# Every almost-Lebesgue measurable set is Lebesgue measurable.

The following problem is from exercise 8 of Tao's introductory measure theory book.

$\textbf{Prove:}$

If for all $\epsilon > 0$ one can find a Lebesgue measurable set $E_{\epsilon}$ such that $m^*(E_{\epsilon} \Delta E) \leq \epsilon$, then $E$ itself must be Lebesgue measurable.

The hint that the book gives is: use the $\epsilon/2^n$ trick to show that $E \subset E_{\epsilon}'$ where $E_{\epsilon}'$ is measurable and $m^*(E_{\epsilon}' \Delta E) \leq \epsilon$; then I should take countable intersections to show that $E$ differs from a Lebesgue measurable set by a null set.

The follow Lemma 10 will probably be useful:

(i) Every open set is Lebesgue measurable.

(ii) Every closed set is Lebesgue measurable.

(iii) Every set of Lebesgue outer measure zero is measurable. (Such sets are called null sets.)

(iv) The empty set is Lebesgue measurable.

(v) If $E \subset {\bf R}^d$ is Lebesgue measurable, then so is its complement ${{\bf R}^d \backslash E}$.

(vi) If ${E_1, E_2, E_3, \ldots \subset {\bf R}^d}$ are a sequence of Lebesgue measurable sets, then the union ${\bigcup_{n=1}^\infty E_n}$ is Lebesgue measurable.

(vii) If ${E_1, E_2, E_3, \ldots \subset {\bf R}^d}$ are a sequence of Lebesgue measurable sets, then the intersection ${\bigcap_{n=1}^\infty E_n}$ is Lebesgue measurable.

I am not sure at all how to follow the hint. Specifically I have been unable to come up with an $E_{\epsilon}'$ which satisfies the properties that I want. I find it very hard to work with $m^*(A \Delta B)$ in general. Does anyone have any tips how to construct $E_{\epsilon}'$? My guess is that we use the fact that $E_{\epsilon}$ is Lebesgue measurable in some way to approximate it from the outside, perhaps by an open set which contains $E$?

Let $\epsilon>0$ be arbitrary. For each $n\in\Bbb N$, there is a measurable set $E_n$ such that $m^*(E_n\Delta E) \le \epsilon/2^n$. We claim that after neglecting a set of measure $0$, in fact $E\subset\bigcup_{n=1}^\infty E_n$. That is, that the set $E-\bigcup E_n$ has measure $0$. Indeed, for every $N\in\Bbb N$,

\begin{align*} m^*(E-\bigcup E_n) &\le m^*(E\cap E_N^c) \\ &\le m^*(E\Delta E_N) \le \epsilon/2^N. \end{align*}

Since this holds for every $N$, our claim is proved. Let $\tilde E = \bigcup E_n-E$. Then $\tilde E \subset \bigcup E_n$ and $m^*(\bigcup E_n\Delta \tilde E) = m^*(\bigcup E_n- \tilde E) \le \sum \epsilon/2^n = \epsilon$. Put $E_\epsilon' = \bigcup E_n$. By outer regularity, $\tilde E$ differs from a measurable set by a null set (pick a sequence $E_{\epsilon}'\searrow \tilde E$ as $\epsilon\to 0$), and is thereby measurable. Since $E$ is measurable if and only if $\tilde E$ is measurable, $E$ is also measurable, and the claim is proved.

This is a problem in Folland stated in terms of premeasures and associated outer measures (defined by taking the infimum over premeasures of countable covers). The construction of the Lebesgue measure from lengths of intervals will be a special case. I have to sleep in the near future so I'll sketch what I did when I did this for homework, and if you need further clarification, I'll come back to this thread tomorrow.

The first step is to show that if $E \subset X$, $\mathscr{A}$ is an algebra on $X$ and $\epsilon > 0$ is given, there is an $A \in \mathscr{A}_\sigma$ such that $E \subset A$ and $\mu^*(A) \leq \mu^*(E) + \epsilon$. Since the outer measure is the infimum, there is a countable cover of $E$ such that $\sum \mu_0(A_i) \leq \mu^*(E) + \epsilon$. If you take $A$ to be the union over this sequence, this is in $\mathscr{A}_\sigma$, contains $E$ by construction, and since the elements $A_i$ need not be disjoint, and premeasurable sets are outer measurable, it follows that $\mu^*(A) \leq \sum \mu_0(A_i) \leq \mu^*(E) + \epsilon$

This doesn't quite finish the argument, but it's an important first step. Since $\epsilon$ is arbitrary, we can find $B_n \in \mathscr{A}_\sigma$ so that $\mu^*(B_n) \leq \mu^*(E) + 1/n$ by the above argument. Now define $B = \cap_n B_n$. $E$ is contained in each $B_i$ and so is contained in $B$. You can do so some set algebra to prove that $\mu^*(B- E) \leq 1/n$ for every $n$, and this shows that in fact the measure is $0$.

This finishes it since in $E \subset B$ implies that $E \Delta B = B - E$, and so this is what you want to prove.

Hope this helps!

• I have been following Tao's book closely so I am unfamiliar with some of the notions you have used here (I don't know what an algebra is). Sep 20, 2017 at 3:31

HINT:

Let $E_{\epsilon}$ a set that approximates $E$. For every $A$ subset we have $$A \cap E \subset (A \cap E_{\epsilon} ) \cup(E \backslash E_{\epsilon})\\ A \backslash E \subset (A\backslash E_{\epsilon}) \cup (E_{\epsilon} \backslash E)$$

We get for the outer measure $$\mu^{\star}(A\cap E)+ \mu^{\star}(A\backslash E) \le \mu^{\star}(A\cap E_{\epsilon}) + \mu^{\star}(A\backslash E_{\epsilon}) +\mu^{\star}(E\backslash E_{\epsilon}) + \mu^{\star}(E_{\epsilon} \backslash E)$$

If $E_{\epsilon}$ is measurable then we have $\mu^{\star}(A\cap E_{\epsilon}) + \mu^{\star}(A\backslash E_{\epsilon}) = \mu^{\star}(A)$. Also, we have $\mu^{\star}(E\backslash E_{\epsilon}) + \mu^{\star}(E_{\epsilon} \backslash E)\le 2 \mu^{\star}(E\Delta E_{\epsilon})\le 2 \epsilon$.

Since $\epsilon>0$ is arbitrary we get $$\mu^{\star}(A\cap E)+ \mu^{\star}(A\backslash E) \le \mu^{\star}(A)$$ and since the opposite inequality holds, we get equality for all $A$. Hence $E$ is measurable.