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Let $X$ be a nonempty set, $\mathcal{X}$ a $\sigma$-algebra of subsets of $X$, and $\mu$ a measure on $\mathcal{X}$ (i.e., $\mu:X\to[0,+\infty]$, $\mu(\phi)=0$, and $\mu$ is countably additive.

Proposition: If $f:X\to R$ is a measurable function, $f=g$ almost everywhere, then $g$ is a measurable function.

Can this proposition be proved in the the measure space is not complete ($\mathcal{X}$ contains all subsets of measure zero)? If not, can someone provide a counterexample?

Thanks.

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  • $\begingroup$ This is in the second case of the only answer in the linked post. $\endgroup$ – awllower Jul 9 '13 at 3:38
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Assume that $f=0$ and $g$ is the characteristic function of a non-measurable set contained in a set of measure zero.

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    $\begingroup$ Nice example, but there is part of it that I don't understand. If your measure space is not complete, how do you prove that there is a non-measurable set contained in a set of measure zero. $\endgroup$ – David Jul 9 '13 at 2:27
  • $\begingroup$ Isn't a complete measure one in which all subsets of sets of measure zero have measure zero (in particular they are measurable)? We are assuming that the measure is not complete. Then, you are assuming that there are subsets of sets of measure zero that are not measurable. (If they were measurable the measure would be zero by additivity and positivity). $\endgroup$ – OR. Jul 9 '13 at 2:32
  • $\begingroup$ Very well written. Thank you for the response. $\endgroup$ – David Jul 9 '13 at 3:08

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