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Section 10.3 Outer measure and null sets

Note: 2 lectures

Subsection 10.3.1 Outer measure and null sets

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 \(\R^n\text{.}\)

Definition 10.3.1.

Let \(S \subset \R^n\) be a subset. Define the outer measure of \(S\) as

\begin{equation*} m^*(S) := \inf\, \sum_{j=1}^\infty V(R_j) , \end{equation*}

where the infimum is taken over all sequences \(\{ R_j \}\) of open rectangles such that \(S \subset \bigcup_{j=1}^\infty R_j\text{,}\) and we are allowing both the sum and the infimum to be \(\infty\text{.}\) See Figure 10.4. In particular, \(S\) is of measure zero or a null set if \(m^*(S) = 0\text{.}\)


Figure 10.4. Outer measure construction, in this case \(S \subset R_1 \cup R_2 \cup R_3 \cup \cdots\text{,}\) so \(m^*(S) \leq V(R_1) + V(R_2)+V(R_3) + \cdots\text{.}\)

An immediate consequence (Exercise 10.3.2) of the definition is that if \(A \subset B\text{,}\) then \(m^*(A) \leq m^*(B)\text{.}\) It is also not difficult to show (Exercise 10.3.13) that we obtain the same number \(m^*(S)\) if we also allow both finite and infinite sequences of rectangles in the definition. It is not enough, however, to allow only finite sequences.

The theory of measures on \(\R^n\) is a very complicated subject. We will only require measure-zero sets and so we focus on these. A set \(S\) is of measure zero if for every \(\epsilon > 0\) there exists a sequence of open rectangles \(\{ R_j \}\) such that

\begin{equation} S \subset \bigcup_{j=1}^\infty R_j \qquad \text{and} \qquad \sum_{j=1}^\infty V(R_j) < \epsilon.\tag{10.2} \end{equation}

If \(S\) is of measure zero and \(S' \subset S\text{,}\) then \(S'\) is of measure zero. We can use the same exact rectangles.

It is sometimes more convenient to use balls instead of rectangles. Furthermore, we can choose balls no bigger than a fixed radius.

Note that the “volume” of \(B_j\) is proportional to \(r_j^n\text{.}\)

Proof.

If \(C\) is a closed cube (rectangle with all sides equal) of side \(s\text{,}\) then \(C\) is contained in a closed ball of radius \(\sqrt{n}\, s\) by Proposition 10.1.14, and therefore in an open ball of radius \(2 \sqrt{n}\, s\text{.}\)

Suppose \(R\) is a rectangle of positive volume. Let \(s > 0\) be a number that is less than the smallest side of \(R\) and also so that \(2\sqrt{n} \, s < \delta\text{.}\) We claim \(R\) is contained in a union of closed cubes \(C_1, C_2, \ldots, C_k\) of sides \(s\) such that

\begin{equation*} \sum_{j=1}^k V(C_j) \leq 2^n V(R) . \end{equation*}

It is clearly true (without the \(2^n\)) if \(R\) has sides that are integer multiples of \(s\text{.}\) So if a side is of length \((\ell+\alpha) s\text{,}\) for \(\ell \in \N\) and \(0 \leq \alpha < 1\text{,}\) then \((\ell+\alpha)s \leq 2\ell s\text{.}\) Increasing the side to \(2\ell s\text{,}\) and then doing the same for every side, we obtain a new larger rectangle of volume at most \(2^n\) times larger, but whose sides are multiples of \(s\text{.}\)

So suppose that \(S\) is a null set and there exist \(\{ R_j \}\) whose union contains \(S\) and such that (10.2) is true. As we have seen above, we can choose closed cubes \(\{ C_k \}\) with \(C_k\) of side \(s_k\) as above that cover all the rectangles \(\{ R_j \}\) and so that

\begin{equation*} \sum_{k=1}^\infty s_k^n = \sum_{k=1}^\infty V(C_k) \leq 2^n \sum_{j=1}^\infty V(R_k) < 2^n \epsilon. \end{equation*}

Covering \(C_k\) with balls \(B_k\) of radius \(r_k = 2\sqrt{n} \, s_k < \delta\) we obtain

\begin{equation*} \sum_{k=1}^\infty r_k^n = \sum_{k=1}^\infty {(2\sqrt{n})}^n s_k^n < {(4\sqrt{n})}^n \epsilon . \end{equation*}

And as \(S \subset\bigcup_{j} R_j \subset \bigcup_{k} C_k \subset \bigcup_{k} B_k\text{,}\) we are finished.

For the other direction, suppose \(S\) is covered by balls \(B_j\) of radii \(r_j\text{,}\) such that \(\sum r_j^n < \epsilon\text{,}\) 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,

\begin{equation*} S \subset \bigcup_{j=1}^\infty R_j \qquad \text{and} \qquad \sum_{j=1}^\infty V(R_j) \leq \sum_{j=1}^\infty 2^n r_j^n < 2^n \epsilon. \qedhere \end{equation*}

The definition of outer measure (not just null sets) could have been done with open balls as well. We leave this generalization to the reader.

Subsection 10.3.2 Examples and basic properties

Example 10.3.3.

The set \(\Q^n \subset \R^n\) of points with rational coordinates is a set of measure zero.

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

\begin{equation*} \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 . \end{equation*}

The example points to a more general result.

Proof.

Suppose

\begin{equation*} S = \bigcup_{j=1}^\infty S_j , \end{equation*}

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

\begin{equation*} S_j \subset \bigcup_{k=1}^\infty R_{j,k} \end{equation*}

and

\begin{equation*} \sum_{k=1}^\infty V(R_{j,k}) < 2^{-j} \epsilon . \end{equation*}

Then

\begin{equation*} S \subset \bigcup_{j=1}^\infty \bigcup_{k=1}^\infty R_{j,k} . \end{equation*}

As \(V(R_{j,k})\) is always positive, the sum over all \(j\) and \(k\) can be done in any manner. In particular, it can be done as

\begin{equation*} \sum_{j=1}^\infty \sum_{k=1}^\infty V(R_{j,k}) < \sum_{j=1}^\infty 2^{-j} \epsilon = \epsilon . \qedhere \end{equation*}

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

Example 10.3.5.

Let \(P := \{ x \in \R^n : x_k = c \}\) for a fixed \(k=1,2,\ldots,n\) and a fixed constant \(c \in \R\text{.}\) Then \(P\) is of measure zero.

Proof: First fix \(s\) and let us prove that

\begin{equation*} P_s := \bigl\{ x \in \R^n : x_k = c, \sabs{x_j} \leq s \text{ for all } j\not=k \bigr\} \end{equation*}

is of measure zero. Given any \(\epsilon > 0\) define the open rectangle

\begin{equation*} R := \bigl\{ x \in \R^n : c-\epsilon < x_k < c+\epsilon, \sabs{x_j} < s+1 \text{ for all } j\not=k \bigr\} . \end{equation*}

It is clear that \(P_s \subset R\text{.}\) Furthermore,

\begin{equation*} V(R) = 2\epsilon {\bigl(2(s+1)\bigr)}^{n-1} . \end{equation*}

As \(s\) is fixed, we make \(V(R)\) arbitrarily small by picking \(\epsilon\) small enough. So \(P_s\) is measure zero.

Next

\begin{equation*} P = \bigcup_{j=1}^\infty P_j \end{equation*}

and a countable union of measure zero sets is measure zero.

Example 10.3.6.

If \(a < b\text{,}\) then \(m^*\bigl([a,b]\bigr) = b-a\text{.}\)

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

Let us prove the other inequality. Suppose \(\bigl\{ (a_j,b_j) \bigr\}\) are open intervals such that

\begin{equation*} [a,b] \subset \bigcup_{j=1}^\infty (a_j,b_j) . \end{equation*}

We wish to bound \(\sum (b_j-a_j)\) from below. Since \([a,b]\) is compact, then finitely many of the open intervals still cover \([a,b]\text{.}\) As throwing out some of the intervals only makes the sum smaller, we only need to consider the finite number of intervals still covering \([a,b]\text{.}\) If \((a_i,b_i) \subset (a_j,b_j)\text{,}\) then we can throw out \((a_i,b_i)\) as well, in other words the intervals that are left have distinct left endpoints, and whenever \(a_j < a_i < b_j\text{,}\) then \(b_j < b_i\text{.}\) Therefore, \([a,b] \subset \bigcup_{j=1}^k (a_j,b_j)\) for some \(k\text{,}\) and we assume that the intervals are sorted such that \(a_1 < a_2 < \cdots < a_k\text{.}\) Since \((a_2,b_2)\) is not contained in \((a_1,b_1)\text{,}\) since \(a_j > a_2\) for all \(j > 2\text{,}\) and since the intervals must contain every point in \([a,b]\text{,}\) we find that \(a_2 < b_1\text{,}\) or in other words \(a_1 < a_2 < b_1 < b_2\text{.}\) Similarly \(a_j < a_{j+1} < b_j < b_{j+1}\) for all \(j\text{.}\) Furthermore, \(a_1 < a\) and \(b_k > b\text{.}\) Thus,

\begin{equation*} m^*\bigl([a,b]\bigr) \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 . \end{equation*}

Proof.

As \(E\) is of measure zero, there exists a sequence of open rectangles \(\{ R_j \}\) such that

\begin{equation*} E \subset \bigcup_{j=1}^\infty R_j \qquad \text{and} \qquad \sum_{j=1}^\infty V(R_j) < \epsilon. \end{equation*}

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

\begin{equation*} \sum_{j=1}^k V(R_j) \leq \sum_{j=1}^\infty V(R_j) < \epsilon. \end{equation*}

The proof that we can choose balls instead of rectangles is left as an exercise.

Example 10.3.8.

So that the reader is not under the impression that there are only few measure zero sets and that these sets are uncomplicated, let us give an uncountable, compact, measure zero subset of \([0,1]\text{.}\) For every \(x \in [0,1]\text{,}\) write its representation in ternary notation

\begin{equation*} x = \sum_{n=1}^\infty d_n 3^{-n} , \qquad \text{where } d_n=0, 1, \text{ or } 2. \end{equation*}

See Section 1.5, in particular Exercise 1.5.4. Define the Cantor set \(C\) as

\begin{equation*} C := \Bigl\{ x \in [0,1] : x = \sum_{n=1}^\infty d_n 3^{-n}, \text{ where } d_n = 0 \text{ or } d_n = 2 \text{ for all } n \Bigr\} . \end{equation*}

That is, \(x\) is in \(C\) if it has a ternary expansion in only \(0\)s and \(2\)s. If \(x\) has two expansions, as long as one of them does not have any \(1\)s, then \(x\) is in \(C\text{.}\) Define \(C_0 := [0,1]\) and

\begin{equation*} C_k := \Bigl\{ x \in [0,1] : x = \sum_{n=1}^\infty d_n 3^{-n}, \text{ where } d_n = 0 \text{ or } d_n = 2 \text{ for all } n=1,2,\ldots,k \Bigr\} . \end{equation*}

Clearly,

\begin{equation*} C = \bigcap_{k=1}^\infty C_k . \end{equation*}

See Figure 10.5.

We leave as an exercise to prove that

  1. Each \(C_k\) is a finite union of closed intervals. It is obtained by taking \(C_{k-1}\text{,}\) and from each closed interval removing the “middle third.”

  2. Each \(C_k\) is closed, and so \(C\) is closed.

  3. \(m^*(C_k) =1 - \sum_{n=1}^k \frac{2^n}{3^{n+1}}\text{.}\)

  4. Hence, \(m^*(C) = 0\text{.}\)

  5. The set \(C\) is in one-to-one correspondence with \([0,1]\text{,}\) in other words, \(C\) is uncountable.


Figure 10.5. Cantor set construction.

Subsection 10.3.3 Images of null sets under differentiable functions

Before we look at images of measure zero sets, let us see what a continuously differentiable function does to a ball.

Proof.

Suppose that \(B\) is open. The ball \(B\) is convex, and so via Proposition 8.4.2, \(\snorm{f(x)-f(y)} \leq M \snorm{x-y}\text{.}\) So if \(\snorm{x-y} < r\text{,}\) then \(\snorm{f(x)-f(y)} < Mr\text{,}\) or in other words, if \(B=B(y,r)\text{,}\) then \(f(B) \subset B\bigl(f(y),M r \bigr)\text{.}\) If \(B\) is closed, then \(\overline{B(y,r)} = B\text{.}\) As \(f\) is continuous, \(f(B) = f\bigl(\overline{B(y,r)}\bigr) \subset \overline{B\bigl(f(y),M r \bigr)}\text{,}\) as \(f(\widebar{A}) \subset \overline{f(A)}\) for any set \(A\text{.}\)

The image of a measure zero set using a continuous map is not necessarily a measure zero set, although this is not easy to show (see the exercises). However, if the mapping is continuously differentiable, then the mapping cannot “stretch” the set that much.

Proof.

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 \(\snorm{f'(x)}\) 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 \(E\) we the balls \(B(x_j,r_{x_j})\text{.}\) Define

\begin{equation*} U' := \bigcup_{j=1}^q B(x_j,r_{x_j}), \qquad K := \bigcup_{j=1}^q C(x_j,r_{x_j}). \end{equation*}

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 \(\snorm{f'(x)}\) is continuous, and therefore there exists an \(M > 0\) such that \(\snorm{f'(x)} \leq M\) for all \(x \in K\text{.}\) So without loss of generality we may replace \(U\) by \(U'\) and from now on suppose that \(\snorm{f'(x)} \leq M\) for all \(x \in U\text{.}\)

At each \(x \in E\text{,}\) take the maximum radius \(\delta_x\) such that \(B(x,\delta_x) \subset U\) (we may assume \(U \not= \R^n\)). Let \(\delta := \inf_{x\in E} \delta_x\text{.}\) We want to show that \(\delta > 0\text{.}\) Take a sequence \(\{ x_j \} \subset E\) so that \(\delta_{x_j} \to \delta\text{.}\) As \(E\) is compact, we can pick the sequence to be convergent to some \(y \in E\text{.}\) Once \(\snorm{x_j-y} < \frac{\delta_y}{2}\text{,}\) then \(\delta_{x_j} > \frac{\delta_y}{2}\) by the triangle inequality. Thus, \(\delta > 0\text{.}\)

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

\begin{equation*} E \subset B_1 \cup B_2 \cup \cdots \cup B_k \qquad \text{and} \qquad \sum_{j=1}^k r_j^n < \epsilon. \end{equation*}

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,

\begin{equation*} f(E) \subset f(B_1) \cup f(B_2) \cup \cdots \cup f(B_k) \subset B_1' \cup B_2' \cup \cdots \cup B_k' \qquad \text{and} \qquad \sum_{j=1}^k {(Mr_j)}^n < M^n \epsilon. \qedhere \end{equation*}

Subsection 10.3.4 Exercises

Exercise 10.3.1.

Finish the proof of Proposition 10.3.7, that is, show that you can use balls instead of rectangles.

Exercise 10.3.2.

If \(A \subset B\text{,}\) then \(m^*(A) \leq m^*(B)\text{.}\)

Exercise 10.3.3.

Suppose \(X \subset \R^n\) is a set such that for every \(\epsilon > 0\text{,}\) there exists a set \(Y\) such that \(X \subset Y\) and \(m^*(Y) \leq \epsilon\text{.}\) Prove that \(X\) is a measure zero set.

Exercise 10.3.4.

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

Exercise 10.3.5.

The closure of a measure zero set can be quite large. Find an example set \(S \subset \R^n\) that is of measure zero, but whose closure \(\widebar{S} = \R^n\text{.}\)

Exercise 10.3.6.

Prove the general case of Proposition 10.3.10 without using compactness:

  1. Mimic the proof to first prove that the proposition holds if \(E\) is relatively compact; a set \(E \subset U\) is relatively compact if the closure of \(E\) in the subspace topology on \(U\) is compact, or in other words if there exists a compact set \(K\) with \(K \subset U\) and \(E \subset K\text{.}\)
    Hint: The bound on the size of the derivative still holds, but you need to use countably many balls in the second part of the proof. Be careful as the closure of \(E\) need no longer be measure zero.

  2. Now prove it for every null set \(E\text{.}\)
    Hint: First show that \(\{ x \in U : \snorm{x-y} \geq \nicefrac{1}{m} \text{ for all } y \notin U \text{ and } \snorm{x} \leq m \}\) is compact for every \(m > 0\text{.}\)

Exercise 10.3.7.

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

Exercise 10.3.8.

Given a closed rectangle \(R \subset \R^n\text{,}\) show that for every \(\epsilon > 0\text{,}\) there exists a number \(s > 0\) and finitely many open cubes \(C_1,C_2,\ldots,C_k\) of side \(s\) such that \(R \subset C_1 \cup C_2 \cup \cdots \cup C_k\) and

\begin{equation*} \sum_{j=1}^k V(C_j) \leq V(R) + \epsilon . \end{equation*}

Exercise 10.3.9.

Show that there exists a number \(k = k(n,r,\delta)\) depending only on \(n\text{,}\) \(r\) and \(\delta\) such the following holds. Given \(B(x,r) \subset \R^n\) and \(\delta > 0\text{,}\) there exist \(k\) open balls \(B_1,B_2,\ldots,B_k\) of radius at most \(\delta\) such that \(B(x,r) \subset B_1 \cup B_2 \cup \cdots \cup B_k\text{.}\) Note that you can find \(k\) that really only depends on \(n\) and the ratio \(\nicefrac{\delta}{r}\text{.}\)

Exercise 10.3.10.

(Challenging)   Prove the statements of Example 10.3.8. That is, prove:

  1. Each \(C_k\) is a finite union of closed intervals, and so \(C\) is closed.

  2. \(m^*(C_k) =1 - \sum_{n=1}^k \frac{2^n}{3^{n+1}}\text{.}\)

  3. \(m^*(C) = 0\text{.}\)

  4. The set \(C\) is in one-to-one correspondence with \([0,1]\text{.}\)

Exercise 10.3.11.

Prove that the Cantor set of Example 10.3.8 contains no interval. That is, whenever \(a < b\text{,}\) there exists a point \(x \notin C\) such that \(a < x < b\text{.}\)
Note a consequence of this statement. While every open set in \(\R\) is a countable disjoint union of intervals, a closed set (even though it is just the complement of an open set) need not be a union of intervals.

Exercise 10.3.12.

(Challenging)   Let us construct the so-called Cantor function or the Devil's staircase. Let \(C\) be the Cantor set and let \(C_k\) be as in Example 10.3.8. Write \(x \in [0,1]\) in ternary representation \(x = \sum_{n=1} d_n 3^{-n}\text{.}\) If \(d_n \not= 1\) for all \(n\text{,}\) then let \(c_n := \frac{d_n}{2}\) for all \(n\text{.}\) Otherwise, let \(k\) be the smallest integer such that \(d_k = 1\text{.}\) Then let \(c_n := \frac{d_n}{2}\) if \(n < k\text{,}\) \(c_k := 1\text{,}\) and \(c_n := 0\) if \(n > k\text{.}\) Then define

\begin{equation*} \varphi(x) := \sum_{n=1}^\infty c_n \, 2^{-n} . \end{equation*}
  1. Prove that \(\varphi\) is continuous and increasing (see Figure 10.5).

  2. Prove that for \(x \notin C\text{,}\) \(\varphi\) is differentiable at \(x\) and \(\varphi'(x) = 0\text{.}\) (Notice that \(\varphi'\) exists and is zero except for a set of measure zero, yet the function manages to climb from \(0\) to \(1\text{.}\))

  3. Define \(\psi \colon [0,1] \to [0,2]\) by \(\psi(x) := \varphi(x) + x\text{.}\) Show that \(\psi\) is continuous, strictly increasing, and bijective.

  4. Prove that while \(m^*(C) = 0\text{,}\) \(m^*\bigl(\psi(C)\bigr) \not= 0\text{.}\) That is, continuous functions need take measure zero sets to measure zero sets. Hint: \(m^*\bigl(\psi([0,1] \setminus C)\bigr) = 1\text{,}\) but \(m^*\bigl([0,2]\bigr) = 2\text{.}\)


Figure 10.6. Cantor function or Devil's staircase (the function \(\varphi\) from the exercise).

Exercise 10.3.13.

Prove that we obtain the same outer measure if we allow both finite and infinite sequences in the definition. That is, define \(\mu^*(S) := \inf\, \sum_{j \in I} V(R_j)\) where the infimum is taken over all countable (finite or infinite) sets of open rectangles \(\{ R_j \}_{j\in I}\) such that \(S \subset \bigcup_{j \in I} R_j\text{.}\) Prove that for every \(S \subset \R^n\text{,}\) \(\mu^*(S) = m^*(S)\text{.}\)

Exercise 10.3.14.

Prove that for any two subsets \(A, B \subset \R^n\text{,}\) we have \(m^*(A \cup B) \leq m^*(A)+m^*(B)\text{.}\)

Exercise 10.3.15.

Suppose \(A, B \subset \R^n\) are such that \(m^*(B)=0\text{.}\) Prove that \(m^*(A \cup B) = m^*(A)\text{.}\)

Exercise 10.3.16.

Suppose \(R_1,R_2,\ldots,R_n\) are pairwise disjoint open rectangles. Prove that \(m^*(R_1 \cup R_2 \cup \cdots \cup R_n) = m^*(R_1)+m^*(R_2)+\cdots+m^*(R_n)\text{.}\)

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