[Math] If $f$ is Lebesgue measurable on $[0,1]$ then there exists a Borel measurable function $g$ such that $f=g$ a.e.

Question

If $f:[0,1]\to\mathbb{R}$ is Lebesgue measurable then there exists a Borel measurable function $g:[0,1]\to\mathbb{R}$ such that $f=g$ a.e.?

Answer 1

Yes. By Lusin's theorem, for every $\varepsilon > 0$, there is a continuous $f_\varepsilon$ such that the measure of $\{x : f(x) \neq f_\varepsilon(x)\}$ is smaller than $\varepsilon$.

Let $g_n$ a sequence of continuous functions with $\lambda(\{x : f(x) \neq g_n(x)\}) < 2^{-n}$. Since the sum of the measures is finite, almost every $x$ lies in at most finitely many exceptional sets, so $g_n \to f$ almost everywhere. The function $g(x) = \limsup_{n\to\infty} g_n(x)$ is Borel measurable, and $g = f$ almost everywhere.