21
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Have you tried Lusin's theorem? $\endgroup$ – Umberto P. Aug 28 '14 at 16:47
  • 1
    $\begingroup$ @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 '14 at 16:49
3
$\begingroup$

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.

$\endgroup$
2
  • 7
    $\begingroup$ 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 '14 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 '14 at 23:24
3
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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 $$ m(A\smallsetminus B)+m(B\smallsetminus A)=0. $$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
8
  • 1
    $\begingroup$ Did you read my attempt? $\endgroup$ – layman Aug 28 '14 at 16:39
  • 1
    $\begingroup$ @user46944: yes. Show that the sets you defined are Borel sets up to null sets. $\endgroup$ – voldemort Aug 28 '14 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 '14 at 16:42
  • 3
    $\begingroup$ @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. $\endgroup$ – Adam Hughes Aug 28 '14 at 17:10
  • 1
    $\begingroup$ @user46944 Almost everywhere, yes. That's all that matters, though. $\endgroup$ – Adam Hughes Aug 28 '14 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.