# Show by example that the uncountable supremum of Borel measurable functions need not be Borel measurable.

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.

• Your proof is fine. Why do you even suspect that it could be wrong? – Kavi Rama Murthy Aug 24 '18 at 23:14
• @KaviRamaMurthy was mostly wondering if there was a simpler/explicit way to do it. But I think most examples will entail picking your favorite A non-Borel. – Drew Brady Aug 24 '18 at 23:37
• Surely, there is no explicit construction of a non-Borel set in $\mathbb R$. – Kavi Rama Murthy Aug 24 '18 at 23:49