Suppose we have two measurable spaces $(E,\mathcal{E})$, $(S,\mathcal{S})$, an index set $\Gamma$ and a family of measurable functions $f_\gamma:E\to S$ for $\gamma\in\Gamma$. If $\Gamma$ is countable then $\sup_\gamma f_\gamma$ is measurable as well. However this can fail if $\Gamma$ is uncountable. Can someone provide an example of such a family and the respective spaces such that the supremum over this uncountable index set is not measurable.


Let $E=[0,1]$, let $\Gamma\subset E$ be the Vitali set (which is uncountable), and for each $\gamma\in\Gamma$, let $$f_{\gamma}(t)=\mathbf{1}_{\{\gamma\}}(t)=\begin{cases}1 &\text{if }t=\gamma,\\ 0 &\text{if }t\neq \gamma. \end{cases}$$

Then $\sup_{\gamma}f_\gamma=\mathbf{1}_{\Gamma}$, which is not a measurable function because $\Gamma$ is not a measurable set.

  • 2
    $\begingroup$ Of course this works with any nonmeasurable set. $\endgroup$ – Carl Mummert Apr 8 '13 at 14:45
  • $\begingroup$ beautiful! It is a very nice answer, since taking the uncountable family $f_\gamma=\mathbf1_\gamma$ for $\gamma\in\Gamma=[0,1]$ is also an example for an uncountable family, such that the supremum is measurable. $\endgroup$ – math Apr 8 '13 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.