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

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.?

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.

Observe that:

• By decomposing $f$ into positive and negative parts we may assume that $f$ is nonnegative everywhere
• For a real valued nonnegative function to be Borel measurable it is sufficient that the preimage of each ray of the form $(q, \infty)$ for $q$ rational be Borel
• Every Lebesgue measurable set is the disjoint union of an $F_{\sigma}$ set and a null set

With these observations the construction is straightforward. Let $f$ be Lebesgue measurable and nonnegative. Let $U_q$ be the preimage of $(q, \infty)$ under $f$ and let $F_q$ be an $F_{\sigma}$ set of full measure contained in $U_q$. Let $g$ agree with $f$ everywhere expect on those points which do not belong to any of the $F_q$, where we may take $g$ to be zero.