Course 3 Book 3A: Measure, Integration and LᵖChapter 1

The Problem of Measure

What a size should be, why Jordan measure is not enough, and sets that cannot be measured.

26 min read · Updated Oct 2, 2026

Read with Tao, An Introduction to Measure Theory, §1.1 "Prologue: The problem of measure" (elementary measure, Jordan measure, connection with the Riemann integral). The book is freely available from the author's website with the publisher's permission. Stein and Shakarchi, Real Analysis, chapter 1, is an optional second voice.

In this chapter · 7 sections

1.1How long is a coastline?
1.2What a size should do
1.3Elementary measure
1.4Jordan measure
1.4.1Where Jordan measure stops
1.4.2Why countable additivity is the right axiom
1.5Sets that cannot be measured
1.6History
1.7Exercises

Book 2B ended with two complaints about the Riemann integral. It doesn't behave well under limits: a sequence of integrable functions can converge to one that has no integral (2A.11 The Riemann Integral), and limits can be passed through integrals only under uniform convergence (2B.5 Uniform Convergence and Arzelà–Ascoli), far too strong a hypothesis for PDE. And the space of Riemann-integrable functions with the distance $\int|f - g|$ is not complete (2B.2 Completeness and Contraction), so Cauchy sequences of functions, which is how solutions of PDE are built, may have nowhere to converge.

The cure, due to Henri Lebesgue in 1902, starts not with functions but with sets. An integral is a way of measuring the size of the region under a graph, and Riemann's integral fails because the notion of size behind it, Jordan measure, can only measure sets that are well approximated by finitely many boxes. This book builds a better notion of size, Lebesgue measure, then the integral that goes with it, and then the function spaces $L^p$ that the rest of the guidebook works in. Book 3A does for functions what 2A.4 The Real Numbers did for numbers: it fills the holes.

This first chapter asks what a notion of size should do, shows how far the elementary theory goes and exactly where it stops, and proves that no notion of size can measure every set. That last fact is not a curiosity. It is why measure theory has to talk about measurable sets at all.

By the end of this chapter you will be able to:

list the properties a notion of size should have, and explain which ones conflict;
compute elementary and Jordan measures, including the area of a polygon by the shoelace formula;
decide whether a bounded set is Jordan measurable, using its boundary;
explain why countable additivity is the right axiom, with examples that Jordan measure can't handle;
construct Vitali's non-measurable set and state the Banach–Tarski theorem accurately.

How long is a coastline?

In the world Data The coast of Britain

In the 1950s the physicist and meteorologist Lewis Fry Richardson, collecting data for a study of the causes of wars, noticed that published lengths of the same land borders disagreed, sometimes by large factors. He measured coastlines and frontiers on maps by walking a pair of dividers along them with a fixed opening $\ell$ , counting the steps. The total length $L(\ell)$ did not settle down as $\ell$ shrank. It kept growing, roughly like a power:

L(\ell) \approx C\,\ell^{\,1 - D},

with an exponent $D$ that depended on the coastline but not on the scale. Richardson's data were published posthumously in 1961. In a 1967 paper in Science, "How long is the coast of Britain?", Benoit Mandelbrot interpreted $D$ as a dimension, between $1$ (a smooth curve) and $2$ (a region), and reported $D \approx 1.25$ for the west coast of Britain from Richardson's measurements.

With $D = 1.25$ , halving the ruler multiplies the measured length by $2^{0.25} \approx 1.19$ . There is no limit: as $\ell \to 0$ the length tends to infinity, while the area enclosed stays perfectly finite. The coastline is too rough to have a length and too thin to have an area. Asking "how big is it?" is not a pedantic question here. It has no answer until we decide what kind of size we mean.

Figure 1.1. Measuring a jagged curve with shorter and shorter rulers (a computed Koch curve standing in for a coastline). Each time the ruler is divided by

3

, the measured length is multiplied by

\tfrac43

, so

L(\ell) \propto \ell^{1 - D}

with

D = \log 4/\log 3 \approx 1.26

, close to Richardson's value for the west coast of Britain. The length has no limit.

For curves like this, the right notion of size is $D$ -dimensional Hausdorff measure, which assigns a finite, non-zero size to the coastline when the exponent is chosen to be its dimension. This guidebook won't need Hausdorff measure in general. What it needs is the case of full dimension: a notion of $d$ -dimensional volume for subsets of $\mathbb{R}^d$ that agrees with length, area and volume on ordinary shapes and behaves well under limits. That is Lebesgue measure.

What a size should do

Let us write down, as Tao does, what we would like. A measure on $\mathbb{R}^d$ would be a function $m$ assigning to each subset $E \subseteq \mathbb{R}^d$ a size $m(E) \in [0, +\infty]$ such that:

(Normalisation) the unit cube has size $1$ : $m([0, 1]^d) = 1$ ;
(Translation invariance) $m(E + x) = m(E)$ for every $x \in \mathbb{R}^d$ ;
(Countable additivity) if $E_1, E_2, \ldots$ are disjoint, then $m\big(\bigcup_n E_n\big) = \sum_n m(E_n)$ .

Additivity says size can be computed piece by piece. Countable additivity says this survives infinitely many pieces, which is what makes a theory compatible with limits. Translation invariance says size doesn't depend on position. Normalisation fixes the units.

All three together are impossible, if $m$ must be defined on every subset (the Vitali construction at the end of this chapter). One of the wishes has to go. The standard choice, Lebesgue's, keeps all three properties but gives up measuring every set: $m$ is defined only on a large class of measurable sets, so large that every set that arises in practice belongs to it. To see why the class has to be so large, start from the small one.

Elementary measure

A box in $\mathbb{R}^d$ is a product of intervals, $B = I_1 \times \cdots \times I_d$ , each interval bounded and possibly open, closed or half-open. Its volume is $|B| = |I_1| \times \cdots \times |I_d|$ , where $|[a, b]| = b - a$ . An elementary set is a finite union of boxes. Every elementary set can be written as a finite union of disjoint boxes (cut along all the hyperplanes containing a face), and the elementary measure

m(E) = |B_1| + \cdots + |B_k| \qquad (E = B_1 \cup \cdots \cup B_k \text{ disjoint})

does not depend on which partition is used (Tao, Lemma 1.1.2; the proof refines two partitions to a common grid and counts). Elementary measure is finitely additive and translation invariant, and the unit cube has measure $1$ .

In the world In use The area of a land parcel

Geographic information systems store a land parcel as a polygon with vertices $(x_1, y_1), \ldots, (x_n, y_n)$ in order around the boundary, and compute its area by the shoelace formula

A = \tfrac12\Big|\sum_{i=1}^n\big(x_iy_{i+1} - x_{i+1}y_i\big)\Big| \qquad (x_{n+1}, y_{n+1}) = (x_1, y_1).

For the pentagon $(0, 0), (4, 0), (5, 3), (2, 5), (-1, 3)$ it gives $A = 21$ . The formula is finite additivity at work. Each term $\tfrac12(x_iy_{i+1} - x_{i+1}y_i)$ is the signed area of the triangle from the origin to one edge, and the signed areas of the triangles outside the polygon cancel (Exercise 1.6). A polygon is a finite union of triangles, and a triangle's area is a limit of elementary sets squeezing it from inside and outside: exactly the next idea, Jordan measure. It works perfectly for parcels with finitely many edges. It has nothing to say about a coastline.

Jordan measure

Most sets are not finite unions of boxes. A disc isn't. But a disc can be squeezed between elementary sets from inside and outside, and its size is whatever number the squeeze pins down.

Definition 1.1 Jordan measure

Let $E \subseteq \mathbb{R}^d$ be bounded. Its Jordan inner measure and Jordan outer measure are

m_{*,(J)}(E) = \sup_{A \subseteq E}m(A), \qquad m^{*,(J)}(E) = \inf_{B \supseteq E}m(B),

the supremum and infimum running over elementary sets $A$ and $B$ . If they are equal, $E$ is Jordan measurable, and the common value is its Jordan measure $m(E)$ .

Figure 1.2. Inner and outer approximations of the unit disc by grid squares of side

\tfrac18

(left) and

\tfrac1{32}

(right). The inner areas

2.56

and

3.01

and outer areas

3.50

and

3.25

squeeze

\pi = 3.14159\ldots

The disc is Jordan measurable because the squares straddling its boundary have total area tending to

0

The figure suggests the general criterion: the gap between outer and inner measure is the total area of the boxes that straddle the boundary.

Proposition 1.2 The boundary criterion

A bounded set $E \subseteq \mathbb{R}^d$ is Jordan measurable if and only if its boundary $\partial E$ has Jordan outer measure zero.

The proof is Tao's Exercise 1.1.18; Exercise 1.7 guides you through it. It gives a large supply of Jordan measurable sets: polygons and polyhedra, balls, regions bounded by finitely many graphs of continuous functions (a graph of a continuous function on a box has Jordan measure zero by uniform continuity, Exercise 1.8). And Jordan measure is the measure behind the Riemann integral. A bounded non-negative function $f$ on $[a, b]$ is Riemann integrable exactly when the region $\{(x, t) : 0 \leq t \leq f(x)\}$ under its graph is Jordan measurable in $\mathbb{R}^2$ , and then $\int_a^b f$ is that region's Jordan measure (Tao, Exercise 1.1.25). The upper and lower sums of 2A.11 The Riemann Integral are outer and inner elementary approximations.

Where Jordan measure stops

Now the sets Jordan can't measure.

Example 1.3 The rationals in

[0, 1]

Let $E = \mathbb{Q} \cap [0, 1]$ . An elementary set inside $E$ can contain no interval of positive length (every interval contains irrationals), so it is a finite set of points and degenerate boxes, of measure $0$ : $m_{*,(J)}(E) = 0$ . An elementary set containing $E$ is a finite union of intervals whose closure contains the closure of $E$ , which is $[0, 1]$ ; so it has measure at least $1$ : $m^{*,(J)}(E) = 1$ . So $E$ is not Jordan measurable. Its boundary is all of $[0, 1]$ , in line with the criterion. This is the set behind Dirichlet's function in 2A.11 The Riemann Integral, which had no Riemann integral for exactly this reason.

The rationals in $[0, 1]$ are countable, so we can list them, $q_1, q_2, q_3, \ldots$ , and they are a countable union of single points, each of size $0$ . Any notion of size that respects countable additivity must give them size $0$ . Jordan measure can't, because it is only finitely additive: it handles finitely many pieces and sees the closure, not the set.

The next example is worse, because the set is open, about as nice as a set can be.

Example 1.4 A small open set that Jordan sees as big

Fix $\varepsilon \in (0, 1)$ . Around the $n$ -th rational $q_n$ in $[0, 1]$ , take the open interval $I_n$ of length $\varepsilon 2^{-n}$ centred at $q_n$ , and let $U = (0, 1) \cap \bigcup_n I_n$ . Then $U$ is open, and its "size" ought to be at most $\sum_n \varepsilon 2^{-n} = \varepsilon$ . But $U$ is dense in $[0, 1]$ , so every elementary set containing it has measure at least $1$ , while its Jordan inner measure is at most $\varepsilon$ (Exercise 1.9). An open set built from intervals of total length $\varepsilon$ is assigned no size at all.

The moral is that the natural operations of analysis, countable unions and limits, lead immediately out of the Jordan measurable sets. The remedy, carried out in 3A.2 Lebesgue Measure, is to allow countably many boxes in the outer approximation. With that one change, $\mathbb{Q} \cap [0, 1]$ gets outer measure $0$ (cover $q_n$ by a box of length $\varepsilon 2^{-n}$ ) and the open set $U$ gets outer measure at most $\varepsilon$ , as they should.

Why countable additivity is the right axiom

Three reasons, all of which recur.

Limits. If $E_1 \subseteq E_2 \subseteq \cdots$ increase to $E$ , countable additivity applied to the disjoint differences $E_{n+1} \setminus E_n$ gives $m(E) = \lim m(E_n)$ : size is continuous along increasing unions. This is the measure-theoretic heart of the monotone convergence theorem (3A.3 The Lebesgue Integral), and through it of every limit theorem in the book.
Probability. For an infinite sequence of fair coin tosses, the event "a head eventually appears" is the disjoint union of "first head at toss $n$ ", with probabilities $2^{-n}$ , so it has probability $\sum 2^{-n} = 1$ . Kolmogorov's 1933 foundation of probability takes countable additivity as an axiom for this reason (3A.4 Measures, Probability and Weights).
Completeness. The function spaces built on a countably additive measure are complete (3A.7 Lᵖ Spaces and Jensen’s Inequality), which is what PDE needs.

Finite additivity alone is too weak for any of these; additivity for uncountable unions is too strong, since every set is the union of its points, and points have size $0$ .

Sets that cannot be measured

Can we have the three wishes for every subset of $\mathbb{R}$ ? No.

Theorem 1.5 Vitali's theorem

There is no function $m$ defined on all subsets of $\mathbb{R}$ , with values in $[0, \infty]$ , that is countably additive, translation invariant, and gives $[0, 1]$ the size $1$ .

Proof. Building the set. Call two numbers in $[0, 1]$ equivalent if their difference is rational. This is an equivalence relation (2A.2 Sets, Functions and Equivalence), and it splits $[0, 1]$ into equivalence classes, each of the form $(x + \mathbb{Q}) \cap [0, 1]$ . There are uncountably many classes. Using the axiom of choice (2A.8 Infinite Sets), choose exactly one representative from each class, and let $V$ be the set of chosen representatives (Figure 1.3).

The translates of $V$ . Let $q_1, q_2, \ldots$ list the rationals in $[-1, 1]$ , and let $V_n = V + q_n$ . Two facts:

The $V_n$ are disjoint. If $v + q_n = w + q_k$ with $v, w \in V$ , then $v - w = q_k - q_n$ is rational, so $v$ and $w$ are equivalent; since $V$ contains one point from each class, $v = w$ , and then $q_n = q_k$ .
$[0, 1] \subseteq \bigcup_n V_n \subseteq [-1, 2]$ . The second inclusion is clear. For the first, every $x \in [0, 1]$ is equivalent to its class's representative $v \in V$ , and $x - v$ is a rational in $[-1, 1]$ , say $q_n$ ; so $x \in V_n$ .

The contradiction. Suppose $m$ had the three properties. It is monotone ( $A \subseteq B$ implies $m(A) \leq m(B)$ , by additivity applied to $B = A \cup (B \setminus A)$ ), and $m([-1, 2]) = 3$ (three translated unit intervals, overlapping in points of size $0$ ). By translation invariance every $V_n$ has the same size $m(V)$ , and by countable additivity

1 = m([0, 1]) \leq \sum_{n=1}^\infty m(V) \leq m([-1, 2]) = 3.

If $m(V) = 0$ , the sum is $0$ , contradicting the left inequality. If $m(V) > 0$ , the sum is infinite, contradicting the right one.

Figure 1.3. The Vitali construction, schematically. Each row is one equivalence class

(x + \mathbb{Q}) \cap [0, 1]

, a countable dense set; the axiom of choice picks one point from each row (circled). Rational translates of the chosen set are disjoint and countably many of them cover

[0, 1]

, which no countably additive, translation-invariant size can accommodate.

The proof used the axiom of choice to pick the representatives, and it has to. Robert Solovay showed in 1970 that, assuming the existence of an inaccessible cardinal is consistent, there are models of set theory with a weaker form of choice (enough for all of analysis in this guidebook) in which every subset of $\mathbb{R}$ is Lebesgue measurable. So non-measurable sets are a consequence of the full axiom of choice, not of anything an analyst ever constructs explicitly. In practice: every set you will meet in this guidebook is measurable, and the reason for the definitions in 3A.2 Lebesgue Measure is to have a theory that is provably consistent, not to exclude sets that actually come up.

History The Banach–Tarski theorem

In 1924 Stefan Banach and Alfred Tarski proved that a solid ball in $\mathbb{R}^3$ can be cut into finitely many pieces (five suffice) which can be moved by rotations and translations and reassembled into two solid balls, each the same size as the original. No stretching is involved. The theorem is not a paradox in the logical sense: the pieces are non-measurable sets, built using the axiom of choice, and volume simply isn't defined for them. What it shows is that in three dimensions even finite additivity, together with invariance under rigid motions, can't be extended to all sets. In one and two dimensions this stronger failure does not happen: Banach showed in 1923 that length and area can be extended to finitely additive, isometry-invariant functions on all bounded subsets of the line and plane. The difference lies in the structure of the group of rigid motions, which in three dimensions contains free subgroups; the relevant property is called amenability.

Where this goes Where measures are needed later

Lebesgue measure on $\mathbb{R}^d$ is built in 3A.2 Lebesgue Measure. On a Riemannian manifold the same construction, done in coordinate charts and glued with a partition of unity (8A.2 Partitions of Unity, 8A.8 Differential Forms and Stokes’ Theorem), gives the Riemannian volume measure $dV$ (9A.1 Riemannian Metrics and Model Spaces). Every integral in Perelman's work, $\int(R + |\nabla f|^2)e^{-f}dV$ and the rest, is an integral against such a measure, weighted by a density (3A.4 Measures, Probability and Weights). Volumes of balls, and how they scale, are the language of noncollapsing: a manifold is κ-noncollapsed at scale $r$ if balls of radius $r$ with bounded curvature have volume at least $\kappa r^n$ (12A.4 κ-Noncollapsing). And Hausdorff measure, the right notion of size for the coastline, returns when one asks how large the singular set of a limit space can be.

History

Measuring areas by exhausting them with simpler figures goes back to Eudoxus and Archimedes. Giuseppe Peano (1887) and Camille Jordan (1892) gave the inner and outer content now called Jordan measure. Émile Borel (1898) measured more general sets using countable covers, and Henri Lebesgue's thesis (1902) defined measure and the integral in their modern form. Giuseppe Vitali's non-measurable set appeared in 1905, Felix Hausdorff's paradoxical decomposition of the sphere in 1914, and the Banach–Tarski theorem in 1924. Hausdorff introduced the measures and dimension named after him in 1918, which is the framework in which Mandelbrot later read Richardson's coastline data.

Recall Where we stand

We want a size for subsets of $\mathbb{R}^d$ that is countably additive, translation invariant and normalised; Vitali's construction shows this is impossible for all subsets, so we will measure only a class of measurable sets. Elementary sets (finite unions of boxes) have an elementary measure; squeezing between elementary sets gives Jordan measure, which is the measure behind the Riemann integral and works for every set with a negligible boundary. It fails for countable sets like $\mathbb{Q} \cap [0, 1]$ and even for some open sets, because it is only finitely additive. 3A.2 Lebesgue Measure fixes this by allowing countably many boxes, which produces Lebesgue measure.

Exercises

Exercise 1.6 The shoelace formula

(a) Verify the area $21$ for the pentagon $(0,0), (4,0), (5,3), (2,5), (-1,3)$ . (b) Show that $\tfrac12(x_1y_2 - x_2y_1)$ is the signed area of the triangle with vertices $0$ , $(x_1, y_1)$ , $(x_2, y_2)$ , positive when the triangle is traversed anticlockwise. (c) For a convex polygon containing the origin, deduce the formula from finite additivity: the polygon is the disjoint union (up to edges, which have measure zero) of the triangles from the origin to its edges.

Solution

(a) The terms $x_iy_{i+1} - x_{i+1}y_i$ are $0, 12, 19, 6, 3$ , summing to $42$ ; half is $21$ . (b) It is half the determinant of the two edge vectors, which is the signed area of the parallelogram they span. (c) All the triangles are anticlockwise when the origin is inside, so the signed areas are positive and add.

Exercise 1.7 The boundary criterion

Let $E$ be bounded. (a) Show that $m^{*,(J)}(E) = m^{*,(J)}(\overline E)$ and $m_{*,(J)}(E) = m_{*,(J)}(E^\circ)$ . (You may use that an elementary set can be enlarged slightly to an open one, and shrunk slightly to a closed one, changing its measure by as little as you like.) (b) Deduce that $m^{*,(J)}(E) - m_{*,(J)}(E) \leq m^{*,(J)}(\partial E)$ , and conversely that if $E$ is Jordan measurable then $\partial E$ has outer measure zero. This is Proposition 1.2.

Exercise 1.8 Graphs are Jordan null

Let $f : [0, 1] \to \mathbb{R}$ be continuous. Show that its graph $\{(x, f(x))\}$ has Jordan outer measure $0$ in $\mathbb{R}^2$ . (Given $\varepsilon$ , use uniform continuity to cover the graph by $N$ rectangles of width $1/N$ and height $\varepsilon$ .) Deduce that the region under the graph of a continuous non-negative function is Jordan measurable.

Solution

Choose $N$ with $|f(x) - f(y)| \leq \varepsilon/2$ when $|x - y| \leq 1/N$ (2B.3 Compactness). Over $[\frac{k-1}N, \frac kN]$ the graph lies in a rectangle of width $\frac1N$ and height $\varepsilon$ , centred at height $f(\frac kN)$ . The $N$ rectangles have total area $\varepsilon$ . The region's boundary consists of the graph, two vertical segments and a horizontal segment, all Jordan null, so Proposition 1.2 applies.

Exercise 1.9 A dense open set

For the set $U$ of Example 1.4, show that (a) $U$ is open and dense in $[0, 1]$ ; (b) every elementary set containing $U$ has measure at least $1$ ; (c) every elementary set contained in $U$ has measure at most $\varepsilon$ . (For (c), a finite union of boxes inside $U$ is covered by finitely many of the $I_n$ , by compactness of its closure, after shrinking it slightly; or prove it after 3A.2 Lebesgue Measure using countable subadditivity.)

Exercise 1.10 Length of a Koch curve

The Koch curve is built from a segment of length $1$ by replacing the middle third of every segment with two sides of an equilateral triangle, repeatedly. (a) Show that after $k$ steps the polygon has $4^k$ segments of length $3^{-k}$ , so length $(4/3)^k$ . (b) If $L(\ell) = C\ell^{1-D}$ for rulers $\ell = 3^{-k}$ , find $D$ . (c) Show that the curve lies within a bounded region, so its area is finite, and in fact (once Lebesgue measure is available) zero.

Solution

(a) Each step multiplies the number of segments by $4$ and divides their length by $3$ . (b) $(4/3)^k = C(3^{-k})^{1-D}$ gives $4^k = C\,3^{kD}$ , so $D = \log 4/\log 3 \approx 1.262$ . (c) After $k$ steps the curve lies in the union of $4^k$ triangles of area proportional to $9^{-k}$ , total $\propto (4/9)^k \to 0$ .

Exercise 1.11 Where the Vitali argument needs countability

Explain why the proof of Theorem 1.5 breaks down if $m$ is only required to be finitely additive. (Banach's 1923 theorem says that on the line, such an $m$ defined on all bounded sets does exist.)

Solution

Finite additivity only bounds finitely many of the $V_n$ at a time: $N\,m(V) \leq 3$ for every $N$ , which forces $m(V) = 0$ , but then nothing contradicts $m([0, 1]) = 1$ , because $[0, 1]$ is covered only by infinitely many $V_n$ , and without countable additivity the sizes of infinitely many pieces need not add up to the size of their union.

Exercise 1.12 Rehearsal: how volume scales

(a) For an elementary set $E \subseteq \mathbb{R}^d$ and $\lambda > 0$ , show $m(\lambda E) = \lambda^dm(E)$ . Deduce the same for Jordan measurable sets. (b) Deduce that the volume of a ball of radius $r$ is $\omega_dr^d$ , where $\omega_d$ is the volume of the unit ball. (c) In a Riemannian manifold, a ball is called κ-noncollapsed if its volume is at least $\kappa r^d$ . Explain why the ratio $\mathrm{vol}\,B(x, r)/r^d$ doesn't change if the whole space is scaled by a factor $\lambda$ (distances multiplied by $\lambda$ , so $r \mapsto \lambda r$ and volumes $\mapsto \lambda^d \times$ ). This scale invariance is what lets Perelman's noncollapsing estimate survive the rescalings of 12A.4 κ-Noncollapsing and 12B.3 The Canonical Neighbourhood Theorem.

Solution

(a) Scaling a box by $\lambda$ multiplies each side length by $\lambda$ , so its volume by $\lambda^d$ ; a disjoint union of boxes scales the same way. If $A \subseteq E \subseteq B$ with $A, B$ elementary, then $\lambda A \subseteq \lambda E \subseteq \lambda B$ , so inner and outer measures scale by $\lambda^d$ . (b) $B(0, r) = rB(0, 1)$ . (c) Both numerator and denominator are multiplied by $\lambda^d$ .

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1.1How long is a coastline?

1.2What a size should do

1.3Elementary measure

1.4Jordan measure

1.4.1Where Jordan measure stops

1.4.2Why countable additivity is the right axiom

1.5Sets that cannot be measured

1.6History

1.7Exercises

How long is a coastline?

What a size should do

Elementary measure

Jordan measure

Where Jordan measure stops

Why countable additivity is the right axiom

Sets that cannot be measured

History

Exercises