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.

  • $\begingroup$ Your proof is fine. Why do you even suspect that it could be wrong? $\endgroup$ – Kavi Rama Murthy Aug 24 '18 at 23:14
  • $\begingroup$ @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. $\endgroup$ – Drew Brady Aug 24 '18 at 23:37
  • $\begingroup$ Surely, there is no explicit construction of a non-Borel set in $\mathbb R$. $\endgroup$ – Kavi Rama Murthy Aug 24 '18 at 23:49

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