# Supremum of uncountably many measurable functions may not be measurable

Given a family of measurable functions ${f_{\alpha}(x)}$ with ${\alpha \in \Lambda}$, shall we have ${g(x) = \sup_{\alpha\in \Lambda} f_{\alpha}(x)}$ measurable? If each ${f_{\alpha}}$ is continuous, then ${g(x)}$ is lower semicontinuous, so is measurable. However, it may not be true in general.

For simplicity, we take ${\mathbb R}$ as the domain of all functions below. Let ${\mathcal B}$ be Borel ${\sigma}$-algebra and ${\mathcal L}$ be its completion with respect to its usual metric on ${\mathbb R}$. We say a function ${f:\mathbb R \mapsto \mathbb R}$ is measurable if ${f^{-1} B \in \mathcal L}$ for any ${B\in \mathcal B}$, and denoted by ${f\in \mathcal L/\mathcal B}$.

Let’s take ${\Lambda \notin \mathcal L}$ be a non-Lebesgue-measurable set and functions be given by

$\displaystyle f_{\alpha}(x) = I_{\{\alpha\}}(x) -1.$

It is trivial to show that ${\{g \ge 0\} = \Lambda \notin \mathcal L}$. Indeed, one can show this by

$\displaystyle \{g \ge 0\} = \cap_{k=1}^{\infty} \cup_{n = k}^{\infty} \{g > - 1/n\} = \cap_{k=1}^{\infty} \cup_{n = k}^{\infty} \cup_{\alpha\in \Lambda} \{f_{\alpha}(x) > - 1/n\} = \Lambda.$

It is also interesting to show that, if ${f_{\alpha}(x) = - |x - \alpha|}$ is given, then the corresponding ${\{g\ge 0\} = \bar \Lambda \in \mathcal L}$.