the probabilities of the individual events; that is
\begin{equation}
\mu([\omega_1,\cdots\omega_n]) = \prod^n_{i=1}p_{\omega_i}.
\end{equation}
This formula defines a content $\mu$ on the semiring $\mathcal{A}$, and our aim is to extend $\mu$ in a unique way to a probability measure on the $\sigma$-algebra $\sigma(\mathcal{A})$ that is generated by $\mathcal{A}$.
Before we do so, we make the following definition. Define the (ultra-)metric $d$ on $\Omega$ by
\begin{equation}
d(\omega,\omega') =
\begin{cases}
2^{-\inf\{n\in\mathbb{N}:\;\omega_n\neq\omega'_n\}}, & \omega\neq\omega', \\
0, & \omega=\omega'.
\end{cases}
\end{equation}
Hence $(\Omega,d)$ is a compact metric space. Clearly,
\begin{equation}
[\omega_1,\cdots\omega_n]=B_{2^{-n}}(\omega)=\{\omega'\in\Omega:\;d(\omega,\omega') <2^{-n}\}.
\end{equation}
The complement of $[\omega_1,\cdots\omega_n]$ is an open set, as it is the union of $(\#E)^n-1$ open balls
\begin{equation}
[\omega_1,\cdots\omega_n]^c=\bigcup_{(\omega'_1,\cdots\omega'_n)\neq(\omega_1,\cdots\omega_n)}[\omega'_1,\cdots\omega'_n].
\end{equation}
Since $\Omega$ is compact, the closed subset $[\omega_1,\cdots\omega_n]$ is compact.
Question: How to show that $\sigma(\mathcal{A})=\mathcal{B}(\Omega,d)$, i.e., the $\sigma$-algebra generated by $\mathcal{A}$ equals the Borel $\sigma$-algebra of $(\Omega,d)$?
My efforts:
Obviously $\mathcal{A}_0\subset\mathcal{B}(\Omega,d)$. For each $n>0$, $\mathcal{A}_n\subset\mathcal{B}(\Omega,d)$ since $\mathcal{A}_n=\{B_{2^{-n}}(\omega):\omega\in\Omega\}$. Therefore, $\mathcal{A}\subset\mathcal{B}(\Omega,d)$. Since $\sigma(\mathcal{A})$ is the smallest $\sigma$-algebra containing $\mathcal{A}$, $\sigma(\mathcal{A})\subset\mathcal{B}(\Omega,d)$.
Next we only need to show $\mathcal{B}(\Omega,d)\subset\sigma(\mathcal{A})$. Let's try to show that each open set $U$ can be written as countable union of sets in $\mathcal{A}$. For each $\omega\in U$, there exists a positive integer $n(\omega)$ such that $B_{2^{-n(\omega)}}(\omega)\subset U$. Thus $U=\bigcup_{\omega\in U}B_{2^{-n(\omega)}}(\omega)$. We need to reduce this possibly uncountable union to countable union.
Then I don't know how to continue.
Best Answer
The cylinder sets are a countable base for $\Omega$, in the product topology, which is the same as the topology induced by $d$, so all open sets of $\Omega$ are in $\sigma(\mathcal{A})$ and so this $\sigma$-algebra contains the Borel sets.
As all cylinder sets are themselves Borel, the reverse inclusion is immediate.