I'm trying to re-construct the product measure over countable product of probability spaces. The original proof is taken from this note. This is to make the ideas clearer to me. Could you have a check on my attempt?
Let $(\Omega_n, \mathcal{F}_n, \mu_n)_n$ be a sequence of probability spaces. Let $\Omega :=\prod_{n =1}^\infty \Omega_n$ and $\bigotimes_n \mathcal{F}_n$ be the smallest $\sigma$-algebra on $\Omega$ such that all projection maps $\pi_n: \Omega \to \Omega_n$ are measurable. Let $$ \mathcal C := \left\{ \prod_n C_n \,\middle\vert\, \exists N \in \mathbb N^*, \forall n \ge N: C_n = \Omega_n \right\}. $$
Then $\mathcal C$ is an algebra on $\Omega$ and $$ \bigotimes_n \mathcal{F}_n = \sigma (\mathcal C). $$
We define $\nu: \mathcal C \to [0, \infty)$ by $$ \nu (C) := \prod_n \mu_n (C_n) \quad \text{for all} \quad C = \prod_n C_n \in \mathcal C. $$
By construction, $\nu$ is finitely additive and $\nu (\emptyset) = 0$. Let's prove that $\nu$ is $\sigma$-additive. Let $(C_n) \subset \mathcal C$ be a decreasing sequence such that $C_n \searrow \emptyset$. It suffices to prove that $\inf_n \nu (C_n) =0$. Let $C_n := \prod_m C_{n,m}$ with $C_{n,m} \in \mathcal F_m$. WLOG, we assume $C_{n,m} = \Omega_m$ for all $m > n$. Assume the contrary that $\inf_n \nu (C_n) = t>0$. This implies $\nu(C_n) \ge t$ for all $n$.
Let $$ f_{n,1} :\Omega_1 \to \mathbb R, \omega \mapsto \int 1_{C_n} (\omega, \omega_2, \ldots, \omega_n) \mathrm d \mu_2 (\omega_2) \cdots \mathrm d \mu_n (\omega_n) \quad \forall n \ge 1. $$
By Fubini's theorem, $$ \nu (C_n) = \int f_{n,1} \mathrm d \mu_1. $$
Because $(C_n)$ is decreasing, $(f_{n,1})_n$ is decreasing. By DCT, we get $$ \int \left ( \inf_{n \ge 1} f_{n,1} \right ) \mathrm d \mu_1 = \inf_{n \ge 1} \int f_{n,1} \mathrm d \mu_1= \inf_{n \ge 1} \nu (C_n) = t. $$
This implies there is $\gamma_1 \in \Omega_1$ such that $$ \inf_{n \ge 1} f_{n,1} (\gamma_1) \ge t. $$
Then
$$ \gamma_1 \in C_{n,1} \quad \forall n \ge 1. $$
$$ C_n (\gamma_1) := \{ (\omega_2, \ldots, \omega_n) \mid (\gamma_1, \omega_2, \ldots, \omega_n) \in C_n \} \in \bigotimes_{i=2}^n \mathcal F_i \quad \forall n \ge 2. $$
Let $$ f_{n,2} :\Omega_2 \to \mathbb R, \omega \mapsto \int 1_{C_n (\gamma_1)} (\omega, \omega_3, \ldots, \omega_n) \mathrm d \mu_3 (\omega_3) \cdots \mathrm d \mu_n (\omega_n) \quad \forall n \ge 2. $$
Then $$ \int f_{n,2} \mathrm d \mu_2 = f_{n,1} (\gamma_1) \ge t \quad \forall n \ge 2. $$
Clearly, $f_{n,2}$ is decreasing. By DCT, $$ \int \left ( \inf_{n \ge 2} f_{n,2} \right ) \mathrm d \mu_2 = \inf_{n \ge 2} \int f_{n,2} \mathrm d \mu_2\ge t. $$
This implies there is $\gamma_2 \in \Omega_2$ such that $$ \inf_{n \ge 2} f_{n,2} (\gamma_2) \ge t. $$
Then
$$ (\gamma_1, \gamma_2) \in C_{n,1} \times C_{n, 2} \quad \forall n \ge 2. $$
$$ C_n (\gamma_1, \gamma_2) := \{ (\omega_3, \ldots, \omega_n) \mid (\gamma_1, \gamma_2,\omega_3, \ldots, \omega_n) \in C_n \} \in \bigotimes_{i=3}^n \mathcal F_i \quad \forall n \ge 3. $$
We repeat above procedure and obtain a sequence $\pmb \gamma := (\gamma_n)$ such that $$ (\gamma_1, \gamma_2, \ldots, \gamma_n) \in C_{n,1} \times C_{n, 2} \times \cdots \times C_{n, n} \quad \forall n \ge 1. $$
As such, $$ \pmb \gamma \in C_n \quad \forall n \ge 1. $$
Hence $\bigcap_n C_n \neq \emptyset$, which is a contradiction. Thus $\nu$ is a pre-measure on $\mathcal C$. The proof is completed by applying Carathéodory's extension theorem.
My definition of $\mathcal C$ does not imply that $\mathcal C$ is an algebra on $\Omega$. Instead, it should be defined as $$ \mathcal C := \left \{ A \times \prod_{i=n+1}^\infty \Omega_i \,\middle\vert\, n \in \mathbb N^*, A \in \bigotimes_{i=1}^n \mathcal F_i \right \}. $$
Then definition of $\nu: \mathcal C \to [0, \infty)$ is adjusted accordingly. If $C = A \times \prod_{i=n+1}^\infty$ with $A \in \bigotimes_{i=1}^n \mathcal F_i$, then $$ \nu (C) := (\otimes_{i=1}^n \mu_i) (C). $$