Alternative proof for Lebesgue outer measure of elementary sets

$\textbf{The Question:}$

Let $E$ be an elementary set$^1$. Then the Lebesgue outer measure$^2$ $m^*(E)$ of $E$ is equal to the elementary measure$^3$ $m(E)$ of $E$. i.e. $$m^*(E)=m(E)$$

$\textbf{Method of book-solution:}$

$m^*(E) \leq m(E)$ follows trivially from the definitions. To show the other direction, the author employed the powerful Heine-Borel theorem to give a rather lengthy proof (Ref: Lemma $1.2.6$, An introduction to measure theory, Terry Tao). I want a simpler proof.

I'm giving my effort in the following section. Any clarification on whether it is correct or incorrect (in the latter case, if it can be improved to achieve a correct solution), would be much appreciated.

$\textbf{My solution:}$

$m^*(E)=inf\{\sum_{n=1}^{\infty}|B_n|:\cup_{n=1}^{\infty}B_n \supset E~; B_1,B_2,\ldots:\mbox{ disjoint boxes}\}$

Since $E$ is elementary, $E$ can be written as $E=\cup_{n=1}^N A_n~; A_1, A_2,\ldots,A_n:\mbox{ disjoint boxes}$.

Choosing $B_i=A_i, i=1(1)N \mbox{ and } B_i = \phi, i>N$, it is easy to see that $$m^*(E) \leq m(E) \cdots (1)$$

Now consider a class of disjoint boxes $\{B_n\}_{n=1}^{\infty}$ such that $\cup_{n=1}^{\infty} B_n \supset E$. By monotonicity of elementary measure, we have $$\sum_{n=1}^{\infty}|B_n| \geq m(E)$$

Taking infimum over $\{B_n\}_{n=1}^{\infty}$, collection of disjoint boxes such that $\cup_{n=1}^{\infty} B_n \supset E$, we have $$m^*(E) \geq m(E) \cdots (2)$$ Combining $(1)$ and $(2)$, we get the result.

$\textbf{Relevant definitions:}$ $^1$Elementary set: Finite union of (disjoint) boxes, where a box in $\mathbb{R}^d$ is a set of the form $I_1 \times I_2 \times \cdots \times I_d$, i.e. Cartesian product of intervals.

$^2$ Lebesgue outer measure: $$m^*(E)=inf\{\sum_{n=1}^{\infty}|B_n|:\cup_{n=1}^{\infty}B_n \supset E~; B_1,B_2,\ldots:\mbox{ disjoint boxes}\}$$

$^3$ Elementary measure: An elementary set $E$ can be written as $E=\cup_{n=1}^N B_n$, where $B_n$'s are disjoint boxes. Then elementary measure of $E$ is defined as $m(E)=\sum_{n=1}^N |B_n|$, where $|B_n|$ denotes the size/volume of the box $B_n$.

• The statement that you have $\Sigma_{n=1}^{\infty} |B_n| \geq m(E)$ because of monotonicity of elementary measure could be argued in a bit more detail. Presumably you're saying that the elementary set must be covered by finitely many of the boxes. That point could be argued in a bit more detail. But I'm not sure about this definition of Lebesgue outer measure. I've never heard of there being any requirement that all the boxes have to be disjoint before. And without that requirement you won't be able to argue that the elementary set is covered by finitely many of the boxes. Dec 8 '16 at 14:29
• Your definition of the Lebesgue outer measure is incorrect. Dec 8 '16 at 14:46

First, your definition of "outer measure" is not the usual one: normally you take the infimum over all coverings by boxes, not just those by disjoint boxes. I'm not sure whether that matters.

Next, there is a big gap in your proof where you say "by monotonicity of elementary measure".

• You have not shown that elementary measure is well defined. A given elementary set might be able to be written as the finite union of boxes in two or more different ways. Would both ways give the same sum of the volumes of the boxes? This requires proof.

• You have not proved that elementary measure is monotone.

• Here is the big one: monotonicity of elementary measure does not, by itself, imply your claim that $\sum |B_n| \ge m(E)$. It would imply that $m\left(\bigcup_{n=1}^\infty B_n\right) \ge m(E)$. But at this point we have no proof that $m\left(\bigcup_{n=1}^\infty B_n\right)$ has anything to do with $\sum_{n=1}^\infty |B_n|$. The defining property of elementary measure is only for finite unions. Getting infinite unions requires more work, and that's basically the entire hard part of proving that Lebesgue measure is countably additive.

In a larger sense, your proof can't be right, because you haven't used "enough" of the structure of $\mathbb{R}$. Consider for instance what would happen if we tried to construct Lebesgue measure on $\mathbb{Q}$. Everything in your argument up to this point is just as true for $\mathbb{Q}$ as for $\mathbb{R}$. We can define elementary measure for finite unions of "boxes" (i.e. intervals) in $\mathbb{Q}$, where the volume of $[a,b]$ is $b-a$ as usual. I expect you could even show that elementary measure is well defined and monotone. But then your claimed inequality $\sum |B_n| \ge m(E)$ is false. If you take $E = [0,1]$, enumerate the rationals in $[0,1]$ as $q_n$, and let $B_n = \{q_n\}$, then $|B_n|=0$ for all $n$ and your claimed inequality reads $0 \ge 1$.

So your proof will somewhere have to use something about $\mathbb{R}$ that isn't true in $\mathbb{Q}$. Usually a completeness or compactness principle is the right thing, e.g. Heine-Borel in Tao's proof. The countable additivity of Lebesgue measure isn't a trivial fact; it relies on fairly deep properties of the real line. So you should not expect it to have a trivial proof.

• Since boxes can be open, closed, or generally $\frac{1}{2^m}$-open, where $0 \leqslant m \leqslant d$, you can write every countable union of boxes as a countable union of disjoint boxes. Thus the definition is equivalent. Dec 8 '16 at 16:37
• @Nate Eldredge: This was extremely helpful. I'm learning the subject for the first time and can't help coming up with dumb questions. Thanks a lot for clearing my doubts. Dec 9 '16 at 4:16

I think I see a flaw in this. The elementary measure of $\bigcup_{n=1}^{\infty} B_n$ is not defined since this set is most likely not an elementary set. This means that $\bigcup_{n=1}^{\infty} B_n\supset E$ does not obviously imply that $\sum_{n\geq 1} |B_n|\geq m(E)$.

• True. It's flawed. Thanks for your help. Dec 9 '16 at 4:22