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.
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?