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Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II

Jiří Lebl

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.

Define the outer measure of a set \(S \subset \R^n\) as
\begin{equation*} m^*(S) \coloneqq \inf\, \sum_{j=1}^\infty V(R_j) , \end{equation*}
where the infimum is taken over all sequences \(\{ R_j \}_{j=1}^\infty\) 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 10.6. In particular, \(S\) is of measure zero or a null set if \(m^*(S) = 0\text{.}\)
Figure 10.6. 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 \(\epsilon > 0\text{,}\) there exists a sequence of open rectangles \(\{ R_j \}_{j=1}^\infty\) 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 \(B_k\) is proportional to \(r_k^n\text{.}\)

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 hence 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 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
\begin{equation*} \sum_{k=1}^m V(C_k) \leq 2^n V(R) . \end{equation*}
Figure 10.7. Covering a rectangle by cubes of total size at most \(2^n V(R)\text{.}\)
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
\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_j) < 2^n \epsilon. \end{equation*}
Covering each \(C_k\) with a ball \(B_k\) of radius \(r_k = 2\sqrt{n} \, s_k < \delta\text{,}\) 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*}
As \(S \subset\bigcup_{j} R_j \subset \bigcup_{k} C_k \subset \bigcup_{k} B_k\) and \({(4\sqrt{n})}^n \epsilon\) 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,
\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 of measure zero.
Proof: The set \(\Q^n\) is countable, so write it as a sequence \(q_1,q_2,\ldots\text{.}\) For each \(q_j\text{,}\) 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 \(j\text{,}\) 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} \qquad \text{and} \qquad \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*}
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
\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.

Suppose \(n \in \N\text{,}\) \(k=1,2,\ldots,n\text{,}\) and \(c \in \R\text{.}\) Then \(P \coloneqq \{ x \in \R^n : x_k = c \}\) is of measure zero. Note that if \(n \geq 2\text{,}\) then \(P\) is uncountable.
Proof: First fix \(s \in \N\) and consider
\begin{equation*} P_s \coloneqq \bigl\{ x \in \R^n : x_k = c \text{ and } \sabs{x_j} \leq s \text{ for all } j\not=k \bigr\} . \end{equation*}
Given any \(\epsilon > 0\) define the open rectangle
\begin{equation*} R \coloneqq \bigl\{ x \in \R^n : c-\epsilon < x_k < c+\epsilon \text{ and } \sabs{x_j} < s+1 \text{ for all } j\not=k \bigr\} . \end{equation*}
Clearly, \(P_s \subset R\text{.}\) Furthermore,
\begin{equation*} V(R) = 2\epsilon {\bigl(2(s+1)\bigr)}^{n-1} . \end{equation*}
As \(s\) is fixed, \(V(R)\) can be arbitrarily small by picking \(\epsilon\) small enough. So \(P_s\) is of measure zero.
Next
\begin{equation*} P = \bigcup_{j=1}^\infty P_j \end{equation*}
and a countable union of measure zero sets is of 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 \(\epsilon > 0\text{,}\) we have \(m^*\bigl([a,b]\bigr) \leq b-a\text{.}\)
The other inequality is harder. Suppose \(\bigl\{ (a_j,b_j) \bigr\}_{j=1}^\infty\) are open intervals such that
\begin{equation*} [a,b] \subset \bigcup_{j=1}^\infty (a_j,b_j) . \end{equation*}
We wish to bound \(\sum_{j=1}^\infty (b_j-a_j)\) from below. Since \([a,b]\) is compact, 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 covering \([a,b]\text{.}\) If \((a_i,b_i) \subset (a_j,b_j)\text{,}\) then we throw out \((a_i,b_i)\) as well. 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 \(a_1 < a_2 < \cdots < a_k\text{.}\) As \((a_2,b_2)\) 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 \(b_k > b\text{.}\) See Figure 10.8 for a sample configuration. As \(b_j-a_j > a_{j+1}-a_j\text{,}\) we obtain
\begin{equation*} \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*}
So \(m^*\bigl([a,b]\bigr) \geq b-a\text{.}\)
Figure 10.8. Open intervals covering \([a,b]\) which satisfy \(a_j < a_{j+1} < b_j < b_{j+1}\) for all \(j\text{.}\)

Proof.

As \(E\) is of measure zero, there exists a sequence of open rectangles \(\{ R_j \}_{j=1}^\infty\) 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, here is an uncountable, compact, measure zero subset of \([0,1]\text{,}\) which contains no intervals. Any \(x \in [0,1]\) can be expanded in ternary:
\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 \coloneqq \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 \(C_0 \coloneqq [0,1]\) and
\begin{equation*} C_k \coloneqq \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 10.9.
We leave as an exercise to prove:
  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 10.9. 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 \(B\) is open. As the ball \(B\) is convex, Proposition 8.4.2 says that \(\snorm{f(x)-f(y)} \leq M \snorm{x-y}\) for all \(x,y \in B\text{.}\) So if \(\snorm{x-y} < r\text{,}\) then \(\snorm{f(x)-f(y)} < Mr\text{.}\) 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, \(f(B) = f\bigl(\overline{B(y,r)}\bigr) \subset \overline{f\bigl(B(y,r)\bigr)} \subset \overline{B\bigl(f(y),M r \bigr)}\text{,}\) 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 takes some work to show (see the exercises). However, if the mapping is continuously differentiable, then it 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 \(E\) with the balls \(B(x_j,r_{x_j})\text{.}\) Define
\begin{equation*} U' \coloneqq \bigcup_{j=1}^q B(x_j,r_{x_j}), \qquad K \coloneqq \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 \(\delta \coloneqq \inf_{x\in E} \delta_x\text{.}\) We want to show that \(\delta > 0\text{.}\) Take a sequence \(\{ x_j \}_{j=1}^\infty\) in \(E\) 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 \(r_1,r_2,\ldots,r_k < \nicefrac{\delta}{2}\) 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*}

Exercises 10.3.4 Exercises

10.3.1.

Finish the proof of Proposition 10.3.7: Show that you can use balls instead of rectangles.

10.3.2.

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

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.

10.3.4.

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

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 \(\widebar{S} = \R^n\text{.}\)

10.3.6.

Prove the general case of Proposition 10.3.10 without using compactness:
  1. Mimic the proof to 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 \(m > 0\text{.}\)

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 \(G \coloneqq \bigl\{ (x,y) \in U \times \R : y = f(x) \bigr\}\) be the graph of \(f\text{.}\) Show that \(G\) is of measure zero.

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*}

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 only depends on \(n\) and the ratio \(\nicefrac{\delta}{r}\text{.}\)

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 \([0,1]\text{.}\)

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.

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 \(c_n \coloneqq \frac{d_n}{2}\) for all \(n\text{.}\) Otherwise, let \(k\) be the smallest integer such that \(d_k = 1\text{.}\) Let \(c_n \coloneqq \frac{d_n}{2}\) if \(n < k\text{,}\) \(c_k \coloneqq 1\text{,}\) and \(c_n \coloneqq 0\) if \(n > k\text{.}\) Define
\begin{equation*} \varphi(x) \coloneqq \sum_{n=1}^\infty c_n \, 2^{-n} . \end{equation*}
  1. Prove that \(\varphi\) is continuous and increasing (see Figure 10.9).
  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 \(\psi(x) \coloneqq \varphi(x) + x\text{.}\) 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 not 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 10.10. Cantor function or Devil’s staircase (the function \(\varphi\) from the 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 \(\mu^*(S) \coloneqq \inf\, \sum_{j \in I} V(R_j)\) 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 \(S \subset \R^n\text{,}\) \(\mu^*(S) = m^*(S)\text{.}\)

10.3.14.

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

10.3.15.

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

10.3.16.

(Challenging)   Suppose \(R_1,R_2,\ldots,R_n\) are pairwise disjoint open rectangles. Prove that \(m^*(R_1 \cup R_2 \cup \cdots \cup R_n) = m^*(R_1)+m^*(R_2)+\cdots+m^*(R_n)\text{.}\) Hint: Some of the exercises above may prove very useful.
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