Prove that every Lebesgue measurable function is equal almost everywhere to a Borel measurable function

Question

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?

Answer 1

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.

Answer 2

Hint: A lebesgue measurable set in $\mathbb{R}$ is a Borel measurable set, upto sets of measure $0$.

Answer 3

Hint.

Fact 1. If $f:\mathbb R\to [0,\infty]$ is Lebesgue measurable, then it expressed as $$ f(x)=\sum_{n\in\mathbb N}a_n\chi_{A_n}, $$ where $A_n$ Lebesgue measurable and $a_n\in [0,\infty]$ - This can be shown approximating $f$ from below by simple functions, and then using Lebesgue's Monotone Convergence Theorem.

Fact 2. If $A$ is Lebesgue measurable, then there exists a Borel set $B$ such that $$ m(A\smallsetminus B)+m(B\smallsetminus A)=0. $$

Answer 4

For any Lebesegue mesurable set $A_i$ exsists a Borel set $C_i$ such that $C_i \subseteq A_i$ and $m(A_i\setminus C_i)=0$. If $A_i$ are disjointed also $C_i$ are disjointed. Is not important that $$\bigcup^{n}_{i=1} {C_i} = \mathbb{R}$$ or, if you prefer, let $C=\bigcup^{n}_{i=1} {C_i}$ then $$ t(x) = \sum^n_{i=i}{\alpha_i\chi_{C_i}(x)}+ 0\chi_{C^c}(x) $$ in the required function ($C^c$ is trivially Borel-measurable).