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Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue measurable and $f=g$ a.e with respect to Lebesgue measure, then $g: \mathbb{R} \rightarrow \mathbb{R}$ is also Lebesgue measurable.

I've been looking up solutions but I can't quite understand why any subset of the measurable set that has measure 0 is also measurable.

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  • $\begingroup$ This comes from the fact that Lebesgue measure is complete. See here for elements of the proof. $\endgroup$ – mathcounterexamples.net Nov 12 '19 at 12:39
  • $\begingroup$ That is a property of complete measures and the Lebesgue measure is a complete measure. If you start with an outer measure and restrict the outer measure (with Caratheodory) to the the $\sigma$-algebra of measurable sets then the result is a complete measure. That process also gives birth to the Lebesgue measure. $\endgroup$ – drhab Nov 12 '19 at 12:44
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Consider $g^{-1}(\{y|y<m\})=\{x|g(x) < m\}=\{x|g(x)<m \text{ and } g(x)=f(x)\} \cup \{x|g(x)<m \text{ and } g(x) \neq f(x)\}$.

The second set in the union has measure $0$ and hence measurable. The first set in the union is $\{x|g(x)<m\} \cap (\mathbb{R}-\ \{x|g(x)\neq f(x)\})$ is measurable.

As complement of measurable set is measurable and intersection of measurable set is measurable?

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