---
title: "Measure Theory"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 9
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/9-measure-theory
---

# Chapter 9 — Measure 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](#def-b3-measure-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](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue)), the $L^p$ spaces of functional analysis, and the whole of modern probability ([Chapter 22](https://one-course.com/books/math/5/en/chapter/22-probability-foundations-and-the-law-of-large-numbers#ch-b3-probability)) are built on the three theorems proved here: Dynkin’s uniqueness lemma, Carathéodory’s extension theorem, and the existence of [Lebesgue measure](#def-b3-measure-lebesgueouter).

## 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) 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](#exo-b3-measure-1)) — each family generates the opens by countable operations, e.g. every [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) 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 π\piπ–λ\lambdaλ 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](#def-b3-measure-lebesgueouter).

**Proposition 9.6.**

Let $\mu$ be a [measure](#def-b3-measure-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from below: $A_n \uparrow A \Rightarrow \mu(A_n) \to \mu(A)$. (d) [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-lebesgueouter): $A_n \downarrow \varnothing$ but $\mu(A_n) =
\infty$. ∎

**Theorem 9.7 (Uniqueness).**

Let $\mu, \nu$ be [measures](#def-b3-measure-measure) 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](#def-b3-measure-measure) $\mu_k(A) = \mu(A \cap
P_k)$ 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from below ([Proposition 9.6](#prop-b3-measure-basics)(c)). It contains the $\pi$-system $\mathcal P$, so Dynkin ([Theorem 9.4](#thm-b3-measure-dynkin)) gives $\mathcal D \supseteq
\sigma(\mathcal P)$: $\mu_k = \nu_k$ everywhere. Finally, for any $A \in \sigma(\mathcal P)$, [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-measure). Moreover every set with $\mu^*(N) = 0$ belongs to $\mathcal M$ (the [measure](#def-b3-measure-measure) is *[complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-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](#def-b3-measure-outer), invariant under translations, and $\lambda^*(I) = \ell(I)$ (the length) for every interval $I$.

**Proof.** [Outer measure](#def-b3-measure-outer): $\varnothing$ 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](#def-b3-measure-outer) $0$: 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#thm-b3-topology-metriccompact)), 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](#def-b3-measure-measure) on $\mathcal B(\R)$ assigning to each interval its length; it is translation-invariant and $\sigma$-finite.

**Proof.** By [Theorem 9.9](#thm-b3-measure-caratheodory) it suffices to show each ray $A = \intoo{-\infty}c$ is $\lambda^*$-measurable (rays generate $\mathcal B$, [Definition 9.2](#def-b3-measure-borel)). 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](#def-b3-measure-measure) 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](#thm-b3-measure-uniqueness) 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $U \supseteq A$ 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from below) and apply the bounded case inside. ∎

**Example 9.14.**

The Cantor set ([Exercise 6.10](https://one-course.com/books/math/5/en/chapter/6-general-topology#exo-b3-topology-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](#def-b3-measure-measure). Conversely, *fat Cantor sets* ([Exercise 9.5](#exo-b3-measure-5)) are nowhere dense with positive [measure](#def-b3-measure-measure): [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) does not see [measure](#def-b3-measure-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](#def-b3-measure-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](#def-b3-measure-measure) equal: check them on a generating $\pi$-system plus $\sigma$-finiteness ([Theorem 9.7](#thm-b3-measure-uniqueness)). To build a [measure](#def-b3-measure-measure): build an [outer measure](#def-b3-measure-outer) 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 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology), 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](#def-b3-measure-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from above ([Proposition 9.6](#prop-b3-measure-basics)(d)) fails without the finiteness assumption. (c) Show that a countable set has [Lebesgue measure](#def-b3-measure-lebesgueouter) zero. Deduce $\lambda(\Q) = 0$ and $\lambda(\intcc01\setminus\Q) =
1$.

**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](#def-b3-measure-lebesgueouter), $A_n = [n, +\infty) \downarrow
\varnothing$, 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](#def-b3-measure-measure) 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](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) invariant — the foundation of [Chapter 22](https://one-course.com/books/math/5/en/chapter/22-probability-foundations-and-the-law-of-large-numbers#ch-b3-probability).)

**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](#exo-b3-measure-1)). The sets $P_k = \intoc{-\infty}k$ increase to $\R$ with $\mu(P_k) \leq 1 < \infty$: [Theorem 9.7](#thm-b3-measure-uniqueness) applies, and $\mu = \nu$ on $\mathcal B(\R)$. Thus the distribution function determines the [measure](#def-b3-measure-measure).

**Exercise 9.4 ★★.**

(Borel–Cantelli, [measure](#def-b3-measure-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 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), has empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) (no interval survives), and

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

a nowhere [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) of [measure](#def-b3-measure-measure) $\frac12$. Deduce a *[meagre](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#rem-b3-complete-meagre)* subset of $\intcc01$ of full [measure](#def-b3-measure-measure) $1$, and an open dense subset of [measure](#def-b3-measure-measure) $< \varepsilon$.

**Solution of Exercise 9.5.**

$K = \bigcap K_n$ is an intersection of finite unions of closed intervals: [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior). The [measure](#def-b3-measure-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K^{(\varepsilon)}$ of [measure](#def-b3-measure-measure) $1 - \frac\varepsilon2$. Then $\bigcup_m
K^{(1/m)}$ is [meagre](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#rem-b3-complete-meagre) (countable union of nowhere [dense sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior)) of [measure](#def-b3-measure-measure) $\geq \sup_m(1 - \frac1{2m}) = 1$: a [meagre set](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#rem-b3-complete-meagre) of full [measure](#def-b3-measure-measure) — and its complement in $\intoo01$ is a dense $G_\delta$ of [measure](#def-b3-measure-measure) $0$ ([topologically](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) fat, metrically null). The complement of $K^{(\varepsilon)}$ in $\intoo01$ is open, dense, of [measure](#def-b3-measure-measure) $\frac\varepsilon2 < \varepsilon$.

**Exercise 9.6 ★★.**

Let $\mu$ be a [measure](#def-b3-measure-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](#thm-b3-measure-uniqueness).)*

**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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-measure) ($\mu(\intoc{-k}k) = 2kc$): [Theorem 9.7](#thm-b3-measure-uniqueness) 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 of Exercise 9.7.**

By regularity ([Theorem 9.13](#thm-b3-measure-regularity)) choose $K
\subseteq A \subseteq U$ with $K$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology)); the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#exo-b3-measure-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](#def-b3-measure-measure) $> \frac32\ell(I)$: they must intersect.)*

**Solution of Exercise 9.8.**

Replacing $A$ by $A \cap [-M, M]$ of positive [measure](#def-b3-measure-measure) (some $M$ works, by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-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](#thm-b3-measure-vitali) — Steinhaus ([Exercise 9.8](#exo-b3-measure-8)) helps: a measurable set of positive [measure](#def-b3-measure-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 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](#exo-b3-measure-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 of Exercise 9.10.**

*Measurable $\Rightarrow$ $\varepsilon$-approximation*: by outer regularity ([Theorem 9.13](#thm-b3-measure-regularity)) 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](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) ([Theorem 9.9](#thm-b3-measure-caratheodory)), and $A = G \setminus
(G\setminus A)$ is measurable ($G$ is Borel). So Lebesgue sets are exactly “Borel modulo null”.

**Exercise 9.11 ★★.**

([Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-lebesgueouter) 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 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-measure) ambient $U = \bigcup A_n$ — [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-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](#def-b3-measure-measure) $\leq \inf\mu(C_N) = 0$. Monotone cases: increasing is [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from below; decreasing with $\mu(A_1) < \infty$ is [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from above — both proved in [Chapter 9](#ch-b3-measure)’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](#def-b3-measure-measure) set. Where did (a) use $\mu(X) < \infty$?

**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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-measure). In (a), finiteness converted “$E_{k,N}\nearrow X$” into “the complements’ [measures](#def-b3-measure-measure) tend to $0$”: [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from above needs a finite start, and on [infinite-measure](#def-b3-measure-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.](https://one-course.com/images/onecourse/chapters/math-5/b3-measure/fig-86d48a949602.svg)

*The Cantor–Vitali staircase: constant on every gap of the Cantor set, yet climbing from $0$ to $1$ [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). 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](#def-b3-measure-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#exo-b3-topology-10)).

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

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

1. Show that each $c_n$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) , nondecreasing, with $c_n(0) = 0$ , $c_n(1) = 1$ , and that $\norm{c_{n+1} - c_n}_\infty \leq  \tfrac12\norm{c_n - c_{n-1}}_\infty$ .
2. Deduce that $(c_n)$ converges uniformly to a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) 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.
3. Show that $c$ is constant on each [connected component](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-components) 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#pb-b3-topology-1) ’s function $g$ , made monotone and global).
4. Deduce that $c$ is differentiable, with $c' = 0$, at every point of $\intcc01\setminus C$: $c' = 0$ *$\lambda$-almost everywhere* ([Example 9.14](#ex-b3-measure-cantor)). 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](#def-b3-measure-measure) set where $c' = 0$; anticipating [Chapter 10](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue), null sets do not affect integrals). Which hypothesis of the $\mathcal C^1$ fundamental theorem is violated?
5. Show that $c(C) = \intcc01$ : the null set $C$ is mapped *onto* a set of full [measure](#def-b3-measure-measure) .

**Part II — The crooked [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity).** Let $h(x) = \frac{c(x) + x}{2}$.

6. Show that $h \colon \intcc01 \to \intcc01$ is a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (strictly increasing, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) , surjective).
7. 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 $1$ .
8. Deduce $\lambda\bigl(h(C)\bigr) = \tfrac12$ : the [homeomorphic](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) image of a null set can have positive [measure](#def-b3-measure-measure) . (Where does this contradict naive intuition about “size”?)

**Part III — A measurable set that is not Borel.**

9. By [Exercise 9.9](#exo-b3-measure-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](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete), [Theorem 9.9](#thm-b3-measure-caratheodory).)*
10. Show that the preimage of a Borel set under a [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) is Borel. *(Good sets principle: $\{B : h^{-1}(B) \in \mathcal B\}$ is a $\sigma$-algebra containing the [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) — mind the direction of the map.)*
11. Conclude that $Z$ is *not* Borel: if it were, $W = (h^{-1})^{-1}(Z)$ would be Borel (apply question 10 to the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $h^{-1}$), hence measurable — contradiction. Therefore $$\boxed{\ \mathcal B(\R) \subsetneq \mathcal L\ }$$ and [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) genuinely enlarges the Borel world.
12. Exhibit a Lebesgue-measurable function $g$ and a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) function $\varphi$ such that $g \circ  \varphi$ is not Lebesgue-measurable: measurability, unlike [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) , does not compose. *(Take $g =  \mathbf 1_Z$ and $\varphi = h^{-1}$, anticipating the definition of measurable functions from [Chapter 10](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue): preimages of Borel sets are Lebesgue sets. Where must one be careful about which $\sigma$-algebra is used on the target?)*

**Part IV — Epilogue.**

13. 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](#def-b3-measure-measure): mass on a null set.** The staircase is the distribution function of a remarkable [measure](#def-b3-measure-measure), which we now build with this chapter’s own tools.

14. (Lebesgue–Stieltjes, existence) Let $F\colon\R\to\R$ be nondecreasing, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), 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](#def-b3-measure-outer) and that $\mu_F^*\bigl(\intoc ab\bigr) = F(b) - F(a)$ *(imitate the compactness argument of [Theorem 9.12](#thm-b3-measure-lebesgue), enlarging each $\intoc{a_k}{b_k}$ to an open interval at $F$-cost $\leq \varepsilon2^{-k}$ — where is [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $F$ used?)*.
15. 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](#thm-b3-measure-caratheodory)’s application)* , so that $\mu_F = \mu_F^*$ restricted to $\mathcal B(\R)$ is a [measure](#def-b3-measure-measure) with $\mu_F(\intoc ab) = F(b) - F(a)$ : the *Lebesgue–Stieltjes [measure](#def-b3-measure-measure)* of $F$ .
16. Apply this to the staircase ($F = c$ extended by $0$ on $\R_-$ and $1$ on $\intco1\infty$): the *Cantor [measure](#def-b3-measure-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](#def-b3-measure-measure) in this position are called *mutually singular*, written $\mu \perp  \lambda$.
17. Show that $\mu$ has no atoms: $\mu(\{x\}) = 0$ for every $x$ *([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $c$)* . An atomless probability [measure](#def-b3-measure-measure) carried by a Lebesgue-null [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) : compare with the only [measures](#def-b3-measure-measure) seen so far.
18. (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](#def-b3-measure-measure) is the law of an infinite sequence of fair coin flips read in ternary — [Chapter 22](https://one-course.com/books/math/5/en/chapter/22-probability-foundations-and-the-law-of-large-numbers#ch-b3-probability) will make this exact.
19. 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](#thm-b3-measure-uniqueness))*.
20. 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)* .
21. 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](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue)): 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*.
22. Show that the ( [topological](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) ) *support* of $\mu$ — the smallest closed set of full [measure](#def-b3-measure-measure) — is exactly $C$ .
23. (Synthesis) The staircase $c$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and nondecreasing yet fails the fundamental theorem of calculus (Part I); the [measure](#def-b3-measure-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](#def-b3-measure-measure) ( $F \leftrightarrow \mu_F$ ), differentiability a.e. corresponds to the “absolutely [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) part”, and $c$ is the standard witness that a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $F$ can carry *no* absolutely [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) part at all.
24. (The sharp modulus of [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)) Let $s =  \frac{\ln 2}{\ln 3}$. Show that $c$ is Hölder [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-hausdorff) dimension of $C$, as later courses will say.)*
25. (Self-similarity characterizes $\mu$) Prove the converse of question 19: if $\nu$ is a probability [measure](#def-b3-measure-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](#thm-b3-measure-uniqueness).)*

**Solution of Problem 9.1.**

**1.** Induction. [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): the three formulas agree at the junctions ($\frac12c_n(1) = \frac12$ and $\frac12 +
\frac12c_n(0) = \frac12$); each piece is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), 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](#ex-b3-measure-cantor)), $c' = 0$ 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (or at least integrable, plus absolute [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) — see [Chapter 10](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue)) derivative; $c$ is not differentiable at points of $C$, and more fundamentally $c$ fails *absolute [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#def-b3-measure-measure) $1$ — the null set $C$ carries, through $c$, the whole interval.

**6.** $h$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and strictly increasing ($x$ is, $c$ is nondecreasing); $h(0) = 0$, $h(1) = 1$, so by the intermediate value theorem $h$ is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) bijection of $\intcc01$; a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) bijection from a [compact space](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) to a [Hausdorff](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-hausdorff) one is a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ([Corollary 6.14](https://one-course.com/books/math/5/en/chapter/6-general-topology#cor-b3-topology-compacthomeo)).

**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](#def-b3-measure-measure) $\frac12\sum\ell = \frac12(1 - \lambda(C)) = \frac12$.

**8.** $h(\intcc01) = \intcc01$ and $h(C)$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ([continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) image), hence measurable, with

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

A [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) can inflate a null set to [measure](#def-b3-measure-measure) $\frac12$: “[topological](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) size” (category, dimension) and “[measure](#def-b3-measure-measure)” are transported by [homeomorphisms](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) very differently — only the former is a [topological](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) 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](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) ([Theorem 9.9](#thm-b3-measure-caratheodory)).

**10.** Let $\varphi$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)): $\mathcal D \supseteq \mathcal B$ — preimages of Borel sets under [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) are Borel.

**11.** If $Z$ were Borel, apply question 10 to the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\varphi = h^{-1}$: $\varphi^{-1}(Z) = h(Z) = W$ 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) does not grant (here $\varphi^{-1}(Z) \notin
\mathcal L$ even though $\varphi$ is a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)).

**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](#def-b3-measure-measure) (separating null sets from Borel sets); Vitali’s $V$ is not in $\mathcal L$ (last gap). [Measure](#def-b3-measure-measure), [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) and cardinality slice $\mathcal P(\R)$ along genuinely different lines.

**14.** [Outer measure](#def-b3-measure-outer): $\mu_F^*(\varnothing) = 0$ (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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#thm-b3-measure-lebesgue) (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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-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](#thm-b3-measure-caratheodory)) 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](#def-b3-measure-measure) to $\mathcal B$: the Lebesgue–Stieltjes [measure](#def-b3-measure-measure) $\mu_F$.

**16.** $\mu(\R) = \lim_n(F(n) - F(-n)) = 1 - 0 = 1$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of the [measure](#def-b3-measure-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](#ex-b3-measure-cantor)): 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $c$: no atoms. So $\mu$ is an atomless probability [measure](#def-b3-measure-measure) carried by a Lebesgue-null [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — neither diffuse-with-density like $\lambda$’s restrictions, nor atomic like counting [measures](#def-b3-measure-measure): 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](#def-b3-measure-measure) $\nu = \frac12\,(x \mapsto \tfrac x3)_*\mu + \frac12\,(x
\mapsto \tfrac{x+2}3)_*\mu$ evaluated at $A$ — probability [measure](#def-b3-measure-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](#def-b3-measure-measure) agreeing on the $\pi$-system of half-open intervals coincide on $\mathcal B$ ([Theorem 9.7](#thm-b3-measure-uniqueness)): $\nu =
\mu$.

**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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) 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](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ch-b3-lebesgue) 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) 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](#def-b3-measure-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](#def-b3-measure-measure) $\mu_F$ (questions 14–15); $F$ an integral of its derivative $\leftrightarrow$ $\mu_F$ has a density (the “absolutely [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)” 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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), with derivative $0$ a.e. — its [measure](#def-b3-measure-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](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) 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](#def-b3-measure-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](#def-b3-measure-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](#def-b3-measure-measure), so [Theorem 9.7](#thm-b3-measure-uniqueness) gives $\nu = \mu$. The staircase [measure](#def-b3-measure-measure) is thus *the* fixed point of the two-map averaging scheme — the measure-level statement of the self-similarity of $C$.