Suppose $(\mathbb{R},\Sigma(m),m)$ is our measure space, where $m$ is Lebesgue measure. Also, suppose $f : \mathbb{R} \to [-\infty, \infty]$ is a Lebesgue measurable function.
The problem:
Prove that $f$ is equal almost everywhere to a Borel measurable function.
My attempt: We only have to prove this assertion for a non-negative Lebesgue measurable function $f$ because the result will follow for all Lebesgue measurable functions.
Furthermore, since any Lebesgue measurable function can be approximated by a sequence of non-negative, monotonically increasing, Lebesgue measurable simple functions $s_{n}$, we only need to prove the claim for an arbitrary Lebesgue measurable simple function.
So, let $s: \mathbb{R} \to [-\infty, \infty] $ be a Lebesgue measurable simple function. We can write $s$ canonically as $$ s(x) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(x)$$ where $\alpha_{i} \in [-\infty, \infty]$, $\bigcup \limits_{i = 1}^{n} A_{i} = \mathbb{R}$, and $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$.
For each $i$, since $A_{i}$ is Lebesgue measurable, we can find a Borel set $B_{i}$ such that $A_{i} \subseteq B_{i}$ and $m(B_{i} \setminus A_{i}) = 0$. Clearly, this implies that $\bigcup \limits_{i = 1}^{n} B_{i} = \mathbb{R}$. Now we just need the $B_{i}$'s to be pairwise disjoint, with each $B_{i}$ still retaining $A_{i}$.
To make them pairwise disjoint, I constructed the following sets:
$\tilde{B_{i}} = [B_{i} \setminus (\bigcup \limits_{j \neq i} B_{j})] \cup A_{i}$. This construction gives us that the $\tilde{B_{i}}$'s are pairwise disjoint (I think….) and $A_{i} \subseteq \tilde{B_{i}}$. But I don't know that $\tilde{B_{i}}$ is necessarily still a Borel set. 🙁 🙁 Am I approaching this problem all wrong?
Best Answer
You just make sure you have countably many Lebesgue sets in your formulation, and from each of the disjoint sets remove sets of measure $0$ to make them Borel.
Almost everywhere equality follows, and that's all the question cares about.