Consider any (possibly infinite) set $X$. We fix on $\overline{\mathbb{R}}^X$ the $\sigma$-algebra product of the Borel $\sigma$-algebra on $\overline{\mathbb{R}}$ (where $\overline{\mathbb{R}}$ is the extended real line with its usual topology).
Question : is $\mathbb{R}^X$ a measurable subset of $\overline{\mathbb{R}}^X$?
For each $Y\subset X$ let $\pi_Y: \overline{\mathbb{R}}^X\to \overline{\mathbb{R}}^Y$ be the natural projection. I assume that $\sigma$-algebra product $\mathcal A$ of the Borel $\sigma$-algebra on $\overline{\mathbb{R}}$ is a $\sigma$-algebra on $\overline{\mathbb{R}}^X$ generated by the set $\pi_{\{x\}}^{-1}(B)$, where $x\in X$ and $B$ is a Borel subset of $\overline{\mathbb{R}}$. If $X$ is countable then $\mathbb R^X=\bigcap_{x\in X} \pi_{\{x\}}^{-1}(\mathbb R)\in\mathcal A$. Now we assume that the set $X$ is uncountable. The construction of a generated $\sigma$-algebra implies that for each $C\in\mathcal A$ there exists a countable subset $Y$ of $X$ such that $C=\pi^{-1}_Y\pi_Y(C)$. The set $\mathbb R^X$ does not have this property, so it does not belong to $\mathcal A$.