11.3: Outer measure and null sets

Outer measure

Before we characterize all Riemann integrable functions, we need to make a slight detour. We introduce a way of measuring the size of sets in ${\mathbb{R}}^n$.

Let $S \subset {\mathbb{R}}^n$ be a subset. Define the outer measure of $S$ as $$m^*(S) := \inf\, \sum_{j=1}^\infty V(R_j) ,$$ where the infimum is taken over all sequences $\{ R_j \}$ of open rectangles such that $S \subset \bigcup_{j=1}^\infty R_j$. In particular $S$ is of measure zero or a null set if $m^*(S) = 0$.

We will only need measure zero sets and so we focus on these. Note that $S$ is of measure zero if for every $\epsilon > 0$ there exist a sequence of open rectangles $\{ R_j \}$ such that $$S \subset \bigcup_{j=1}^\infty R_j \qquad \text{and} \qquad \sum_{j=1}^\infty V(R_j) < \epsilon.$$ Furthermore, if $S$ is measure zero and $S' \subset S$, then $S'$ is of measure zero. We can in fact use the same exact rectangles.

The set ${\mathbb{Q}}^n \subset {\mathbb{R}}^n$ of points with rational coordinates is a set of measure zero.

Proof: The set ${\mathbb{Q}}^n$ is countable and therefore let us write it as a sequence $q_1,q_2,\ldots$. For each $q_j$ find an open rectangle $R_j$ with $q_j \in R_j$ and $V(R_j) < \epsilon 2^{-j}$. Then $${\mathbb{Q}}^n \subset \bigcup_{j=1}^\infty R_j \qquad \text{and} \qquad \sum_{j=1}^\infty V(R_j) < \sum_{j=1}^\infty \epsilon 2^{-j} = \epsilon .$$

In fact, the example points to a more general result.

A countable union of measure zero sets is of measure zero.

Suppose $$S = \bigcup_{j=1}^\infty S_j$$ where $S_j$ are all measure zero sets. Let $\epsilon > 0$ be given. For each $j$ there exists a sequence of open rectangles $\{ R_{j,k} \}_{k=1}^\infty$ such that $$S_j \subset \bigcup_{k=1}^\infty R_{j,k}$$ and $$\sum_{k=1}^\infty V(R_{j,k}) < 2^{-j} \epsilon .$$ Then $$S \subset \bigcup_{j=1}^\infty \bigcup_{k=1}^\infty R_{j,k} .$$ As $V(R_{j,k})$ is always positive, the sum over all $j$ and $k$ can be done in any order. In particular, it can be done as $$\sum_{j=1}^\infty \sum_{k=1}^\infty V(R_{j,k}) < \sum_{j=1}^\infty 2^{-j} \epsilon = \epsilon . \qedhere$$

The next example is not just interesting, it will be useful later.

[mv:example:planenull] Let $P := \{ x \in {\mathbb{R}}^n : x^k = c \}$ for a fixed $k=1,2,\ldots,n$ and a fixed constant $c \in {\mathbb{R}}$. Then $P$ is of measure zero.

Proof: First fix $s$ and let us prove that $$P_s := \{ x \in {\mathbb{R}}^n : x^k = c, \left\lvert {x^j} \right\rvert \leq s \text{ for all $j\not=k$} \}$$ is of measure zero. Given any $\epsilon > 0$ define the open rectangle $$R := \{ x \in {\mathbb{R}}^n : c-\epsilon < x^k < c+\epsilon, \left\lvert {x^j} \right\rvert < s+1 \text{ for all $j\not=k$} \}$$ It is clear that $P_s \subset R$. Furthermore $$V(R) = 2\epsilon {\bigl(2(s+1)\bigr)}^{n-1} .$$ As $s$ is fixed, we can make $V(R)$ arbitrarily small by picking $\epsilon$ small enough.

Next we note that $$P = \bigcup_{j=1}^\infty P_j$$ and a countable union of measure zero sets is measure zero.

If $a < b$, then $m^*([a,b]) = b-a$.

Proof: In the case of ${\mathbb{R}}$, open rectangles are open intervals. Since $[a,b] \subset (a-\epsilon,b+\epsilon)$ for all $\epsilon > 0$. Hence, $m^*([a,b]) \leq b-a$.

Let us prove the other inequality. Suppose that $\{ (a_j,b_j) \}$ are open intervals such that $$[a,b] \subset \bigcup_{j=1}^\infty (a_j,b_j) .$$ We wish to bound $\sum (b_j-a_j)$ from below. Since $[a,b]$ is compact, then there are only finitely many open intervals that still cover $[a,b]$. As throwing out some of the intervals only makes the sum smaller, we only need to take the finite number of intervals still covering $[a,b]$. If $(a_i,b_i) \subset (a_j,b_j)$, then we can throw out $(a_i,b_i)$ as well. Therefore we have $[a,b] \subset \bigcup_{j=1}^k (a_j,b_j)$ for some $k$, and we assume that the intervals are sorted such that $a_1 < a_2 < \cdots < a_k$. Note that since $(a_2,b_2)$ is not contained in $(a_1,b_1)$ we have that $a_1 < a_2 < b_1 < b_2$. Similarly $a_j < a_{j+1} < b_j < b_{j+1}$. Furthermore, $a_1 < a$ and $b_k > b$. Thus, $$m^*([a,b]) \geq \sum_{j=1}^k (b_j-a_j) \geq \sum_{j=1}^{k-1} (a_{j+1}-a_j) + (b_k-a_k) = b_k-a_1 > b-a .$$

[mv:prop:compactnull] Suppose $E \subset {\mathbb{R}}^n$ is a compact set of measure zero. Then for every $\epsilon > 0$, there exist finitely many open rectangles $R_1,R_2,\ldots,R_k$ such that $$E \subset R_1 \cup R_2 \cup \cdots \cup R_k \qquad \text{and} \qquad \sum_{j=1}^k V(R_j) < \epsilon.$$

Find a sequence of open rectangles $\{ R_j \}$ such that $$E \subset \bigcup_{j=1}^\infty R_j \qquad \text{and} \qquad \sum_{j=1}^\infty V(R_j) < \epsilon.$$ By compactness, finitely many of these rectangles still contain $E$. That is, there is some $k$ such that $E \subset R_1 \cup R_2 \cup \cdots \cup R_k$. Hence $$\sum_{j=1}^k V(R_j) \leq \sum_{j=1}^\infty V(R_j) < \epsilon. \qedhere$$

The image of a measure zero set using a continuous map is not necessarily a measure zero set. However if we assume the mapping is continuously differentiable, then the mapping cannot “stretch” the set too much. The proposition does not require compactness, and this is left as an exercise.

[prop:imagenull] Suppose $U \subset {\mathbb{R}}^n$ is an open set and $f \colon U \to {\mathbb{R}}^n$ is a continuously differentiable mapping. If $E \subset U$ is a compact measure zero set, then $f(E)$ is measure zero.

As FIXME: distance to boundary, did we do that? We should!

FIXME: maybe this closed/open rectangle bussiness should be addressed above

Let $\epsilon > 0$ be given.

FIXME: Let $\delta > 0$ be the distance to boundary

Let us “fatten” $E$ a little bit. Using compactness, there exist finitely many open rectangles $T_1,T_2,\ldots,T_k$ such that $$E \subset T_1 \cup T_2 \cup \cdots \cup T_k \qquad \text{and} \qquad V(T_1) + V(T_2) + \cdots + V(T_k) < \epsilon .$$ Since a closed rectangle has the same volume as an open rectangle with the same sides, so we could take $R_j$ to be the closure of $T_j$, Furthermore a closed rectangle can be written as finitely many small rectangles. Consequently for some $\ell$ there exist finitely many closed rectangles $R_1,R_2,\ldots,R_n$ of side at most $\frac{\sqrt{n}\delta}{2}$. such that $$E \subset R_1 \cup R_2 \cup \cdots \cup R_\ell \qquad \text{and} \qquad V(R_1) + V(R_2) + \cdots + V(R_\ell) < \epsilon .$$ Let $$E' := R_1 \cup R_2 \cup \cdots \cup R_\ell$$

It is left as an exercise (see Exercise

As $f$ is continuously differentiable, the function that takes $x$ to $\left\lVert {Df(x)} \right\rVert$ is continuous, therefore $\left\lVert {Df(x)} \right\rVert$ achieves a maximum on $E$. Thus there exists some $C > 0$ such that $\left\lVert {Df(x)} \right\rVert \leq C$ on $E$.

FIXME

FIXME: may need the fact that the derivative exists AND is continuous on a FATTER E which is still comapact and of size $\epsilon$.

FIXME: Then use the whole lipschitz thing we have.

so we can assume that on $E$

FIXME:

FIXME: Cantor set, fat cantor set, can be done in ${\mathbb{R}}^n$

FIXME: maybe too much

FIXME

Exercises

FIXME:

If $A \subset B$ then $m^*(A) \leq m^*(B)$.

Show that if $R \subset {\mathbb{R}}^n$ is a closed rectangle then $m^*(R) = V(R)$.

Prove a version of without using compactness:
a) Mimic the proof to first prove that the proposition holds only if $E$ is relatively compact; a set $E \subset U$ is relatively compact if the closure of $E$ in the subspace topology on $E$ is compact, or in other words if there exists a compact set $K$ with $K \subset U$ and $E \subset K$.
Hint: The bound on the size of the derivative still holds, but you may need to use countably many rectangles. Be careful as the closure of $E$ need no longer be measure zero.
b) Now prove it for any null set $E$.
Hint: First show that $\{ x \in U : d(x,y) \geq \nicefrac{1}{M} \text{ for all\)y U\(and } d(0,x) \leq M \}$ is a compact set for any $M > 0$.

Let $U \subset {\mathbb{R}}^n$ be an open set and let $f \colon U \to {\mathbb{R}}$ be a continuously differentiable function. Let $G := \{ (x,y) \in U \times {\mathbb{R}}: y = f(x) \}$ be the graph of $f$. Show that $f$ is of measure zero.