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UPDATE: I added an answer based off the hints provided by copper.hat. It may, however, need some adjustment.

I'm trying to solve another question from Stein and Shakarchi's analysis text.

Basically, I'm trying to prove that $m_{\star}(E)=m_{\star}^{R}(E)$ for every $E\subset\mathbb{R}^{d}$ where $m_{\star}$ is the exterior measure taken with closed cubes and $m_{\star}^{R}(E)$ is the exterior measure taken with closed rectangles.

Here's how I've gotten started:

Proof (beginning): Let $E$ be a subset of $\mathbb{R^{d}}$. Let's assume for now that any rectangle $R\subset\mathbb{R^{d}}$ can be decomposed into a countable number of almost disjoint closed cubes: $R_{j}=\bigcup_{k=1}^{\infty}Q_{k}^{j}$. If the decomposition is finite up to $N$, then simply put $\emptyset$ for the remaining sets. Then we compute the following (the steps which require justification will be addressed below): \begin{eqnarray} m_{\star}^{R}(E) &=& \inf\limits_{\bigcup_{j=1}^{\infty}R_{j}\supset E}\sum\limits_{j=1}^{\infty}|R_{j}|\\ &=& \inf\limits_{\bigcup_{j=1}^{\infty}R_{j}\supset E}\sum\limits_{j=1}^{\infty}\sum\limits_{k=1}^{\infty}Q_{k}^{j}\\ &=& \inf\limits_{\bigcup_{j=1}^{\infty}|R_{j}|\supset E}\sum\limits_{i=1}^{\infty}|Q_{i}|, \end{eqnarray} where the collection $\{Q_{i}\}$ comes from the union of the decompositions of each $R_{j}$

\begin{eqnarray} \phantom{m_{\star}^{R}(E)} &=& \inf\limits_{\bigcup_{n=1}^{\infty}Q_{n}\supset E}\sum\limits_{n=1}^{\infty}|Q_{n}|,\end{eqnarray} where the collection $\{Q_{n}\}$ comes from the standard definition of exterior measure \begin{eqnarray} \phantom{m_{\star}^{R}(E)} &=& m_{\star}(E)\phantom{\text{udontseeme}} \end{eqnarray}

Now, we must justify three things: $(1)$ that each $R_{j}$ can in fact be decomposed into a countable union of almost disjoint closed cubes; $(2)$ that the volume of each $R_{j}$ is equal to the sum of the volumes of each $Q_{k}^{j}$; and $(3)$ that the family of cubes $\{Q_{i}\}$ coincides with the family of cubes $\{Q_{n}\}$. First observe that (3) follows immediately from (1) and (2), for if the cover $\{R_{j}\}$ is infimal (in the sense of volume), and each $R_{j}$ is decomposable into almost disjoint closed cubes such that the volume of each $R_{j}$ is equal to the sum of the volumes of each $Q_{k}^{j}$, it must be that the cover $\{Q_{i}\}$ is also infimal, and so is equal to $\{Q_{n}\}$.

To prove (2) and (3) I found to be a little difficult. I started writing out solutions to them, but I keep getting tangled up in the concepts and the constant effort to be rigorous. Any suggestions are welcomed, including a completely different way to go out proving this!

Thanks again ~~

P.S. sorry I don't know what LaTeX commands you need to use to align when posting here. :p

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Maybe you should start by showing the exterior measures coincide on rectangles first? – copper.hat May 9 '12 at 16:30
I'm not sure I understand your hint; isn't that exactly what the problem asks us to do? – Taylor Martin May 9 '12 at 17:25
No, the problem is to show they coincide on all subsets. I'm suggesting you start on rectangles. Also, you have that $m_*^R E \leq m_* E$ for free, since the $\inf$ is taken over a larger set of objects (cubes are rectangles too). – copper.hat May 9 '12 at 17:28
Ah, that makes more sense. My next question would be then what is the most logical way to decompose a rectangle into a countable union of almost disjoint cubes? The book recommends using a proved Lemma that says that the volume of a rectangle is the FINITE sum of the volumes of its almost disjoint component rectangles (if it can be written as such). How does one extend this to a countable sum? Since I doubt it is possible to write an arbitrary rectangle as a countable union of almost disjoint cubes. – Taylor Martin May 9 '12 at 17:37
Actually, I think I have a more direct solution thanks to your hint now. It is relatively easy to show that the cubic exterior measure of a rectangle is equal to its volume, which is in turn equal to the rectangular outer measure. I will post this when I get a chance to be on my laptop again (on phone atm). So the measures coincide on rectangles and cubes. So I just need to think about how to get to the conclusion from there for a minute – Taylor Martin May 9 '12 at 18:08
up vote 0 down vote accepted

Here is a tentative solution. Thank you copper.hat for the very helpful hint. There is one interesting thing to note here. The text suggests applying Lemma 1.1 which proves that a finite decomposition of a rectangle into sub rectangles (or cubes) has additive volume if the decomposition is almost disjoint (and sub additive in general by extension). I had difficulty extending this to countable decompositions, so I applied copper.hat's suggestion (second lemma that I prove in the solution), and an auxiliary lemma proved in the book that says the (cubic) exterior measure of a set which can be decomposed into almost disjoint closed cubes has additive measure. Obviously if I could apply the book's hint, I would not need to pass from the rectangular measure to the cubic measure, and could do the computation at the end more directly. But in any case, it's not really much different either way.

Problem Statement: If $E\subset\mathbb{R}^{d}$, then $m_{\star}(E) = m_{\star}^{R}(e)$.

PROOF: It is clear $m_{\star}(E)\geq m_{\star}^{R}(E)$, since any cover of $E$ by cubes is also a cover of $E$ by rectangles. To show the reverse inequality, we need two lemmas.

LEMMA #1: Every closed rectangle can be decomposed into a countable union of almost disjoint closed cubes.

PROOF: Obviously this is true for $\mathbb{R}^{1}$. To prove this in general, let $R = I_{1}\times\ldots\times I_{d}$ be any cube in $\mathbb{R}^{d}$ ($d\geq2$). Take the smallest interval (call it $I_{0}$ say) and construct closed cubes of side length $I_{0}$, so that each has a volume of $\left|I_{0}\right|^{d}$. For each interval composing $R$, compute the value $N_{j}=\frac{|I_{j}|}{|I_{0}}$, and discard the remainder (note that for the minimal interval, the quotient will be exactly $1$). Then precisely $N_{j}$ cubes can be exactly inscribed along each $I_{j}$, and this embedding of cubes will be pairwise disjoint (it helps to draw a picture in $\mathbb{R}^{3}$ to get the general idea; a picture in $\mathbb{R}^{2}$ tends to obscure the general situation). Note that the total number of cubes embedded is $\sum_{j=1}^{d}N_{j}$, which is finite. What remains is exactly $(d)$ rectangles and the union of these remaining rectangles unioned with the union of the embedded cubes gives back $R$ exactly. (Note that some of the $d$ remaining rectangles may have $0$ volume, since we may have $|I_{j}|=k|I_{0}|$ for some positive integer k; in fact, this occurs for at least one interval, namely the smallest interval). We can continue this construction indefinitely, applying the procedure to each of the remaining rectangles at each step of the decomposition (note that the volumes of the remaining rectangles goes to zero in the limit). Since the total number of cubes at the end of the construction is a countable union of a countable union of finite unions of cubes (mouthfull), the total number of cubes embedded is countable. By construction, they are also all almost disjoint from each other. Finally, let $Q=\bigcup_{k=1}^{\infty}$ where $\{Q_{k}\}_{k=1}^{\infty}$ is some enumeration of the embedded cubes, and put $x\in Q$. Then clearly, $x\in R$ as well, since each embedded cube is completely contained in $R$. Now put $x\in R$. Then $x\in Q_{k}$ for some index $k$ since it is contained in some subrectangle of $R$ (whether on the boundary of $R$ or the interior, it does not matter since each $Q_{k}$ is closed), which after going a large enough number of steps in the construction, is eventually contained in some $Q_{k}$. Therefore, $R=\cup_{k=1}^{\infty}Q_{k}$. (Note: there are probably more elgant ways to prove this; a geometric-series-ish construction comes to mind). QED.

LEMMA #2: If $R\subset\mathbb{R}^{d}$ is any closed rectangle (or closed cube), then $m_{\star}^{R}(R)=|R|=m_{\star}(R)$.

PROOF: The RHS equivalence follows from the computations given in the text (in particular, example 4). As for the LHS equivalence, note immediately that $m_{\star}^{R}(R)\leq|R|$ since $R$ covers itself. To prove the reverse inequality, we apply the "$\epsilon$-fattening" trick as in example 2. Let $\epsilon>0$ be given and suppose $\{R_{j}\}_{j=1}^{\infty}$ is a cover of $R$ by closed rectangles. Let $\{S_{j}\}_{j=1}^{\infty}$ be a collection of open rectangles such that for each $j=1,2,\ldots$ we have that $|S_{j}|\leq(1+\epsilon)|R_{j}$ and each $S_{j}\supset R_{j}$. (Note that the indexing sets coincide). Then it is clear $R\subset\bigcup_{j=1}^{\infty}S_{j}$, and since $R$ is compact, we can extract a finite number of $S_{j}$ such that $R\subset\bigcup_{j=1}^{N}S_{j}$. Consequently, \begin{equation*} |R|\leq\sum_{j=1}^{N}|S_{j}|\leq(1+\epsilon)\sum_{j=1}^{N}|R_{j}|\leq(1+\epsilon)\sum_{j=1}^{\infty}|R_{j}|. \end{equation*} Since both $\epsilon$ and the cover $\{R_{j}\}_{j=1}^{\infty}$ were arbitrary, we conclude $|R|\leq m_{\star}^{R}(R)$ as desired. QED

Returning to our proof that $m_{\star}(E)\leq m_{\star}^{R}(E)$ (and thus proving they are equal), choose any $\epsilon>0$ and let $\{R_{j}\}_{j=1}^{\infty}$ be any cover of $E$ by closed rectangles such that $m_{\star}^{R}(E)\geq\sum_{j=1}^{\infty}|R_{j}|-\epsilon$. Then by the first lemma, $R_{j}=\cup_{k=1}^{\infty}Q_{k}^{j}$ for each $j=1,2,\ldots$ This means $E\subset\bigcup_{j=1}^{\infty}R_{j}=\bigcup_{j=1}^{\infty}\bigcup_{k=1}^{\infty}Q_{k}^{j}$. And so, we compute: \begin{align*} m_{\star}^{R}(E)+\epsilon &\geq \sum\limits_{j=1}^{\infty}|R_{j}| &\text{Consequence of definition of outer rectangular measure.}\\ &= \sum\limits_{j=1}^{\infty}m_{\star}(R_{j}) &\text{Consequence of the second lemma.} \\ &= \sum\limits_{j=1}^{\infty}m_{\star}\left(\bigcup_{k=1}^{\infty}Q_{k}^{j}\right) &\text{Consequence of the first lemma.} \\ &= \sum\limits_{j=1}^{\infty}\sum\limits_{k=1}^{\infty}|Q_{k}^{j}| &\text{Consequence of property 5 of outer cubic measure from text}\\ &= \sum\limits_{n=1}^{\infty}|Q_{n}^{\star}| &\text{Just notation to collapse the double sum into a single sum} \\ &\geq \inf\limits_{\{Q_{j}\}_{j=1}^{\infty}\supset E}\sum\limits_{j=1}^{\infty}|Q_{j}|\\ &= m_{\star}(E). \end{align*} Since $\epsilon$ is arbitrary, the conclusion follows and $m_{\star}^{R}=m_{\star}(E)$. QED.

NOTE: The "computations from the text" refer to proofs that $m_{\star}(R) = |R|$ and $m_{\star}(Q) = |Q|$ for both open and closed cubes and rectangles. I only need to prove $m_{\star}^{R}(R) = |R|$ for closed rectangles (and consequently cubes) for purposes of the problem. Finally, "property 5 from the text" refers to the fact that if a set is decomposable into an almost disjoint union of cubes, its outer measure becomes additive.

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