I heard that given family of probability spaces $(\Omega_{\alpha}, \mathcal{F}_{\alpha}, \mu_{\alpha})_{\alpha \in A}$, there exist product measure $\mu$ on product sigma-algebra (smallest sigma-algebra containing sets of the form $ \prod_{i \in I}E_i \times \prod_{\alpha \in A \setminus I} \Omega_\alpha $ for any finite $I \subseteq A$). This product measure satisfies $$ \mu\left(\prod_{i \in I}E_i \times \prod_{\alpha \in A \setminus I} \Omega_{\alpha}\right) = \prod_{i\in I}\mu_i(E_i) $$
However, I found example that seems to deny that existence. The question is: what is wrong with it.
Observe that $\mu(\prod_{\alpha\in A} \Omega_\alpha) = 1$, so $\mu$ is probability measure. This implies we have continuity from below as well as from above.
Let $A = \mathbb{N}$. Then \begin{align} \mu\left(\prod_{n\in \mathbb{N}} E_n\right) & = \mu\left(\bigcap_{k \in \mathbb{N}} \prod_{n=1}^k E_n \times \prod_{n>k}\Omega_n\right) \\ & = \lim_{k\to\infty} \mu\left(\prod_{n=1}^k E_n \times \prod_{n>k} \Omega_n\right) \\ & = \lim_{k\to\infty} \prod_{n=1}^k \mu_n(E_n) = \prod_{n=1}^\infty \mu_n(E_n). \end{align} Of course such infinite products are measurable, as they are countable intersections of cylinders.
Next choose any non-trivial probability space $(\Omega_0, \mathcal{F}_0, \mu_0)$ and pick $E \in \mathcal{F}_0$ such that $\mu_0(E) \in (0, 1)$. By $\mu$ denote product measure on $\Omega_0^{\mathbb N}$, where each probability space was the same as initial one. Denote $$ E_n = \prod_{k\leq n} \Omega_0 \times \prod_{k > n} E. $$ We have $E_n \subseteq E_{n+1}$ for all $n$, and therefore $$ \mu\left(\bigcup_{n\in \mathbb{N}} E_n\right) = \lim_{n\to\infty}\mu(E_n) = \lim_{n\to\infty}\prod_{k > n}\mu_0(E) = 0, $$ but on the other hand $$ \mu\left(\bigcup_{n\in \mathbb{N}} E_n\right) = \mu\left(\prod_{n\in\mathbb{N}} \Omega_0\right) = 1. $$
Let $\omega_0\in\Omega_0-E$.
Then for element $\omega:=(\omega_0,\omega_0,\omega_0\dots)\in\prod_{n\in\mathbb N}\Omega_0$ for every $n$ we have: $$\omega\notin\Omega_0\times\Omega_0\times\cdots\Omega_0\times E\times E\times\cdots=E_n$$ Showing that: $$\bigcup_{n\in\mathbb N}E_n\neq\prod_{n\in\mathbb N}\Omega_0$$