Outer measure and null sets

Suppose $R$ is a rectangle of positive volume. Let $s > 0$ be a number less than the smallest side of $R$ and such that $2\sqrt{n} \, s < \delta\text{.}$ If each side of $R$ is an integer multiple of $s\text{,}$ then $R$ is contained in a union of closed cubes $C_1, C_2, \ldots, C_m$ of side $s$ such that $\sum_{k=1}^m V(C_k) = V(R)\text{.}$ So suppose the sides of $R$ are not integer multiples of $s\text{.}$ Consider a side of length $(\ell+\alpha) s\text{,}$ for an integer $\ell$ and $0 \leq \alpha < 1\text{.}$ As $s$ is less than the smallest side, $\ell \geq 1\text{,}$ and so $(\ell+\alpha)s \leq 2\ell s\text{.}$ Increasing this side to $2\ell s\text{,}$ and similarly increasing every side of $R\text{,}$ we obtain a new larger rectangle of volume at most $2^n$ times larger, whose sides are multiples of $s\text{.}$ See Figure 10.7. Thus $R$ is contained in a union of closed cubes $C_1, C_2, \ldots, C_m$ of side $s$ such that

So suppose that $S$ is a null set and there exist open rectangles $\{ R_j \}_{j=1}^\infty$ whose union contains $S$ and such that (10.2) is true. Choose closed cubes $\{ C_k \}_{k=1}^\infty$ with $C_k$ of side $s_k$ as above that cover all the rectangles $\{ R_j \}_{j=1}^\infty$ and so that

Covering each $C_k$ with a ball $B_k$ of radius $r_k = 2\sqrt{n} \, s_k < \delta\text{,}$ we obtain

As $S \subset\bigcup_{j} R_j \subset \bigcup_{k} C_k \subset \bigcup_{k} B_k$ and ${\bigl(4\sqrt{n}\bigr)}^n \epsilon$ can be arbitrarily small, the forward direction follows.

For the other direction, suppose $S$ is covered by balls $B_j$ of radii $r_j\text{,}$ such that $\sum_{j=1}^\infty r_j^n < \epsilon\text{,}$ as in the statement of the proposition. Each $B_j$ is contained in an open cube $R_j$ of side $2r_j\text{.}$ So $V(R_j) = {(2 r_j)}^n = 2^n r_j^n\text{.}$ Therefore,

Suppose

where $S_j$ are all measure zero sets. Let $\epsilon > 0$ be given. For each $j\text{,}$ there exists a sequence of open rectangles $\{ R_{j,k} \}_{k=1}^\infty$ such that

Then

All $V(R_{j,k})$ are nonnegative, so the sum over all $j$ and $k$ can be done by summing first over the $k$ and then over the $j\text{,}$ see Exercise 2.6.15. In particular,

As $E$ is of measure zero, there exists a sequence of open rectangles $\{ R_j \}_{j=1}^\infty$ such that

By compactness, there are finitely many of these rectangles that still contain $E\text{.}$ That is, there is some $k$ such that $E \subset R_1 \cup R_2 \cup \cdots \cup R_k\text{.}$ Hence

We prove the proposition for a compact $E$ and leave the general case as an exercise. Suppose $E$ is compact and of measure zero. First, we will replace $U$ by a smaller open set to make $\bnorm{f'(x)}$ bounded. At each point $x \in E$ pick an open ball $B(x,r_x)$ such that the closed ball $C(x,r_x) \subset U\text{.}$ By compactness, we only need to take finitely many points $x_1,x_2,\ldots,x_q$ to cover $E$ with the balls $B(x_j,r_{x_j})\text{.}$ Define

We have $E \subset U' \subset K \subset U\text{.}$ The set $K\text{,}$ being a finite union of compact sets, is compact. The function that takes $x$ to $\bnorm{f'(x)}$ is continuous, and therefore there exists an $M > 0$ such that $\bnorm{f'(x)} \leq M$ for all $x \in K\text{.}$ So without loss of generality, we may replace $U$ by $U'$ and from now on suppose that $\bnorm{f'(x)} \leq M$ for all $x \in U\text{.}$

Given $\epsilon > 0\text{,}$ there exist balls $B_1,B_2,\ldots,B_k$ of radii $r_1,r_2,\ldots,r_k < \nicefrac{\delta}{2}$ such that

We can assume that each ball contains a point of $E$ and so the balls are contained in $U\text{.}$ Suppose $B_1', B_2', \ldots, B_k'$ are the balls of radius $Mr_1, Mr_2, \ldots, Mr_k$ from Lemma 10.3.9, such that $f(B_j) \subset B_j'$ for all $j\text{.}$ Then,