# Show that if f is Lebesgue measurable and f = g almost everywhere with respect to Lebesgue measure, then g is Lebesgue measurble.

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.

• This comes from the fact that Lebesgue measure is complete. See here for elements of the proof. – mathcounterexamples.net Nov 12 '19 at 12:39
• 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. – drhab Nov 12 '19 at 12:44

Consider $$g^{-1}(\{y|y.
The second set in the union has measure $$0$$ and hence measurable. The first set in the union is $$\{x|g(x) is measurable.