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}.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s