If $f$ is Lebesgue measurable, prove that there is a Borel measurable function $g$ such that $f=g$ except, possibly, on a Borel set of measure zero.
Suppose $f$ is Lebesgue measurable. Then there exists an increasing sequence of simple functions $\varphi_n$ such that $\lim_{n\to\infty}\varphi_n=f$. Then $\int \lim_{n\to\infty}\varphi_n=\int f$. By the Monotone Convergence Theorem, we can interchange the integral and limit signs. We get $\lim_{n\to\infty}\int \varphi_n$.
A Borel measurable function is a function whose preimage is a Borel set. A Borel set is the smallest sigma algebra containing open sets. A Borel measurable set is necessarily Lebesgue measurable. How to continue?