University Mathematics — Year 3 · Bachelor Year 3

9Measure Theory

How long is a subset of $\R$ ? The naive answer — assign to every set a translation-invariant length extending that of intervals — is impossible: Vitali’s construction, at the end of this chapter, produces a set with no consistent length. Measure theory is the disciplined retreat: restrict attention to a rich class of measurable sets, on which a countably additive length exists and is unique. The rewards are immense — Lebesgue’s integral (Chapter 10), the $L^p$ spaces of functional analysis, and the whole of modern probability (Chapter 22) are built on the three theorems proved here: Dynkin’s uniqueness lemma, Carathéodory’s extension theorem, and the existence of Lebesgue measure.

9.1 $\sigma$ -algebras

Definition 9.1

A $\sigma$ -algebra on a set $X$ is a family $\mathcal A$ of subsets containing $\varnothing$ , stable under complement and under countable unions (hence countable intersections, set differences, and it contains $X$ ). The pair $(X, \mathcal A)$ is a measurable space; members of $\mathcal A$ are measurable sets. For any family $\mathcal E$ of subsets, $\sigma(\mathcal E)$ denotes the smallest $\sigma$ -algebra containing $\mathcal E$ (the intersection of all of them — an intersection of $\sigma$ -algebras is one).

Definition 9.2

The Borel $\sigma$ -algebra of a topological space is $\mathcal B(X) = \sigma(\{\text{open sets}\})$ . On $\R$ : $\mathcal B(\R)$ is also generated by the open intervals, by the closed intervals, by the rays $(-\infty, a]$ , and by rays with rational endpoints (Exercise 9.1) — each family generates the opens by countable operations, e.g. every open set of $\R$ is a countable union of open intervals with rational data.

Definition 9.3

A $\pi$ -system is a family stable under finite intersections. A $\lambda$ -system (Dynkin class) is a family $\mathcal D$ with: $X \in \mathcal D$ ; $A, B \in \mathcal D$ , $A \subseteq B$ $\Rightarrow$ $B \setminus A \in \mathcal D$ ; and $A_n \uparrow A$ , $A_n \in \mathcal D$ $\Rightarrow$ $A \in \mathcal D$ .

Theorem 9.4 (Dynkin’s $\pi$ – $\lambda$ lemma)

If a $\lambda$ -system $\mathcal D$ contains a $\pi$ -system $\mathcal P$ , then $\mathcal D \supseteq \sigma(\mathcal P)$ .

Proof. Let $\mathcal D_0$ be the smallest $\lambda$ -system containing $\mathcal P$ (intersection of all such); it suffices to show $\mathcal D_0$ is a $\sigma$ -algebra, since then $\sigma(\mathcal P) \subseteq \mathcal D_0 \subseteq \mathcal D$ . A $\lambda$ -system stable under finite intersections is a $\sigma$ -algebra: complements ( $X \setminus A = X \setminus A$ with $A \subseteq X$ ), finite unions ( $A \cup B = X \setminus ((X\setminus A)\cap(X\setminus B))$ ), and countable unions via $\bigcup_{k \leq n}A_k \uparrow \bigcup_kA_k$ . So we prove $\mathcal D_0$ is a $\pi$ -system, in two steps. Let

\mathcal D_1 = \{A \in \mathcal D_0 : A \cap P \in \mathcal D_0 \ \forall P \in \mathcal P\}.

$\mathcal D_1$ is a $\lambda$ -system (all three axioms are verified by intersecting with $P$ : e.g. $(B\setminus A)\cap P = (B \cap P)\setminus(A \cap P)$ , a proper difference within $\mathcal D_0$ ) and contains $\mathcal P$ ( $\pi$ -system): $\mathcal D_1 = \mathcal D_0$ . Now let

\mathcal D_2 = \{A \in \mathcal D_0 : A \cap D \in \mathcal D_0\ \forall D \in \mathcal D_0\}.

By the previous step, $\mathcal D_2 \supseteq \mathcal P$ ; and $\mathcal D_2$ is a $\lambda$ -system by the same verification: $\mathcal D_2 = \mathcal D_0$ , which says exactly that $\mathcal D_0$ is stable under intersections. ∎

9.2 Measures

Definition 9.5

A measure on $(X, \mathcal A)$ is a map $\mu \colon \mathcal A \to [0, +\infty]$ with $\mu(\varnothing) = 0$ that is $\sigma$ -additive: for pairwise disjoint $(A_n)_{n\in\N}$ ,

\mu\Bigl(\bigsqcup_n A_n\Bigr) = \sum_n \mu(A_n).

$(X, \mathcal A, \mu)$ is a measure space; $\mu$ is finite if $\mu(X) < \infty$ , a probability measure if $\mu(X) = 1$ , $\sigma$ -finite if $X$ is a countable union of sets of finite measure. Examples: counting measure on $(\N, \mathcal P(\N))$ ; the Dirac mass $\delta_a(A) = \mathbf 1_{a \in A}$ ; and, the object of this chapter, Lebesgue measure.

Proposition 9.6

Let $\mu$ be a measure. (a) Monotonicity: $A \subseteq B \Rightarrow \mu(A) \leq \mu(B)$ . (b) Countable subadditivity: $\mu(\bigcup A_n) \leq \sum\mu(A_n)$ . (c) Continuity from below: $A_n \uparrow A \Rightarrow \mu(A_n) \to \mu(A)$ . (d) Continuity from above: $A_n \downarrow A$ with $\mu(A_1) < \infty$ $\Rightarrow \mu(A_n) \to \mu(A)$ .

Proof. (a) $B = A \sqcup (B\setminus A)$ . (b) Disjointify: $B_n = A_n \setminus \bigcup_{k<n}A_k$ are disjoint with the same union, and $\mu(B_n) \leq \mu(A_n)$ . (c) $A = \bigsqcup_n (A_n \setminus A_{n-1})$ ( $A_0 = \varnothing$ ): the partial sums of $\sum\mu(A_n\setminus A_{n-1})$ are $\mu(A_n)$ . (d) Apply (c) to $A_1 \setminus A_n \uparrow A_1 \setminus A$ and subtract from $\mu(A_1)$ — finiteness makes the subtraction legitimate. Counterexample without it: $A_n = [n, \infty)$ for Lebesgue measure: $A_n \downarrow \varnothing$ but $\mu(A_n) = \infty$ . ∎

Theorem 9.7 (Uniqueness)

Let $\mu, \nu$ be measures on $\sigma(\mathcal P)$ , $\mathcal P$ a $\pi$ -system, with $\mu = \nu$ on $\mathcal P$ . If there are sets $P_k \in \mathcal P$ with $P_k \uparrow X$ and $\mu(P_k) < \infty$ , then $\mu = \nu$ on all of $\sigma(\mathcal P)$ .

Proof. Fix $k$ and consider the finite measures $\mu_k(A) = \mu(A \cap P_k)$ and $\nu_k(A) = \nu(A \cap P_k)$ on $\sigma(\mathcal P)$ : they agree on $\mathcal P$ , since $P \cap P_k \in \mathcal P$ ( $\pi$ -system), and they assign $X$ the same finite value $\mu(P_k)$ . The class $\mathcal D = \{A : \mu_k(A) = \nu_k(A)\}$ is a $\lambda$ -system: $X \in \mathcal D$ ; proper differences by subtraction (finite values); increasing limits by continuity from below (Proposition 9.6(c)). It contains the $\pi$ -system $\mathcal P$ , so Dynkin (Theorem 9.4) gives $\mathcal D \supseteq \sigma(\mathcal P)$ : $\mu_k = \nu_k$ everywhere. Finally, for any $A \in \sigma(\mathcal P)$ , continuity from below along $A \cap P_k \uparrow A$ gives $\mu(A) = \lim_k\mu_k(A) = \lim_k\nu_k(A) = \nu(A)$ . ∎

9.3 Outer measures and Carathéodory’s theorem

Definition 9.8

An outer measure on $X$ is a map $\mu^* \colon \mathcal P(X) \to [0, \infty]$ with $\mu^*(\varnothing) = 0$ , monotone, and countably subadditive. A set $A$ is $\mu^*$ -measurable (Carathéodory) if it splits every set additively:

\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \setminus A) \qquad \text{for every } E \subseteq X

( $\leq$ always holds by subadditivity; the content is $\geq$ ).

Theorem 9.9 (Carathéodory)

The $\mu^*$ -measurable sets form a $\sigma$ -algebra $\mathcal M$ , and $\mu^*\restriction_{\mathcal M}$ is a measure. Moreover every set with $\mu^*(N) = 0$ belongs to $\mathcal M$ (the measure is complete).

Proof. $\mathcal M$ contains $\varnothing$ and is stable under complement (the defining condition is symmetric in $A$ , $X\setminus A$ ). Finite unions: let $A, B \in \mathcal M$ and $E$ arbitrary; splitting $E$ by $A$ , then each piece by $B$ :

\mu^*(E) = \mu^*(E\cap A\cap B) + \mu^*(E\cap A\setminus B) + \mu^*(E\cap B\setminus A) + \mu^*(E\setminus(A\cup B)).

The first three pieces cover $E \cap (A \cup B)$ , so subadditivity gives $\mu^*(E) \geq \mu^*(E\cap(A\cup B)) + \mu^*(E\setminus(A\cup B))$ : $A \cup B \in \mathcal M$ . By induction, finite unions; with complements, finite disjointness manipulations are available.

Additivity on $\mathcal M$ : for disjoint $A, B \in \mathcal M$ and any $E$ : $\mu^*(E\cap(A\sqcup B)) = \mu^*(E\cap A) + \mu^*(E \cap B)$ (split by $A$ ); by induction,

\mu^*\Bigl(E \cap \bigsqcup_{k\leq n}A_k\Bigr) = \sum_{k\leq n}\mu^*(E\cap A_k). \tag{$*$}

Countable unions: let $(A_k) \subseteq \mathcal M$ be disjoint (suffices, by disjointification within the algebra $\mathcal M$ ), $A = \bigsqcup A_k$ , $E$ arbitrary. Using $\bigsqcup_{k\leq n}A_k \in \mathcal M$ and monotonicity:

\mu^*(E) = \mu^*\Bigl(E\cap\bigsqcup_{k\leq n}A_k\Bigr) + \mu^*\Bigl(E\setminus\bigsqcup_{k\leq n}A_k\Bigr) \geq \sum_{k \leq n}\mu^*(E\cap A_k) + \mu^*(E\setminus A)

by ( $*$ ). Let $n \to \infty$ and use countable subadditivity backwards:

\mu^*(E) \geq \sum_{k}\mu^*(E\cap A_k) + \mu^*(E\setminus A) \geq \mu^*(E \cap A) + \mu^*(E\setminus A) \geq \mu^*(E):

all inequalities are equalities. This proves both $A \in \mathcal M$ and, taking $E = A$ , countable additivity of $\mu^*$ on $\mathcal M$ .

Null sets: if $\mu^*(N) = 0$ , then for any $E$ : $\mu^*(E \cap N) + \mu^*(E\setminus N) \leq 0 + \mu^*(E)$ : $N \in \mathcal M$ . ∎

9.4 Lebesgue measure on $\R$

Definition 9.10

The Lebesgue outer measure of $A \subseteq \R$ is

\lambda^*(A) = \inf\Bigl\{\sum_{n} (b_n - a_n) : A \subseteq \bigcup_n \intoo{a_n}{b_n}\Bigr\}

(countable covers by open intervals).

Lemma 9.11

$\lambda^*$ is an outer measure, invariant under translations, and $\lambda^*(I) = \ell(I)$ (the length) for every interval $I$ .

Proof. Outer measure: $\varnothing$ is covered by arbitrarily small intervals; monotonicity is clear; subadditivity: given covers of each $A_n$ within $\varepsilon 2^{-n}$ of the infimum, their union covers $\bigcup A_n$ with total length $\leq \sum \lambda^*(A_n) + \varepsilon$ . Translation invariance: translate the covers.

Length: it suffices to treat $I = \intcc ab$ (other types differ by endpoints, which have outer measure $0$ : cover by tiny intervals; then squeeze $\intcc{a+\varepsilon}{b - \varepsilon} \subseteq \intoo ab$ -type comparisons). $\lambda^*(\intcc ab) \leq b - a$ : cover by $\intoo{a-\varepsilon}{b+\varepsilon}$ . Conversely let $\intcc ab \subseteq \bigcup_n\intoo{a_n}{b_n}$ : by compactness (Borel–Lebesgue, Theorem 6.16), finitely many intervals suffice, say $I_1, \dots, I_N$ . We show $\sum_{k\leq N}(b_k - a_k) \geq b - a$ by induction on $N$ : choose $I_{k_1} \ni a$ ; if $b_{k_1} > b$ done ( $b_{k_1} - a_{k_1} > b - a$ ); else the segment $\intcc{b_{k_1}}b$ is covered by the remaining $N - 1$ intervals, and induction gives $\sum_{k \neq k_1}(b_k - a_k) \geq b - b_{k_1}$ , while $b_{k_1} - a_{k_1} > b_{k_1} - a$ : sum up. ∎

Theorem 9.12 (Lebesgue measure)

Every Borel set of $\R$ is $\lambda^*$ -measurable. The restriction $\lambda$ of $\lambda^*$ to the $\sigma$ -algebra $\mathcal L = \mathcal M_{\lambda^*} \supseteq \mathcal B(\R)$ (the Lebesgue $\sigma$ -algebra) is the unique measure on $\mathcal B(\R)$ assigning to each interval its length; it is translation-invariant and $\sigma$ -finite.

Proof. By Theorem 9.9 it suffices to show each ray $A = \intoo{-\infty}c$ is $\lambda^*$ -measurable (rays generate $\mathcal B$ , Definition 9.2). Let $E \subseteq \R$ with $\lambda^*(E) < \infty$ and $\bigcup I_n \supseteq E$ a cover with $\sum\ell(I_n) \leq \lambda^*(E) + \varepsilon$ . Each $I_n$ splits into the two intervals $I_n' = I_n \cap A$ and $I_n'' = I_n\setminus A$ (an interval minus a ray is an interval) with $\ell(I_n') + \ell(I_n'') = \ell(I_n)$ ; the $I_n'$ cover $E \cap A$ and the $I_n''$ cover $E \setminus A$ (enlarge each to an open interval of length $\ell + \varepsilon2^{-n}$ to stay within the definition), so

\lambda^*(E\cap A) + \lambda^*(E\setminus A) \leq \sum_n\bigl(\ell(I_n') + \ell(I_n'')\bigr) + 2\varepsilon \leq \lambda^*(E) + 3\varepsilon .

Uniqueness: two measures agreeing with length on the $\pi$ -system of intervals $\intoc ab$ (finite on them) agree on $\sigma(\text{intervals}) = \mathcal B$ by Theorem 9.7 with $P_k = \intoc{-k}k$ . $\sigma$ -finiteness: $\R = \bigcup(-k, k]$ . ∎

Theorem 9.13 (Regularity)

For every $A \in \mathcal L$ :

\lambda(A) = \inf\{\lambda(U) : U \supseteq A \text{ open}\} = \sup\{\lambda(K) : K \subseteq A \text{ compact}\}.

Proof. Outer: a cover $\bigcup I_n$ with $\sum\ell(I_n) \leq \lambda(A) + \varepsilon$ is an open set $U \supseteq A$ with $\lambda(U) \leq \lambda(A) + \varepsilon$ (subadditivity); if $\lambda(A) = \infty$ the statement is trivial. Inner: first let $A$ be bounded, $A \subseteq [-M, M]$ . Choose an open $U \supseteq ([-M,M]\setminus A)$ with $\lambda(U) \leq \lambda([-M,M]\setminus A) + \varepsilon$ ; then $K = [-M, M]\setminus U$ is compact, $K \subseteq A$ , and

\lambda(K) = \lambda([-M,M]) - \lambda([-M,M]\cap U) \geq \lambda([-M,M]) - \bigl(\lambda([-M,M]) - \lambda(A) + \varepsilon\bigr) = \lambda(A) - \varepsilon .

For general $A$ : $\lambda(A) = \lim_M\lambda(A\cap[-M,M])$ (continuity from below) and apply the bounded case inside. ∎

Example 9.14

The Cantor set (Exercise 6.10) has $\lambda(C) = 0$ : $C \subseteq C_n$ , a union of $2^n$ intervals of length $3^{-n}$ , so $\lambda(C) \leq (2/3)^n \to 0$ . An uncountable null set — cardinality does not see measure. Conversely, fat Cantor sets (Exercise 9.5) are nowhere dense with positive measure: topology does not see measure either. The weekend problem pushes this interplay to its striking conclusion: there are Lebesgue-measurable sets that are not Borel.

Theorem 9.15 (Vitali)

There is no measure on all subsets of $\R$ that is translation-invariant and assigns to each interval its length. In particular $\mathcal L \neq \mathcal P(\R)$ : non-measurable sets exist.

Proof. Suppose $\mu$ were one. On $\intcc01$ , consider the equivalence $x \sim y \iff x - y \in \Q$ ; by the axiom of choice pick one representative in $\intcc01$ per class: a set $V$ . For $q \in \Q\cap\intcc{-1}1$ , the translates $V + q$ are pairwise disjoint (two points of $V$ differing by a rational would be equivalent yet distinct representatives) and

\intcc01 \subseteq \bigsqcup_{q \in \Q\cap\intcc{-1}1}(V + q) \subseteq \intcc{-1}2 :

the first inclusion since every $x \in \intcc01$ differs from its representative $v$ by a rational $q = x - v \in \intcc{-1}1$ . Monotonicity and $\sigma$ -additivity give

1 \leq \sum_{q}\mu(V + q) \leq 3, \qquad\text{with } \mu(V + q) = \mu(V) \text{ for all } q .

An infinite sum of the constant $\mu(V)$ is $0$ or $\infty$ : both bounds cannot hold. So no such $\mu$ exists — and $V \notin \mathcal L$ , since $\lambda$ on $\mathcal L$ has all the properties used. ∎

Remark 9.16

In $\R^3$ the failure is more dramatic: the Banach–Tarski paradox decomposes a ball into five pieces that reassemble, by rotations and translations, into two balls of the same radius — so not even a finitely additive rotation-invariant volume on all subsets of $\R^3$ exists. The pieces are, of course, non-measurable. Measurability is not bureaucratic caution; it is the boundary of coherence.

Method 9.17

The good sets principle: to prove that all sets of $\sigma(\mathcal E)$ have a property, show that the good sets form a $\sigma$ -algebra (or a $\lambda$ -system, if the property is measure-theoretic and $\mathcal E$ is a $\pi$ -system — then Dynkin) containing $\mathcal E$ . Almost every proof of this chapter and the next is an instance. To prove two measures equal: check them on a generating $\pi$ -system plus $\sigma$ -finiteness (Theorem 9.7). To build a measure: build an outer measure by covers and quote Carathéodory.

9.5 Exercises

Exercise 9.1 ★

(a) Show that $\{A \subseteq X : A$ or $X\setminus A$ is countable $\}$ is a $\sigma$ -algebra: the one generated by singletons. (b) Show that $\mathcal B(\R)$ is generated by each of: open intervals; closed intervals; rays $\intoc{-\infty}a$ ; rays with $a \in \Q$ . (c) Is the family of finite disjoint unions of intervals $\intoc ab$ a $\sigma$ -algebra? An algebra (stable under complement and finite unions)?

Solution

Solution of Exercise 9.1.

(a) Complementation swaps the two defining cases. A countable union of countable sets is countable; if one member is co-countable, the union is co-countable: stability holds. It contains the singletons, and any $\sigma$ -algebra containing them contains all countable sets (countable unions) and their complements: it is $\sigma(\{\text{singletons}\})$ .

(b) Write $\mathcal B = \sigma(\text{opens})$ . Every open subset of $\R$ is a countable union of open intervals with rational endpoints (around each rational point of the open set, a rational-radius interval inside it), so opens $\in \sigma(\text{open intervals}) \subseteq \sigma(\text{rational data})$ . Conversions: $\intoo ab = \bigcup_n\intcc{a + \frac1n}{b - \frac1n}$ ; $\intcc ab = \bigcap_n \intoo{a - \frac1n}{b + \frac1n}$ ; $\intoc{-\infty}a = \bigcap_n \intoo{-\infty}{a + \frac1n}$ and conversely $\intoo ab = \intoo{-\infty}b \setminus \intoc{-\infty}a$ ; rational rays: $\intoc{-\infty}a = \bigcap_{q \in \Q,\, q > a}\intoc{-\infty}q$ . Each family generates the others by countable operations: all four generate $\mathcal B$ .

(c) With finite endpoints only, the family is not even an algebra: the complement of $\intoc01$ contains unbounded rays. Allowing infinite endpoints ( $\intoc{-\infty}b$ , $\intoo a{+\infty}$ ) it becomes an algebra (complements and finite unions of such unions are such), but not a $\sigma$ -algebra: $\{0\} = \bigcap_n\intoc{-\frac1n}0$ is not a finite union of nondegenerate intervals.

Exercise 9.2 ★

(a) Prove inclusion–exclusion for a finite measure: $\mu(A\cup B) = \mu(A) + \mu(B) - \mu(A\cap B)$ , and the three-set version. (b) Give an example showing that continuity from above (Proposition 9.6(d)) fails without the finiteness assumption. (c) Show that a countable set has Lebesgue measure zero. Deduce $\lambda(\Q) = 0$ and $\lambda(\intcc01\setminus\Q) = 1$ .

Solution

Solution of Exercise 9.2.

(a) $A \cup B = A \sqcup (B \setminus (A\cap B))$ , so $\mu(A\cup B) = \mu(A) + \mu(B) - \mu(A\cap B)$ (finiteness permits the subtraction). Three sets: apply the two-set formula twice,

\mu(A\cup B\cup C) = \sum\mu(A) - \sum\mu(A\cap B) + \mu(A\cap B\cap C)

(sums over the obvious index sets).

(b) For Lebesgue measure, $A_n = [n, +\infty) \downarrow \varnothing$ , but $\lambda(A_n) = \infty \not\to 0$ .

(c) A point lies in an interval of length $\varepsilon$ : $\lambda(\{x\}) = 0$ ; countable subadditivity kills countable sets. Hence $\lambda(\Q \cap \intcc01) = 0$ and, by additivity, $\lambda(\intcc01\setminus\Q) = 1$ : the irrationals carry all the length.

Exercise 9.3 ★★

Let $\mu, \nu$ be probability measures on $\mathcal B(\R)$ with $\mu(\intoc{-\infty}t) = \nu(\intoc{-\infty}t)$ for all $t \in \R$ . Show $\mu = \nu$ . (This makes the distribution function $F(t) = \mu(\intoc{-\infty}t)$ a complete invariant — the foundation of Chapter 22.)

Solution

Solution of Exercise 9.3.

The rays $\intoc{-\infty}t$ form a $\pi$ -system (the intersection of two is the smaller) generating $\mathcal B(\R)$ (Exercise 9.1). The sets $P_k = \intoc{-\infty}k$ increase to $\R$ with $\mu(P_k) \leq 1 < \infty$ : Theorem 9.7 applies, and $\mu = \nu$ on $\mathcal B(\R)$ . Thus the distribution function determines the measure.

Exercise 9.4 ★★

(Borel–Cantelli, measure version) Let $(A_n)$ be measurable with $\sum_n\mu(A_n) < \infty$ , and $\limsup A_n = \bigcap_N\bigcup_{n\geq N}A_n$ (the points belonging to infinitely many $A_n$ ). Show $\mu(\limsup A_n) = 0$ . Application: for almost every $x \in \intcc01$ , only finitely many $n$ satisfy $\abs{x - p/q_n} \leq 4^{-n}$ for the $n$ -th rational $p/q_n$ of an enumeration of $\Q\cap\intcc01$ .

Solution

Solution of Exercise 9.4.

For every $N$ , $\limsup A_n \subseteq \bigcup_{n \geq N}A_n$ , so $\mu(\limsup A_n) \leq \sum_{n\geq N}\mu(A_n)$ , the tail of a convergent series: let $N \to \infty$ . Application: with $A_n = \{x \in \intcc01 : \abs{x - r_n} \leq 4^{-n}\}$ ( $r_n$ the $n$ -th rational), $\lambda(A_n) \leq 2\cdot4^{-n}$ is summable: $\lambda(\limsup A_n) = 0$ , i.e. almost every $x$ belongs to only finitely many $A_n$ . (Yet every $x$ is a limit of rationals: the point is the speed $4^{-n}$ .)

Exercise 9.5 ★★

(Fat Cantor set) Repeat the Cantor construction on $\intcc01$ , but at step $n$ remove from each of the $2^{n-1}$ intervals a centered open interval of length $4^{-n}$ only. Show that the resulting $K = \bigcap K_n$ is compact, has empty interior (no interval survives), and

\lambda(K) = 1 - \sum_{n\geq1}2^{n-1}4^{-n} = \tfrac12 :

a nowhere dense set of measure $\frac12$ . Deduce a meagre subset of $\intcc01$ of full measure $1$ , and an open dense subset of measure $< \varepsilon$ .

Solution

Solution of Exercise 9.5.

$K = \bigcap K_n$ is an intersection of finite unions of closed intervals: compact. At stage $n$ there remain $2^n$ intervals of common length $\ell_n \leq 2^{-n}$ (each stage halves and shrinks); an interval $I \subseteq K$ would lie inside a single stage- $n$ interval for every $n$ , forcing $\ell(I) = 0$ : empty interior. The measure removed is $\sum_{n\geq1}2^{n-1}\cdot 4^{-n} = \frac12\sum_{n\geq1}2^{-n} = \frac12$ , all removals being disjoint open intervals: $\lambda(K) = \frac12$ .

Variant: removing central intervals of length $\varepsilon 4^{-n}$ leaves a nowhere dense compact $K^{(\varepsilon)}$ of measure $1 - \frac\varepsilon2$ . Then $\bigcup_m K^{(1/m)}$ is meagre (countable union of nowhere dense sets) of measure $\geq \sup_m(1 - \frac1{2m}) = 1$ : a meagre set of full measure — and its complement in $\intoo01$ is a dense $G_\delta$ of measure $0$ (topologically fat, metrically null). The complement of $K^{(\varepsilon)}$ in $\intoo01$ is open, dense, of measure $\frac\varepsilon2 < \varepsilon$ .

Exercise 9.6 ★★

Let $\mu$ be a measure on $\mathcal B(\R)$ , invariant under translations, with $c = \mu(\intoc01) < \infty$ . Show $\mu = c\,\lambda$ on $\mathcal B(\R)$ . (Compute $\mu$ on dyadic intervals by dividing $\intoc01$ into $2^n$ translates, then invoke Theorem 9.7.)

Solution

Solution of Exercise 9.6.

Cutting $\intoc01$ into $2^n$ translates of $\intoc0{2^{-n}}$ : $c = 2^n\,\mu(\intoc0{2^{-n}})$ , so $\mu(\intoc0{2^{-n}}) = c\,2^{-n} = c\,\lambda(\intoc0{2^{-n}})$ . By translation invariance and additivity, $\mu = c\lambda$ on every interval $\intoc ab$ with $b - a$ a dyadic rational and any $a$ ; a general $\intoc ab$ is an increasing union of such ( $b_k \uparrow b$ dyadic steps from $a$ ), and continuity from below extends the equality. The intervals $\intoc ab$ form a $\pi$ -system generating $\mathcal B(\R)$ , with $\intoc{-k}k \uparrow \R$ of finite measure ( $\mu(\intoc{-k}k) = 2kc$ ): Theorem 9.7 gives $\mu = c\lambda$ on $\mathcal B(\R)$ .

Exercise 9.7 ★★

(Approximation) Let $A \in \mathcal L$ with $\lambda(A) < \infty$ and $\varepsilon > 0$ . Show there is a finite union of intervals $B$ with $\lambda(A\,\triangle\,B) < \varepsilon$ ( $\triangle$ = symmetric difference). (Regularity: squeeze $K \subseteq A \subseteq U$ and use the structure of the open $U$ as a countable union of intervals, plus compactness of $K$ .)

Solution

Solution of Exercise 9.7.

By regularity (Theorem 9.13) choose $K \subseteq A \subseteq U$ with $K$ compact, $U$ open, $\lambda(U\setminus K) < \varepsilon$ (both approximations within $\varepsilon/2$ , and $\lambda(U \setminus K) = \lambda(U\setminus A) + \lambda(A \setminus K)$ ). Write $U$ as a countable disjoint union of open intervals $(I_n)$ (the components of the open set); the compact $K$ is covered by finitely many, $K \subseteq B = I_1\cup\dots\cup I_N \subseteq U$ . Then $A \setminus B \subseteq A\setminus K \subseteq U\setminus K$ and $B \setminus A \subseteq U \setminus A \subseteq U\setminus K$ : $\lambda(A\,\triangle\,B) \leq 2\lambda(U\setminus K)$ — start from $\varepsilon/2$ to land below $\varepsilon$ .

Exercise 9.8 ★★★

(Steinhaus) Let $A \in \mathcal L$ with $\lambda(A) > 0$ . Show that $A - A = \{x - y : x, y \in A\}$ contains an interval around $0$ . (Reduce to $\lambda(A) < \infty$ ; by Exercise 9.7-style regularity, find an interval $I$ with $\lambda(A \cap I) > \frac34\ell(I)$ ; then for $\abs t < \frac12\ell(I)$ , the sets $A\cap I$ and $(A\cap I) + t$ both sit in an interval of length $\frac32\ell(I)$ and have total measure $> \frac32\ell(I)$ : they must intersect.)

Solution

Solution of Exercise 9.8.

Replacing $A$ by $A \cap [-M, M]$ of positive measure (some $M$ works, by continuity from below), assume $0 < \lambda(A) < \infty$ . Take $U \supseteq A$ open with $\lambda(U) < \frac43\lambda(A)$ and decompose $U = \bigsqcup_nI_n$ into disjoint open intervals: $\lambda(A) = \sum_n\lambda(A\cap I_n)$ . If every $n$ had $\lambda(A\cap I_n) \leq \frac34\ell(I_n)$ , summing would give $\lambda(A) \leq \frac34\lambda(U) < \lambda(A)$ : some interval $I$ satisfies $\lambda(A\cap I) > \frac34\ell(I)$ . Set $B = A \cap I$ and let $\abs t < \frac12\ell(I)$ : both $B$ and $B + t$ lie in the interval $I \cup (I + t)$ , of length $< \frac32\ell(I)$ . If they were disjoint: $\lambda(B) + \lambda(B + t) = 2\lambda(B) > \frac32\ell(I)$ would exceed the containing interval’s measure — impossible. So $B \cap (B + t) \neq \varnothing$ : some $x \in B$ writes $x = y + t$ with $y \in B$ , and $t = x - y \in A - A$ . Hence $\intoo{-\frac{\ell(I)}2}{\frac{\ell(I)}2} \subseteq A - A$ .

Exercise 9.9 ★★★

Show that every $A \in \mathcal L$ with $\lambda(A) > 0$ contains a non-measurable subset. (Intersect $A$ with the translates $V + q$ of Vitali’s set: if all $A \cap (V+q)$ were measurable, each would be null by the argument of Theorem 9.15 — Steinhaus (Exercise 9.8) helps: a measurable set of positive measure inside $V + q$ would give $(V+q) - (V+q) \supseteq$ an interval, contradicting that this difference set meets $\Q$ only at $0$ ; conclude with subadditivity.)

Solution

Solution of Exercise 9.9.

The Vitali translates $(V + q)_{q\in\Q}$ partition $\R$ (every real is equivalent to exactly one representative). Suppose all the sets $B_q = A \cap (V + q)$ were measurable. Any two elements of $V + q$ differ by an irrational or zero (two distinct representatives are inequivalent), so $B_q - B_q$ meets $\Q$ only in $\{0\}$ : it contains no interval, and Steinhaus (Exercise 9.8) forces $\lambda(B_q) = 0$ . Then $\lambda(A) \leq \sum_{q}\lambda(B_q) = 0$ , contradicting $\lambda(A) > 0$ . So some $B_q \subseteq A$ is non-measurable.

Exercise 9.10 ★★

Show that $A \subseteq \R$ with $\lambda^*(A) < \infty$ is Lebesgue-measurable iff for every $\varepsilon > 0$ there is an open $U \supseteq A$ with $\lambda^*(U \setminus A) < \varepsilon$ , iff there is a $G_\delta$ set $G \supseteq A$ with $\lambda^*(G\setminus A) = 0$ . (So Lebesgue sets are Borel sets modulo null sets.)

Solution

Solution of Exercise 9.10.

Measurable $\Rightarrow$ $\varepsilon$ -approximation: by outer regularity (Theorem 9.13) pick open $U \supseteq A$ with $\lambda(U) \leq \lambda(A) + \varepsilon$ ; measurability allows the subtraction $\lambda(U\setminus A) = \lambda(U) - \lambda(A) \leq \varepsilon$ . $\varepsilon$ -version $\Rightarrow$ $G_\delta$ -version: take $U_n$ with $\lambda^*(U_n\setminus A) < \frac1n$ and $G = \bigcap U_n$ : a $G_\delta$ with $\lambda^*(G\setminus A) \leq \lambda^*(U_n\setminus A) \to 0$ . $G_\delta$ -version $\Rightarrow$ measurable: $G\setminus A$ is $\lambda^*$ -null, hence measurable by completeness (Theorem 9.9), and $A = G \setminus (G\setminus A)$ is measurable ( $G$ is Borel). So Lebesgue sets are exactly “Borel modulo null”.

Exercise 9.11 ★★

(Continuity along monotone limits, and its sharpness) (a) Show that for measurable sets, $\mu(\liminf A_n) \leq \liminf\mu(A_n)$ (Fatou for sets), and that if $\mu\bigl(\bigcup A_n\bigr) < \infty$ , also $\limsup\mu(A_n) \leq \mu(\limsup A_n)$ . (b) Exhibit, for Lebesgue measure on $\R$ , a sequence with $\mu(A_n) = 1$ for all $n$ yet $\mu(\limsup A_n) = 0$ : the finiteness hypothesis in the second inequality is not decorative. (c) Deduce: if $\sum\mu(A_n) < \infty$ then $\mu(\limsup A_n) = 0$ (Borel–Cantelli again), and if the $A_n$ increase or decrease (with $\mu(A_1) < \infty$ in the decreasing case), $\mu(\lim A_n) = \lim\mu(A_n)$ .

Solution

Solution of Exercise 9.11.

(a) $\liminf A_n = \bigcup_N\bigcap_{n\geq N}A_n$ is an increasing union of the sets $B_N = \bigcap_{n\geq N}A_n$ , so $\mu(\liminf A_n) = \lim_N\mu(B_N)$ (continuity from below); and $\mu(B_N) \leq \inf_{n \geq N}\mu(A_n)$ , whose limit is $\liminf\mu(A_n)$ . For the $\limsup$ : apply the same to the complements inside the finite-measure ambient $U = \bigcup A_n$ — continuity from above on the decreasing $C_N = \bigcup_{n\geq N}A_n$ requires $\mu(C_1) \leq \mu(U) < \infty$ , and gives $\mu(\limsup A_n) = \lim\mu(C_N) \geq \limsup\mu(A_n)$ .

(b) The moving interval $A_n = \intcc n{n+1}$ : every point belongs to at most two of the $A_n$ and to none eventually, so $\limsup A_n = \varnothing$ ; yet $\mu(A_n) = 1$ . Thus $\limsup\mu(A_n) = 1 > 0 = \mu(\limsup A_n)$ : without a finite-measure envelope, the second inequality of (a) fails — the mass escapes to infinity, where no fixed set can catch it.

(c) If $\sum\mu(A_n) < \infty$ : $\mu(C_N) \leq \sum_{n\geq N}\mu(A_n) \to 0$ and $\limsup A_n = \bigcap C_N$ has measure $\leq \inf\mu(C_N) = 0$ . Monotone cases: increasing is continuity from below; decreasing with $\mu(A_1) < \infty$ is continuity from above — both proved in Chapter 9’s basic properties; the counterexample $A_n = \intco n\infty$ (decreasing to $\varnothing$ with $\mu \equiv \infty$ ) shows the finiteness is again essential.

Exercise 9.12 ★★★

(Egorov’s theorem) Let $\mu(X) < \infty$ and $f_n \to f$ pointwise, all measurable (real-valued). For $k, N \geq 1$ set

E_{k,N} = \bigcap_{n \geq N}\Bigl\{x : \abs{f_n(x) - f(x)} \leq \tfrac1k\Bigr\} .

(a) Show that for fixed $k$ , $E_{k,N} \nearrow X$ as $N \to \infty$ , and deduce $N_k$ with $\mu(X \setminus E_{k,N_k}) \leq \varepsilon2^{-k}$ . (b) Conclude Egorov’s theorem: for every $\varepsilon > 0$ there is a measurable $A$ with $\mu(X\setminus A) \leq \varepsilon$ such that $f_n \to f$ uniformly on $A$ — pointwise convergence is uniform convergence off an arbitrarily small set. (c) Show the theorem fails on $(\R, \lambda)$ : the moving bumps $f_n = \mathbf 1_{\intcc n{n+1}}$ converge pointwise to $0$ but uniformly on no complement of a finite-measure set. Where did (a) use $\mu(X) < \infty$ ?

Solution

Solution of Exercise 9.12.

(a) The sets $E_{k,N}$ increase with $N$ (fewer constraints), and every $x$ eventually satisfies $\abs{f_n(x) - f(x)} \leq \frac1k$ for all $n \geq N(x)$ (pointwise convergence): $\bigcup_NE_{k,N} = X$ . Continuity from below: $\mu(E_{k,N}) \to \mu(X) < \infty$ , so $\mu(X\setminus E_{k,N}) \to 0$ ; choose $N_k$ accordingly.

(b) Let $A = \bigcap_kE_{k,N_k}$ : $\mu(X\setminus A) \leq \sum_k\varepsilon2^{-k} = \varepsilon$ . On $A$ : for every $k$ , all $n \geq N_k$ satisfy $\sup_A\abs{f_n - f} \leq \frac1k$ — exactly uniform convergence on $A$ .

(c) For the moving bump, uniform convergence on $A$ requires $A$ to eventually avoid every $\intcc n{n+1}$ — more precisely $\sup_A\abs{f_n} < \frac12$ forces $A \cap \intcc n{n+1}$ to be empty for large $n$ , so $X \setminus A$ contains a tail $\bigcup_{n\geq n_0}\intcc n{n+1}$ , of infinite measure. In (a), finiteness converted “ $E_{k,N}\nearrow X$ ” into “the complements’ measures tend to $0$ ”: continuity from above needs a finite start, and on infinite-measure spaces the escape to infinity is precisely what it cannot see.

9.6 Problem: the Cantor–Vitali staircase and a measurable set that is not Borel

The Cantor–Vitali staircase: constant on every gap of the Cantor set, yet climbing from 0 to 1 continuously. Its derivative vanishes almost everywhere — all the climbing happens on a null set. — The Cantor–Vitali staircase: constant on every gap of the Cantor set, yet climbing from $0$ to $1$ continuously. Its derivative vanishes almost everywhere — all the climbing happens on a null set.

Problem 9.1

Weekend problem — the devil’s staircase, and $\mathcal B(\R) \subsetneq \mathcal L$

We construct the Cantor–Vitali function (devil’s staircase), use it to transport measure pathologically, and conclude with a theorem that no soft argument gives: there exist Lebesgue-measurable sets that are not Borel. Notation: $C$ is the Cantor set, $C_n$ its $n$ -th stage ( $2^n$ intervals of length $3^{-n}$ ), and every $x \in C$ has ternary digits $x = \sum 2b_n3^{-n}$ , $b_n \in \{0,1\}$ (Exercise 6.10).

Part I — The staircase. Define $c_0(x) = x$ and $c_{n+1}$ from $c_n$ by

c_{n+1}(x) = \begin{cases} \tfrac12\,c_n(3x) & 0 \leq x \leq \tfrac13,\\[2pt] \tfrac12 & \tfrac13 \leq x \leq \tfrac23,\\[2pt] \tfrac12 + \tfrac12\,c_n(3x - 2) & \tfrac23 \leq x \leq 1. \end{cases}

Show that each $c_n$ is continuous, nondecreasing, with $c_n(0) = 0$ , $c_n(1) = 1$ , and that $\norm{c_{n+1} - c_n}_\infty \leq \tfrac12\norm{c_n - c_{n-1}}_\infty$ .
Deduce that $(c_n)$ converges uniformly to a continuous nondecreasing $c$ with $c(0) = 0$ , $c(1) = 1$ (the Cantor–Vitali function), which satisfies the same self-similar relations as the $c_{n+1}$ above.
Show that $c$ is constant on each connected component of $\intcc01\setminus C$ , and that for $x = \sum_n 2b_n3^{-n} \in C$ , $c(x) = \sum_n b_n2^{-n}$ : the staircase reads Cantor digits in binary (Problem 6.1’s function $g$ , made monotone and global).
Deduce that $c$ is differentiable, with $c' = 0$ , at every point of $\intcc01\setminus C$ : $c' = 0$ $\lambda$ -almost everywhere (Example 9.14). Conclude that the fundamental theorem of calculus fails for $c$ :

$c(1) - c(0) = 1 \neq 0 = \int_0^1 c'(t)\,\dd t$

(the integral being over the full-measure set where $c' = 0$ ; anticipating Chapter 10, null sets do not affect integrals). Which hypothesis of the $\mathcal C^1$ fundamental theorem is violated?
Show that $c(C) = \intcc01$ : the null set $C$ is mapped onto a set of full measure.

Part II — The crooked homeomorphism. Let $h(x) = \frac{c(x) + x}{2}$ .

Show that $h \colon \intcc01 \to \intcc01$ is a homeomorphism (strictly increasing, continuous, surjective).
Show that $\lambda\bigl(h(\intcc01\setminus C)\bigr) = \tfrac12$ : on each gap of length $\ell$ , $h$ acts as an affine map of slope $\tfrac12$ , and the gaps have total length $1$ .
Deduce $\lambda\bigl(h(C)\bigr) = \tfrac12$ : the homeomorphic image of a null set can have positive measure. (Where does this contradict naive intuition about “size”?)

Part III — A measurable set that is not Borel.

By Exercise 9.9, choose a non-measurable $W \subseteq h(C)$ . Show that $Z = h^{-1}(W) \subseteq C$ is Lebesgue-measurable. (It is a subset of a null set; completeness, Theorem 9.9.)
Show that the preimage of a Borel set under a continuous map is Borel. (Good sets principle: $\{B : h^{-1}(B) \in \mathcal B\}$ is a $\sigma$ -algebra containing the open sets — mind the direction of the map.)
Conclude that $Z$ is not Borel: if it were, $W = (h^{-1})^{-1}(Z)$ would be Borel (apply question 10 to the continuous $h^{-1}$ ), hence measurable — contradiction. Therefore

$\boxed{\ \mathcal B(\R) \subsetneq \mathcal L\ }$

and completeness genuinely enlarges the Borel world.
Exhibit a Lebesgue-measurable function $g$ and a continuous function $\varphi$ such that $g \circ \varphi$ is not Lebesgue-measurable: measurability, unlike continuity, does not compose. (Take $g = \mathbf 1_Z$ and $\varphi = h^{-1}$ , anticipating the definition of measurable functions from Chapter 10: preimages of Borel sets are Lebesgue sets. Where must one be careful about which $\sigma$ -algebra is used on the target?)

Part IV — Epilogue.

Sort the following classes by strict inclusion and justify each strictness with an example from this chapter and its problem: countable sets; Borel null sets; Lebesgue null sets; Borel sets; Lebesgue sets; arbitrary sets.

Part V — The Cantor measure: mass on a null set. The staircase is the distribution function of a remarkable measure, which we now build with this chapter’s own tools.

(Lebesgue–Stieltjes, existence) Let $F\colon\R\to\R$ be nondecreasing, continuous, bounded. On half-open intervals define $\rho\bigl(\intoc ab\bigr) = F(b) - F(a)$ and, for $A \subseteq \R$ ,

$\mu_F^*(A) = \inf\Bigl\{\sum_k\bigl(F(b_k) - F(a_k)\bigr) : A \subseteq \bigcup_k\intoc{a_k}{b_k}\Bigr\} .$

Show that $\mu_F^*$ is an outer measure and that $\mu_F^*\bigl(\intoc ab\bigr) = F(b) - F(a)$ (imitate the compactness argument of Theorem 9.12, enlarging each $\intoc{a_k}{b_k}$ to an open interval at $F$ -cost $\leq \varepsilon2^{-k}$ — where is continuity of $F$ used?).
Show that every Borel set is $\mu_F^*$ -measurable in the Carathéodory sense (as in the Lebesgue case, it suffices to test half-lines; follow the proof of Theorem 9.9’s application), so that $\mu_F = \mu_F^*$ restricted to $\mathcal B(\R)$ is a measure with $\mu_F(\intoc ab) = F(b) - F(a)$ : the Lebesgue–Stieltjes measure of $F$ .
Apply this to the staircase ( $F = c$ extended by $0$ on $\R_-$ and $1$ on $\intco1\infty$ ): the Cantor measure $\mu$ . Show $\mu(\R) = 1$ , that every gap of the Cantor set is $\mu$ -null ( $c$ is constant there), and conclude

$\mu(C) = 1, \qquad \lambda(C) = 0 :$

$\mu$ and $\lambda$ live on disjoint carriers ( $C$ and its complement). Two measures in this position are called mutually singular, written $\mu \perp \lambda$ .
Show that $\mu$ has no atoms: $\mu(\{x\}) = 0$ for every $x$ (continuity of $c$ ). An atomless probability measure carried by a Lebesgue-null compact: compare with the only measures seen so far.
(Coin tossing in disguise) For a word $(\varepsilon_1, \dots, \varepsilon_m) \in \{0,1\}^m$ , let $C_{\varepsilon}$ be the set of $x \in C$ whose ternary digits satisfy $b_i(x) = \varepsilon_i$ for $i \leq m$ (one of the $2^m$ Cantor pieces of depth $m$ ). Show $\mu(C_\varepsilon) = 2^{-m}$ (the staircase climbs $2^{-m}$ across that piece: use Part I, question 3). The Cantor measure is the law of an infinite sequence of fair coin flips read in ternary — Chapter 22 will make this exact.
Prove the self-similarity: for every Borel $A$ ,

$\mu(A) = \tfrac12\,\mu(3A) + \tfrac12\,\mu(3A - 2) ,$

where $3A - 2 = \{3x - 2 : x \in A\}$ (check it on the generating intervals $\intoc ab$ via the self-similar relations of $c$ , then invoke uniqueness, Theorem 9.7).
Show that the reflection $s(x) = 1 - x$ preserves $\mu$ : $\mu(s(A)) = \mu(A)$ (via $c(1 - x) = 1 - c(x)$ , which follows from the symmetry of the construction — prove it).
Compute the first two moments of $\mu$ , i.e. of a random point $X$ with law $\mu$ (the integrals may be handled as limits of sums over the depth- $m$ pieces, anticipating Chapter 10): symmetry gives $\int x\,\dd\mu = \frac12$ , and self-similarity gives

$\int x^2\,\dd\mu = \frac38, \qquad\text{hence}\qquad \operatorname{Var}(X) = \frac18 .$

Compare with the uniform law on $\intcc01$ (variance $\frac1{12}$ ): the Cantor mass, pushed to the edges, spreads more.
Show that the (topological) support of $\mu$ — the smallest closed set of full measure — is exactly $C$ .
(Synthesis) The staircase $c$ is continuous and nondecreasing yet fails the fundamental theorem of calculus (Part I); the measure $\mu_c$ is a probability, atomless, singular with respect to $\lambda$ . Explain in a short paragraph how these are two faces of one phenomenon, and state the general moral: nondecreasing functions correspond to measures ( $F \leftrightarrow \mu_F$ ), differentiability a.e. corresponds to the “absolutely continuous part”, and $c$ is the standard witness that a continuous $F$ can carry no absolutely continuous part at all.
(The sharp modulus of continuity) Let $s = \frac{\ln 2}{\ln 3}$ . Show that $c$ is Hölder continuous of exponent $s$ :

$\abs{c(x) - c(y)} \leq 4\,\abs{x - y}^{s} \qquad (x, y \in \intcc01),$

and that no exponent $t > s$ can work, even locally. Deduce the measure-theoretic form: for every $x$ and every $r \in \intoc01$ ,

$\mu\bigl(\intcc{x - r}{x + r}\bigr) \leq 8\,r^{s} .$

(Compare a depth- $m$ triadic grid with the scale of $\abs{x - y}$ ; question 18 gives the climb across each piece. The exponent $s$ is the Hausdorff dimension of $C$ , as later courses will say.)
(Self-similarity characterizes $\mu$ ) Prove the converse of question 19: if $\nu$ is a probability measure on $\mathcal B(\R)$ carried by $\intcc01$ and satisfying

$\nu(A) = \tfrac12\,\nu(3A) + \tfrac12\,\nu(3A - 2) \qquad (A \in \mathcal B(\R)),$

then $\nu = \mu$ . (Iterate the relation $m$ times to spread $\nu$ over the $2^m$ depth- $m$ Cantor pieces, estimate $\nu(\intoc ab)$ against the count of pieces inside $\intoc ab$ , and let $m \to \infty$ ; finish with Theorem 9.7.)

Solution

Solution of Problem 9.1.

1. Induction. Continuity: the three formulas agree at the junctions ( $\frac12c_n(1) = \frac12$ and $\frac12 + \frac12c_n(0) = \frac12$ ); each piece is continuous. Monotonicity and the boundary values are inherited. For the contraction estimate: on $[0,\frac13]$ , $\abs{c_{n+1} - c_n}(x) = \frac12\abs{c_n - c_{n-1}}(3x) \leq \frac12\norm{c_n - c_{n-1}}_\infty$ ; on the middle third the difference is $0$ ; on the right third, the same as the left.

2. $\norm{c_{n+1} - c_n}_\infty \leq 2^{-n}\norm{c_1 - c_0}_\infty$ : the series of increments converges uniformly, so $c_n \to c$ uniformly; $c$ is continuous, nondecreasing, $c(0) = 0$ , $c(1) = 1$ (all preserved by uniform limits), and passing to the limit in the defining recursion shows $c$ itself satisfies the three self-similar identities.

3. By the middle identity, $c \equiv \frac12$ on $\intcc{\frac13}{\frac23}$ , the first gap. Every gap of $C$ is the image of the first gap under a composition of the two affine contractions $x \mapsto \frac x3$ , $x\mapsto\frac{x + 2}3$ ; the identities transport constancy accordingly (with values the dyadic rationals). For the digit formula, take $x = \sum_n 2b_n3^{-n} \in C$ : if $b_1 = 0$ then $x \in [0,\frac13]$ and $c(x) = \frac12c(3x)$ with $3x$ having digits $(b_2, b_3, \dots)$ ; if $b_1 = 1$ then $x \in [\frac23, 1]$ and $c(x) = \frac12 + \frac12c(3x - 2)$ , same shift. By induction, the first $N$ binary digits of $c(x)$ are $b_1, \dots, b_N$ for every $N$ : $c(x) = \sum_nb_n2^{-n}$ .

4. Off $C$ , $c$ is locally constant: differentiable with derivative $0$ . Since $\lambda(C) = 0$ (Example 9.14), $c' = 0$ almost everywhere. Yet $c(1) - c(0) = 1$ : the fundamental theorem in its $\mathcal C^1$ form requires $c$ to be differentiable everywhere with continuous (or at least integrable, plus absolute continuity — see Chapter 10) derivative; $c$ is not differentiable at points of $C$ , and more fundamentally $c$ fails absolute continuity: it climbs on a null set.

5. Given $y = \sum_n\beta_n2^{-n} \in \intcc01$ ( $\beta_n \in \{0,1\}$ ), the point $x = \sum_n2\beta_n3^{-n} \in C$ has $c(x) = y$ by question 3: $c(C) = \intcc01$ , a set of measure $1$ — the null set $C$ carries, through $c$ , the whole interval.

6. $h$ is continuous, and strictly increasing ( $x$ is, $c$ is nondecreasing); $h(0) = 0$ , $h(1) = 1$ , so by the intermediate value theorem $h$ is a continuous bijection of $\intcc01$ ; a continuous bijection from a compact space to a Hausdorff one is a homeomorphism (Corollary 6.14).

7. On a gap $(u, v)$ (length $\ell$ ), $c$ is constant, so $h$ is affine with slope $\frac12$ : $h((u,v))$ is an interval of length $\frac\ell2$ . The gaps are disjoint and $h$ is injective: the images are disjoint, of total measure $\frac12\sum\ell = \frac12(1 - \lambda(C)) = \frac12$ .

8. $h(\intcc01) = \intcc01$ and $h(C)$ is compact (continuous image), hence measurable, with

\lambda\bigl(h(C)\bigr) = 1 - \lambda\bigl(h(\intcc01\setminus C)\bigr) = 1 - \tfrac12 = \tfrac12 .

A homeomorphism can inflate a null set to measure $\frac12$ : “topological size” (category, dimension) and “measure” are transported by homeomorphisms very differently — only the former is a topological invariant.

9. $Z = h^{-1}(W) \subseteq h^{-1}(h(C)) = C$ has $\lambda^*(Z) \leq \lambda(C) = 0$ : a null set, hence Lebesgue-measurable by completeness (Theorem 9.9).

10. Let $\varphi$ be continuous and $\mathcal D = \{B : \varphi^{-1}(B) \in \mathcal B\}$ . Preimages commute with complements and countable unions, so $\mathcal D$ is a $\sigma$ -algebra; it contains the open sets (continuity): $\mathcal D \supseteq \mathcal B$ — preimages of Borel sets under continuous maps are Borel.

11. If $Z$ were Borel, apply question 10 to the continuous $\varphi = h^{-1}$ : $\varphi^{-1}(Z) = h(Z) = W$ would be Borel, hence Lebesgue-measurable — contradicting the choice of $W$ . So $Z \in \mathcal L \setminus \mathcal B$ : Lebesgue’s $\sigma$ -algebra strictly contains Borel’s.

12. $g = \mathbf 1_Z$ is Lebesgue-measurable ( $Z \in \mathcal L$ ) and $\varphi = h^{-1}$ is continuous, but $(g\circ\varphi)^{-1}(\{1\}) = \varphi^{-1}(Z) = W$ is not measurable: $g \circ \varphi$ is not Lebesgue-measurable. The care needed: “Lebesgue-measurable function” means preimages of Borel sets land in $\mathcal L$ ; composing requires preimages of Lebesgue sets to be Lebesgue, which continuity does not grant (here $\varphi^{-1}(Z) \notin \mathcal L$ even though $\varphi$ is a homeomorphism).

13. The chains, with witnesses for strictness:

\{\text{countable}\} \subsetneq \{\text{Borel null}\} \subsetneq \{\text{Lebesgue null}\} \subsetneq \mathcal L \subsetneq \mathcal P(\R), \qquad \{\text{Borel null}\} \subsetneq \mathcal B \subsetneq \mathcal L .

Witnesses: $C$ is Borel, null, uncountable (first gap); $Z$ is Lebesgue null but not Borel (second and, inside $\mathcal B \subsetneq \mathcal L$ , sixth); a fat Cantor set is Borel, nowhere dense, of positive measure (separating null sets from Borel sets); Vitali’s $V$ is not in $\mathcal L$ (last gap). Measure, topology and cardinality slice $\mathcal P(\R)$ along genuinely different lines.

14. Outer measure: $\mu_F^*(\varnothing) = 0$ (cover by a vanishing interval), monotonicity is clear, and countable subadditivity follows by splicing $\varepsilon2^{-k}$ -optimal covers, exactly as for $\lambda^*$ . The one-interval cover gives $\mu_F^*(\intoc ab) \leq F(b) - F(a)$ . Conversely let $\intoc ab \subseteq \bigcup_k\intoc{a_k}{b_k}$ . By continuity of $F$ , pick $b_k' > b_k$ with $F(b_k') \leq F(b_k) + \varepsilon2^{-k}$ , and $a' \in \intoo ab$ with $F(a') \leq F(a) + \varepsilon$ . The compact $\intcc{a'}b$ is covered by the open $\intoo{a_k}{b_k'}$ : finitely many suffice, and the chaining argument of Theorem 9.12 (walk from $a'$ to $b$ through overlapping intervals, telescoping $F$ -increments, monotonicity absorbing overlaps) yields $F(b) - F(a') \leq \sum_k(F(b_k') - F(a_k)) \leq \sum_k(F(b_k) - F(a_k)) + \varepsilon$ . Let $\varepsilon \to 0$ : equality. (Continuity of $F$ is what allowed opening the intervals at arbitrarily small $F$ -cost.)

15. It suffices to prove each half-line $H_t = \intoc{-\infty}t$ Carathéodory-measurable, since the measurable sets form a $\sigma$ -algebra (Theorem 9.9) and half-lines generate $\mathcal B$ . Given $A$ and an $\varepsilon$ -optimal cover $(\intoc{a_k}{b_k})$ of $A$ : each interval splits as $\intoc{a_k}{t\wedge b_k} \cup \intoc{t \vee a_k}{b_k}$ (one piece possibly empty), with $F$ -costs adding up exactly to $F(b_k) - F(a_k)$ ; the first pieces cover $A \cap H_t$ , the second $A \setminus H_t$ . Hence $\mu_F^*(A\cap H_t) + \mu_F^*(A\setminus H_t) \leq \mu_F^*(A) + \varepsilon$ , and the reverse inequality is subadditivity. Restricting the resulting measure to $\mathcal B$ : the Lebesgue–Stieltjes measure $\mu_F$ .

16. $\mu(\R) = \lim_n(F(n) - F(-n)) = 1 - 0 = 1$ (continuity of the measure along $\intoc{-n}n$ ). On a gap $\intoo uv$ of $C$ , $c$ is constant, so every half-open subinterval is $\mu$ -null and so is the gap (countable union); outside $\intcc01$ , $c$ is constant too. Hence $\mu(\R\setminus C) = 0$ , $\mu(C) = 1$ , while $\lambda(C) = 0$ (Example 9.14): each of $\mu, \lambda$ is carried by a set the other declares null — mutually singular.

17. $\mu(\{x\}) = \lim_{\delta\downarrow0} \mu(\intoc{x-\delta}x) = \lim(c(x) - c(x-\delta)) = 0$ by continuity of $c$ : no atoms. So $\mu$ is an atomless probability measure carried by a Lebesgue-null compact — neither diffuse-with-density like $\lambda$ ’s restrictions, nor atomic like counting measures: a third species.

18. The piece $C_\varepsilon$ spans a ternary interval $I_\varepsilon$ of length $3^{-m}$ , and Part I question 3 shows that across $I_\varepsilon$ the staircase climbs exactly $2^{-m}$ (the first $m$ binary digits of $c$ are frozen to $\varepsilon$ , the rest sweep everything). Hence $\mu(C_\varepsilon) = \mu(I_\varepsilon) = c(\text{right end}) - c(\text{left end}) = 2^{-m}$ : depth- $m$ digit cylinders all have mass $2^{-m}$ , the law of $m$ fair coins.

19. The right-hand side defines the Borel measure $\nu = \frac12\,(x \mapsto \tfrac x3)_*\mu + \frac12\,(x \mapsto \tfrac{x+2}3)_*\mu$ evaluated at $A$ — probability measure. Pointwise, with $c$ globally extended, one checks case by case ( $x \leq 0$ ; the three thirds; $x \geq 1$ ) the identity

c(x) = \tfrac12\,c(3x) + \tfrac12\,c(3x - 2),

e.g. on $\intcc{1/3}{2/3}$ : $\frac12\cdot1 + \frac12\cdot0 = \frac12 = c(x)$ . Evaluating $\nu$ on $\intoc ab$ therefore gives $c(b) - c(a) = \mu(\intoc ab)$ , and two finite measures agreeing on the $\pi$ -system of half-open intervals coincide on $\mathcal B$ (Theorem 9.7): $\nu = \mu$ .

20. By induction on $n$ : $c_0(1-x) = 1 - c_0(x)$ , and if $c_n$ has the symmetry, then for $x \in \intcc0{1/3}$ : $c_{n+1}(1 - x) = \frac12 + \frac12c_n(3(1-x) - 2) = \frac12 + \frac12c_n(1 - 3x) = \frac12 + \frac12(1 - c_n(3x)) = 1 - c_{n+1}(x)$ ; the middle third mirrors around $\frac12$ ; the right third is the left case reflected. In the limit $c(1-x) = 1 - c(x)$ . Pushforward: $(s_*\mu)(\intoc ab) = \mu(\intco{1-b}{1-a}) = c(1-a) - c(1-b)$ ( $\mu$ atomless, question 17, so boundary conventions cost nothing) $= (1 - c(a)) - (1 - c(b)) = \mu(\intoc ab)$ : $s_*\mu = \mu$ by uniqueness.

21. Let $X \sim \mu$ (integrals of continuous functions against $\mu$ exist as limits of Riemann-type sums over the depth- $m$ pieces, each of mass $2^{-m}$ , with sampling error $\leq \operatorname{osc} \leq \norm{f'}_\infty3^{-m}$ ; Chapter 10 will systematize this). Symmetry: $1 - X \sim X$ , so $\E X = \frac12$ . Self-similarity: $X$ has the law of $\frac Y3$ with probability $\frac12$ and of $\frac{Y+2}3$ with probability $\frac12$ , $Y \sim \mu$ , so

\E X^2 = \frac12\,\frac{\E Y^2}9 + \frac12\, \frac{\E Y^2 + 4\E Y + 4}{9} = \frac{2\E X^2 + 6}{18},

whence $\E X^2 = \frac38$ and $\operatorname{Var}X = \frac38 - \frac14 = \frac18$ . The uniform law has variance $\frac1{12} < \frac18$ : Cantor mass hugs the endpoints.

22. $C$ is closed and $\mu(C) = 1$ . If an open $I$ meets $C$ at $x$ , the depth- $m$ pieces containing $x$ shrink to $x$ , so some $I_\varepsilon \subseteq I$ and $\mu(I) \geq 2^{-m} > 0$ : no smaller closed set can carry $\mu$ . Points off $C$ have gap neighborhoods of $\mu$ -measure $0$ . Hence $\operatorname{supp}\mu = C$ exactly.

23. One phenomenon, two dialects. Part I says the growth of $c$ is invisible to its derivative: $c' = 0$ a.e., all the climbing concentrated on the null set $C$ . Part V says the associated measure $\mu_c$ puts all its mass on that same null set: $\mu_c \perp \lambda$ , so no density $f \geq 0$ can satisfy $\mu_c(A) = \int_Af\,\dd\lambda$ — a density forces vanishing on $\lambda$ -null sets. The dictionary: nondecreasing bounded $F \leftrightarrow$ finite measure $\mu_F$ (questions 14–15); $F$ an integral of its derivative $\leftrightarrow$ $\mu_F$ has a density (the “absolutely continuous” case); and in general $F'$ , which exists a.e. for monotone $F$ (Lebesgue’s differentiation theorem, beyond this chapter), recovers only the density part. The staircase is the extreme: continuous, with derivative $0$ a.e. — its measure is purely singular, and the fundamental theorem of calculus, far from failing by accident, fails by the exact amount $\mu_c(\R) = 1$ of singular mass.

24. Fix $x < y$ in $\intcc01$ and choose $m \geq 0$ with $3^{-(m+1)} < y - x \leq 3^{-m}$ . Across any triadic interval $\intcc{k3^{-m}}{(k+1)3^{-m}}$ the staircase climbs at most $2^{-m}$ : such an interval is either one of the $2^m$ pieces of $C_m$ , where the climb is exactly $2^{-m}$ (question 18), or it is contained in the closure of a single gap of some stage $\leq m$ , where $c$ is constant. Since $y - x \leq 3^{-m}$ , the interval $\intcc xy$ meets at most two consecutive depth- $m$ triadic intervals, so

c(y) - c(x) \leq 2\cdot2^{-m} = 2\bigl(3^{-m}\bigr)^{s} < 2\bigl(3(y - x)\bigr)^{s} = 4\,(y - x)^{s},

using $3^{s} = 2$ and $3^{-m} < 3(y - x)$ . Optimality: the endpoints $u < v$ of a depth- $m$ Cantor piece satisfy $v - u = 3^{-m}$ and $c(v) - c(u) = 2^{-m} = (v - u)^{s}$ ; a Hölder bound $\abs{c(v) - c(u)} \leq K(v - u)^{t}$ with $t > s$ would force $K \geq 2^{-m}3^{mt} = (3^{t}/2)^{m} \to \infty$ (since $3^{t} > 3^{s} = 2$ ), and every subinterval of $\intcc01$ contains such pieces, so the failure is local as well. Measure form: $\mu\bigl(\intcc{x-r}{x+r}\bigr) \leq c(x + r) - c(x - r) \leq 4(2r)^{s} = 4\cdot2^{s}r^{s} \leq 8r^{s}$ (values of $c$ extended to $\R$ as in question 16; $\mu$ has no atoms, question 17).

25. Write $S_0(x) = \frac x3$ and $S_1(x) = \frac{x+2}3$ ; the hypothesis says $\nu = \frac12(S_0)_*\nu + \frac12(S_1)_*\nu$ . Iterating $m$ times,

\nu = 2^{-m}\sum_{w \in \{0,1\}^m}(S_w)_*\nu, \qquad S_w = S_{w_1}\circ\dots\circ S_{w_m}.

Since $\nu$ is carried by $\intcc01$ and $S_w(\intcc01) = I_w$ , the depth- $m$ Cantor piece indexed by $w$ , each $(S_w)_*\nu$ is a probability measure carried by $I_w$ ; the $2^m$ pieces are pairwise disjoint closed intervals of length $3^{-m}$ . Fix $\intoc ab$ and let $N_m$ be the number of pieces $I_w \subseteq \intoc ab$ . A piece meeting $\intoc ab$ without being contained in it must contain $a$ or $b$ , and a point lies in at most one piece, so

2^{-m}N_m \leq \nu(\intoc ab) \leq 2^{-m}N_m + 2\cdot2^{-m}.

The same double inequality holds for $\mu$ (question 19 gives the identical iteration), with the same count $N_m$ . Hence $\abs{\nu(\intoc ab) - \mu(\intoc ab)} \leq 2^{-m+1} \to 0$ : $\nu$ and $\mu$ agree on the $\pi$ -system of half-open intervals, and both are probability measures, so Theorem 9.7 gives $\nu = \mu$ . The staircase measure is thus the fixed point of the two-map averaging scheme — the measure-level statement of the self-similarity of $C$ .

9.1 σ\sigmaσ-algebras

9.2 Measures

9.3 Outer measures and Carathéodory’s theorem

9.4 Lebesgue measure on R\RR

9.5 Exercises

9.6 Problem: the Cantor–Vitali staircase and a measurable set that is not Borel

9.1 $\sigma$ -algebras

9.4 Lebesgue measure on $\R$