$m_*^R(E)=m_*(E)$ where $m_*$ is an outer measure.

Question

The following problem is chapter 1 Exercise 15 in Stein's Real analysis. Let $E\subset\mathbb{R}^d$ and $E\subset\bigcup_{j=1}^\infty R_j$ where $R_j$ are closed rectangles in $\mathbb{R}^d$. Define $m_*^R(E)=$inf$\Sigma_{j=1}^\infty |R_j|$ where $|R_j|$ denote its volume. Need to show $m_*(E)=m_*^R(E)$ for any $E\subset\mathbb{R}^d$ (Note that $m_*(E)$=inf$\Sigma_{j=1}^\infty |Q_j|$ where $E\subset\bigcup_{j}^\infty Q_j$ with $Q_j$ are closed cubes in $\mathbb{R}^d$) There is one hint: (Lemma 1.1) If a ractangle is the almost disjoint union of finitely many other rectangles, say $R=\bigcup_{k=1}^N R_k$, then $|R|=\Sigma_{k=1}^N |R_k|$. $m_*^R(E)\leq m_*(E)$ is clear from the definition. The problem is the reverse inequality. How could I use that given hint? Thanks in advanced.

Answer 1

As you have noted, that $m_*^R(E)\leq m_*(E)$ is clear because squares are indeed a type of rectangle. For the reverse direction, the intuition is to find a way to cover any rectangle from above, by squares, in such a way so that the "wasted" area (i.e. non-overlapping, or with minimal spillage over the boundary) is minimized. Here is one way to do this. Choose given a rectangle $R$, choose rational numbers $q_1,\cdots,q_d$ slightly larger than the side lengths of $R$, say so that $\prod q_i<|R|+\epsilon$. Then, there is a rectangle $R'$ with the side lengths $q_1,\cdots,q_d$ whose volume is within $\epsilon$ to $|R|$. Take a common denominator $l$ of $q_1,\cdots,q_d$, then $R'$ can be covered by a finite grid $\{Q_i\}$ of almost disjoint cubes with side length $l$. By the hint, $\sum |Q_i| = |R'|$, in particular $\sum |Q_i|<|R|+\epsilon$. Given some $E\subset \mathbb{R}^d$, suppose you pick an almost minimizing sequence of rectangles $R_1,R_2,\cdots,$ so that $\sum|R_j|\leq m_*^R(E)+\epsilon$. Now repeat the previous procedure for each rectangle $R_j$: given fixed $j$, choose finitely many cubes $Q_{j,k}$ so that $\sum_{k} |Q_{j,k}|<|R_j|+\frac{\epsilon}{2^j}$, then $ \sum |Q_{j,k}|<\sum |R_j|+\epsilon \leq m_*^R(E)+2\epsilon$. As $Q_{j,k}$ is a countable collection of cubes, and $\epsilon$ was arbitrary, we see that the reverse inequality also holds.