I'd like someone to check this demonstration.
(Cohn, 2.1.2) Show that the supremum of an uncountable family of $[-\infty, +\infty]$ valued Borel measurable functions on $\mathbb{R}$ can fail to be Borel measurable.
Let $A \subset \mathbb{R}$ be a non-Borel set. Let $f_\alpha(x) = \mathbb{1}_{\{\alpha\}}(x)$, for $\alpha \in A$. Then $f_{\alpha}$ is trivially Borel measurable. But now $\{x : \sup_\alpha f_{\alpha}(x) \leq 1/2\} = \cap_\alpha \{ x : f_{\alpha}(x) \leq 1/2\} = \mathbb{R} \setminus A$, which isn't Borel, else $A$ would be. Thus $\sup_\alpha f_\alpha$ is not Borel, even though $\{f_\alpha\}$ are.