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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.

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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.

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    $\begingroup$ 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

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