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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?

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  • $\begingroup$ Have you tried Lusin's theorem? $\endgroup$
    – Umberto P.
    Aug 28, 2014 at 16:47
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    $\begingroup$ @UmbertoP. Lusin's theorem allows me to find a continuous function that is equal to my measurable function except on a set of measure less than $\epsilon$. I will have to think about how to possibly apply that. $\endgroup$
    – layman
    Aug 28, 2014 at 16:49

4 Answers 4

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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.

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    $\begingroup$ And to be extra clear in case anyone stumbles across this page in the future, since for any Lebesgue measurable set $A$, we can find a Borel set $B$ such that $A \subseteq B$ and $m(B \setminus A) = 0$, the same holds true for $A^{c}$ since it is Lebesgue measurable. But the Borel set $B'$ that we find for $A^{c}$ has the property that $(B')^{c}$ is a Borel set contained in $A$ with $m(A \setminus (B')^{c}) = 0$. So for our simple function, we can reduce each Lebesgue measurable set $A_{i}$ to a Borel set $B_{i}$, and the resulting simple function is equal a.e. to $s$. $\endgroup$
    – layman
    Aug 28, 2014 at 23:14
  • $\begingroup$ Also, the following argument is useful: If $A$ is Lebesgue measurable, then so is $A^{c}$. Find a Borel set $B$ such that $A^{c} \subseteq B$ and $m(B \setminus A^{c}) = 0$. Then $B^{c}$ is a Borel set, and $B^{c} \subseteq A$. But then $m(A \setminus B^{c}) = m(A \cap (B^{c})^{c}) = m(A \cap B) = m(B \setminus A^{c}) = 0$. This shows that for any Lebesgue measurable set $E$, we can find a borel set $B$ with $B \subseteq E$ and $m(E \setminus B) = 0$. $\endgroup$
    – layman
    Aug 28, 2014 at 23:24
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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 $$ m(A\smallsetminus B)+m(B\smallsetminus A)=0. $$

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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).

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

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    $\begingroup$ Did you read my attempt? $\endgroup$
    – layman
    Aug 28, 2014 at 16:39
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    $\begingroup$ @user46944: yes. Show that the sets you defined are Borel sets up to null sets. $\endgroup$
    – voldemort
    Aug 28, 2014 at 16:40
  • $\begingroup$ Well, they are Lebesgue measurable, but that means they are contained in Borel sets with the measure of the set difference being $0$. But that gets me right back to where I started. The new Borel sets are not necessarily pairwise disjoint, and if I try to make them pairwise disjoint, then my construction yields sets that are not necessarily Borel. Then I can keep cycling like that forever. :( $\endgroup$
    – layman
    Aug 28, 2014 at 16:42
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    $\begingroup$ @user46944 You just make sure you have countably many Lebesgue sets, and from each of the disjoint sets remove sets of measure 0 to make them Borel. $\endgroup$ Aug 28, 2014 at 17:10
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    $\begingroup$ @user46944 Almost everywhere, yes. That's all that matters, though. $\endgroup$ Aug 28, 2014 at 23:00

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