∫

∑

∂

∞

√

∏

RA-1

6-10 hours

Advanced

Measure Foundations & Outer Measure

Master the construction of outer measure via rectangle coverings, understand the Carathéodory measurability criterion, and explore the fundamental properties of measurable sets that form the foundation of modern measure theory.

Learning Objectives

By the end of this course, you will be able to:

Define outer measure using countable rectangle coverings and understand its fundamental properties

Apply the Carathéodory criterion to determine measurability of sets

Prove that measurable sets form a σ-algebra and that outer measure is countably additive on measurable sets

Understand regularity properties: approximation by open and closed sets

Identify null sets and understand their role in measure theory

Connect outer measure to geometric intuition via rectangle coverings

Prerequisites

Before starting this course, you should be familiar with:

Basic set theory: unions, intersections, complements, countable collections
Topology of ℝⁿ: open sets, closed sets, compact sets
Sequences and limits in ℝⁿ
Basic properties of rectangles and their volumes

Core Concepts

Fundamental definitions and properties

Definition 1.1: Outer Measure

Let $E \subseteq \mathbb{R}^n$ . The outer measure of $E$ , denoted $m^*(E)$ , is defined as:

m^*(E) = \inf\left\{\sum_{j=1}^{\infty} |Q_j| : E \subseteq \bigcup_{j=1}^{\infty} Q_j, \text{ where } Q_j \text{ are open rectangles}\right\}

where $|Q_j|$ denotes the volume (Lebesgue measure) of the rectangle $Q_j$ .

Note: We take the infimum over all possible countable coverings of $E$ by open rectangles. If no such covering exists (which cannot happen), we define $m^*(E) = \infty$ .

Definition 1.2: Measurable Set (Carathéodory Criterion)

A set $E \subseteq \mathbb{R}^n$ is said to be measurable (in the sense of Carathéodory) if for every set $A \subseteq \mathbb{R}^n$ , we have:

m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)

where $E^c = \mathbb{R}^n \setminus E$ denotes the complement of $E$ .

Definition 1.3: Null Set

A set $E \subseteq \mathbb{R}^n$ is called a null set (or a set of measure zero) if $m^*(E) = 0$ .

A property is said to hold almost everywhere (abbreviated a.e.) if it holds except possibly on a null set.

Theorem 1.1: Properties of Outer Measure

Outer measure has the following fundamental properties:

Non-negativity: $m^*(E) \geq 0$ for all $E$ , and $m^*(\emptyset) = 0$ .
Monotonicity: If $E \subseteq F$ , then $m^*(E) \leq m^*(F)$ .
Countable Subadditivity: For any countable collection $\{E_j\}_{j=1}^{\infty}$ of sets,

m^*\left(\bigcup_{j=1}^{\infty} E_j\right) \leq \sum_{j=1}^{\infty} m^*(E_j)

Translation Invariance: For any

x \in \mathbb{R}^n

m^*(E + x) = m^*(E)

, where

E + x = \{y + x : y \in E\}

Proof of Theorem 1.1:

Proof:

(1) Non-negativity follows immediately from the definition, as volumes are non-negative and the infimum of non-negative numbers is non-negative. The empty set can be covered by a single rectangle of arbitrarily small volume, so $m^*(\emptyset) = 0$ .

(2) Monotonicity: If $E \subseteq F$ , then any covering of $F$ is also a covering of $E$ . Therefore, the infimum over coverings of $E$ is at most the infimum over coverings of $F$ .

(3) Countable Subadditivity: Given $\epsilon > 0$ , for each $j$ , choose a covering $\{Q_{j,k}\}_{k=1}^{\infty}$ of $E_j$ such that

\sum_{k=1}^{\infty} |Q_{j,k}| < m^*(E_j) + \frac{\epsilon}{2^j}

Then $\{Q_{j,k} : j, k \in \mathbb{N}\}$ is a countable covering of $\bigcup_{j=1}^{\infty} E_j$ , and

m^*\left(\bigcup_{j=1}^{\infty} E_j\right) \leq \sum_{j,k} |Q_{j,k}| < \sum_{j=1}^{\infty} m^*(E_j) + \epsilon

Since $\epsilon > 0$ was arbitrary, the result follows.

(4) Translation Invariance: If $\{Q_j\}$ covers $E$ , then $\{Q_j + x\}$ covers $E + x$ , and $|Q_j + x| = |Q_j|$ . The result follows from the definition. ∎

∎

Theorem 1.2: Measurable Sets Form a σ-Algebra

The collection $\mathcal{M}$ of all measurable subsets of $\mathbb{R}^n$ forms a σ-algebra. That is:

$\mathbb{R}^n \in \mathcal{M}$
If $E \in \mathcal{M}$ , then $E^c \in \mathcal{M}$
If $E_1, E_2, \ldots \in \mathcal{M}$ , then $\bigcup_{j=1}^{\infty} E_j \in \mathcal{M}$

Proof of Theorem 1.2:

Proof:

(1) For any $A$ , we have $m^*(A) = m^*(A \cap \mathbb{R}^n) + m^*(A \cap \emptyset) = m^*(A) + 0$ , so $\mathbb{R}^n$ is measurable.

(2) The measurability condition is symmetric in $E$ and $E^c$ , so if $E$ is measurable, so is $E^c$ .

(3) For countable unions, we use induction and the fact that finite unions of measurable sets are measurable (which follows from De Morgan's laws and the complement property). The countable case then follows from continuity properties. ∎

∎

Theorem 1.3: Countable Additivity on Measurable Sets

If $E_1, E_2, \ldots$ are disjoint measurable sets, then:

m^*\left(\bigcup_{j=1}^{\infty} E_j\right) = \sum_{j=1}^{\infty} m^*(E_j)

This property is called countable additivity.

Proof of Theorem 1.3:

Proof:

By subadditivity, we already have $m^*(\bigcup E_j) \leq \sum m^*(E_j)$ . For the reverse inequality, we use the measurability of the sets and induction. For finite unions, measurability gives additivity. The countable case follows by taking limits. ∎

∎

Corollary 1.1: Lebesgue Measure

The restriction of outer measure to the σ-algebra $\mathcal{M}$ of measurable sets is called Lebesgue measure, denoted $m$ . That is, for measurable $E$ , we define:

m(E) = m^*(E)

Lebesgue measure is countably additive on measurable sets.

Example 1.1: Outer Measure of a Point

Problem: Compute the outer measure of a single point $\{a\}$ in $\mathbb{R}^n$ .

Solution:

For any $\epsilon > 0$ , we can cover the point $\{a\}$ by a single open rectangle $Q$ centered at $a$ with volume $|Q| < \epsilon$ .

Therefore, $m^*(\{a\}) \leq \epsilon$ for all $\epsilon > 0$ , which implies $m^*(\{a\}) = 0$ .

Since a point has outer measure zero, it is a null set and is measurable.

Example 1.2: Outer Measure of a Countable Set

Problem: Show that any countable set $E = \{a_1, a_2, \ldots\}$ in $\mathbb{R}^n$ has outer measure zero.

Solution:

For each $j$ , cover the point $a_j$ by an open rectangle $Q_j$ with volume $|Q_j| < \frac{\epsilon}{2^j}$ .

Then $\{Q_j\}_{j=1}^{\infty}$ covers $E$ , and

\sum_{j=1}^{\infty} |Q_j| < \sum_{j=1}^{\infty} \frac{\epsilon}{2^j} = \epsilon

Therefore, $m^*(E) \leq \epsilon$ for all $\epsilon > 0$ , so $m^*(E) = 0$ .

This shows that countable sets are null sets and hence measurable.

Example 1.3: Outer Measure of an Interval

Problem: Show that for an interval $I = [a, b]$ in $\mathbb{R}$ , we have $m^*(I) = b - a$ .

Solution:

First, note that $I$ can be covered by itself (as an open rectangle by taking $(a - \delta, b + \delta)$ for small $\delta > 0$ ), so $m^*(I) \leq b - a + 2\delta$ for any $\delta > 0$ , hence $m^*(I) \leq b - a$ .

For the reverse inequality, suppose $\{Q_j\}$ is a covering of $I$ by open intervals. By compactness of $I$ , finitely many $Q_{j_1}, \ldots, Q_{j_k}$ cover $I$ . Then

b - a \leq \sum_{i=1}^k |Q_{j_i}| \leq \sum_{j=1}^{\infty} |Q_j|

Taking the infimum over all coverings, we get $m^*(I) \geq b - a$ .

Therefore, $m^*(I) = b - a$ .

Theorem 1.4: Regularity of Measurable Sets

Let $E$ be a measurable set. Then:

Outer Regularity: For any $\epsilon > 0$ , there exists an open set $G$ such that $E \subseteq G$ and $m(G \setminus E) < \epsilon$ .
Inner Regularity: For any $\epsilon > 0$ , there exists a closed set $F$ such that $F \subseteq E$ and $m(E \setminus F) < \epsilon$ .

Proof of Theorem 1.4:

Proof:

(1) By definition of outer measure, for any $\epsilon > 0$ , there exists a covering $\{Q_j\}$ of $E$ by open rectangles such that $\sum |Q_j| < m(E) + \epsilon$ . Let $G = \bigcup Q_j$ . Then $G$ is open, $E \subseteq G$ , and

m(G \setminus E) \leq m(G) - m(E) \leq \sum |Q_j| - m(E) < \epsilon

(2) Apply (1) to $E^c$ to get an open set $G$ containing $E^c$ with $m(G \setminus E^c) < \epsilon$ . Then $F = G^c$ is closed, $F \subseteq E$ , and $m(E \setminus F) = m(G \setminus E^c) < \epsilon$ . ∎

∎

Remark 1.1: Borel Sets are Measurable

All Borel sets (sets in the σ-algebra generated by open sets) are measurable. This follows from the fact that open sets are measurable and measurable sets form a σ-algebra.

However, there exist non-Borel measurable sets. The construction of such sets typically requires the Axiom of Choice.

Example 1.4: Cantor Set has Measure Zero

Problem: Show that the Cantor set $C$ has outer measure zero.

Solution:

The Cantor set is constructed by removing the middle third of $[0,1]$ , then removing the middle third of each remaining interval, and so on.

At the $n$ -th step, we remove $2^{n-1}$ intervals, each of length $\frac{1}{3^n}$ . The total length removed is:

\sum_{n=1}^{\infty} 2^{n-1} \cdot \frac{1}{3^n} = \frac{1}{3} \sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n = \frac{1}{3} \cdot \frac{1}{1 - 2/3} = 1

Since the total length removed is 1, and the Cantor set is what remains, we have $m^*(C) = 0$ .

Note: The Cantor set is uncountable but has measure zero, demonstrating that measure and cardinality are independent concepts.

Lemma 1.1: Approximation by Rectangles

For any set $E \subseteq \mathbb{R}^n$ and any $\epsilon > 0$ , there exists a countable collection of open rectangles $\{Q_j\}$ such that:

$E \subseteq \bigcup_{j=1}^{\infty} Q_j$
$\sum_{j=1}^{\infty} |Q_j| \leq m^*(E) + \epsilon$

Proof of Lemma 1.1:

Proof:

This follows directly from the definition of outer measure as an infimum. For any $\epsilon > 0$ , there exists a covering $\{Q_j\}$ such that

\sum_{j=1}^{\infty} |Q_j| < m^*(E) + \epsilon

If $m^*(E) = \infty$ , the result is trivial. If $m^*(E) < \infty$ , the definition guarantees such a covering exists. ∎

∎

Theorem 1.5: Open Sets are Measurable

Every open set in $\mathbb{R}^n$ is measurable.

Proof of Theorem 1.5:

Proof:

Let $G$ be an open set. We need to show that for any $A \subseteq \mathbb{R}^n$ ,

m^*(A) \geq m^*(A \cap G) + m^*(A \cap G^c)

The reverse inequality always holds by subadditivity, so we only need to prove this direction.

Since $G$ is open, we can write $G = \bigcup_{j=1}^{\infty} Q_j$ as a countable union of disjoint open rectangles (this follows from the structure of open sets in $\mathbb{R}^n$ ).

For any $\epsilon > 0$ , cover $A$ by rectangles $\{R_k\}$ with $\sum |R_k| < m^*(A) + \epsilon$ . Then

m^*(A \cap G) + m^*(A \cap G^c) \leq \sum_{j,k} |R_k \cap Q_j| + \sum_k |R_k \cap G^c| \leq \sum_k |R_k| < m^*(A) + \epsilon

Since $\epsilon > 0$ was arbitrary, the result follows. ∎

∎

Corollary 1.2: Closed Sets are Measurable

Every closed set in $\mathbb{R}^n$ is measurable, as it is the complement of an open set.

Example 1.5: Computing Outer Measure of a Union

Problem: Let $E_1 = [0, 1]$ and $E_2 = [2, 3]$ in $\mathbb{R}$ . Compute $m^*(E_1 \cup E_2)$ .

Solution:

Since $E_1$ and $E_2$ are disjoint intervals, they are measurable. Therefore,

m^*(E_1 \cup E_2) = m(E_1 \cup E_2) = m(E_1) + m(E_2) = 1 + 1 = 2

This demonstrates countable additivity on disjoint measurable sets.

Example 1.6: Outer Measure is Not Countably Additive

Problem: Show that outer measure is not countably additive on all sets by constructing a counterexample.

Solution:

Using the Axiom of Choice, one can construct a non-measurable set $V$ (a Vitali set). For such a set, if we take countably many disjoint translates $V_1, V_2, \ldots$ that cover an interval, we would have:

m^*\left(\bigcup V_j\right) < \sum m^*(V_j)

This violates countable additivity. This is why we restrict to measurable sets to get a proper measure.

Definition 1.4: Inner Measure

For a bounded set $E \subseteq \mathbb{R}^n$ , the inner measure is defined as:

m_*(E) = \sup\{m(K) : K \subseteq E, K \text{ compact}\}

For unbounded sets, we define $m_*(E) = \lim_{n \to \infty} m_*(E \cap [-n, n]^n)$ .

Theorem 1.6: Characterization of Measurability via Inner and Outer Measure

A bounded set $E$ is measurable if and only if $m_*(E) = m^*(E)$ .

Proof of Theorem 1.6:

Proof:

If $E$ is measurable, regularity gives us compact sets $K$ and open sets $G$ such that $K \subseteq E \subseteq G$ and $m(G \setminus K) < \epsilon$ for any $\epsilon > 0$ . This implies $m_*(E) = m^*(E)$ .

Conversely, if $m_*(E) = m^*(E)$ , then for any $\epsilon > 0$ , there exist compact $K$ and open $G$ with $K \subseteq E \subseteq G$ and $m(G \setminus K) < \epsilon$ . This approximation property implies measurability. ∎

∎

Theorem 1.7: Borel Sets are Measurable

The σ-algebra $\mathcal{B}$ of Borel sets (generated by open sets) is contained in the σ-algebra $\mathcal{M}$ of measurable sets. That is, every Borel set is measurable.

Proof of Theorem 1.7:

Proof:

Since open sets are measurable (Theorem 1.5), and measurable sets form a σ-algebra (Theorem 1.2), the σ-algebra generated by open sets (the Borel σ-algebra) must be contained in $\mathcal{M}$ .

More formally, let $\mathcal{B}$ be the smallest σ-algebra containing all open sets. Since all open sets are in $\mathcal{M}$ and $\mathcal{M}$ is a σ-algebra, we have $\mathcal{B} \subseteq \mathcal{M}$ . ∎

∎

Corollary 1.3: Gδ and Fσ Sets are Measurable

Every $G_\delta$ set (countable intersection of open sets) and every $F_\sigma$ set (countable union of closed sets) is measurable.

Example 1.7: Computing Outer Measure of a Rectangle

Problem: Compute the outer measure of a closed rectangle $R = [a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_n, b_n]$ in $\mathbb{R}^n$ .

Solution:

The rectangle $R$ can be covered by itself (as an open rectangle by taking slightly larger intervals), so $m^*(R) \leq \prod_{i=1}^n (b_i - a_i)$ .

For the reverse inequality, note that $R$ is compact. Any covering of $R$ by open rectangles has a finite subcover. Using the structure of rectangles, we can show that the sum of volumes in any covering is at least the volume of $R$ .

Therefore, $m^*(R) = \prod_{i=1}^n (b_i - a_i)$ , which is the expected volume of the rectangle.

Since $R$ is a closed set, it is measurable, so $m(R) = m^*(R) = \prod_{i=1}^n (b_i - a_i)$ .

Example 1.8: Outer Measure of a Dense Set

Problem: Let $E = \mathbb{Q} \cap [0, 1]$ be the rationals in $[0,1]$ . Compute $m^*(E)$ .

Solution:

Since $E$ is countable, by Example 1.2, we have $m^*(E) = 0$ .

This is interesting because $E$ is dense in $[0,1]$ (every point in $[0,1]$ is a limit point of $E$ ), yet it has measure zero. This demonstrates that measure and topological density are independent concepts.

The complement $[0,1] \setminus E$ (the irrationals in $[0,1]$ ) has measure 1, even though it is nowhere dense in a certain sense.

Theorem 1.8: Continuity from Above and Below

Let $\{E_j\}$ be a sequence of measurable sets.

If $E_1 \supseteq E_2 \supseteq \cdots$ and $m(E_1) < \infty$ , then

m\left(\bigcap_{j=1}^{\infty} E_j\right) = \lim_{j \to \infty} m(E_j)

E_1 \subseteq E_2 \subseteq \cdots

, then

m\left(\bigcup_{j=1}^{\infty} E_j\right) = \lim_{j \to \infty} m(E_j)

Proof of Theorem 1.8:

Proof:

(1) Write $E_1 = \bigcap_{j=1}^{\infty} E_j \cup \bigcup_{j=1}^{\infty} (E_j \setminus E_{j+1})$ . Since the sets $E_j \setminus E_{j+1}$ are disjoint,

m(E_1) = m\left(\bigcap E_j\right) + \sum_{j=1}^{\infty} m(E_j \setminus E_{j+1})

Rearranging and using that $m(E_j) = m(E_1) - \sum_{k=1}^{j-1} m(E_k \setminus E_{k+1})$ , we get the result.

(2) Write $\bigcup_{j=1}^{\infty} E_j = E_1 \cup \bigcup_{j=1}^{\infty} (E_{j+1} \setminus E_j)$ . Since these sets are disjoint,

m\left(\bigcup E_j\right) = m(E_1) + \sum_{j=1}^{\infty} m(E_{j+1} \setminus E_j) = \lim_{j \to \infty} m(E_j)

∎

Example 1.9: Application of Continuity from Below

Problem: Let $E_n = (0, 1 - \frac{1}{n})$ for $n \geq 2$ . Compute $m(\bigcup_{n=2}^{\infty} E_n)$ .

Solution:

Note that $E_2 \subseteq E_3 \subseteq \cdots$ and $\bigcup_{n=2}^{\infty} E_n = (0, 1)$ .

We have $m(E_n) = 1 - \frac{1}{n}$ , so $\lim_{n \to \infty} m(E_n) = 1$ .

By continuity from below (Theorem 1.8),

m((0, 1)) = m\left(\bigcup_{n=2}^{\infty} E_n\right) = \lim_{n \to \infty} m(E_n) = 1

This confirms that the measure of $(0,1)$ is 1, as expected.

Lemma 1.2: Vitali Covering Lemma (Simplified)

Let $E$ be a measurable set with $m(E) < \infty$ , and let $\mathcal{V}$ be a Vitali covering of $E$ (a collection of closed balls such that every point of $E$ is contained in balls of arbitrarily small radius).

Then for any $\epsilon > 0$ , there exists a finite disjoint subcollection $\{B_1, \ldots, B_N\}$ such that

m\left(E \setminus \bigcup_{i=1}^N B_i\right) < \epsilon

Proof of Lemma 1.2:

Proof:

This is a simplified version of the Vitali covering lemma. The key idea is to use a greedy algorithm: at each step, select the largest ball that is disjoint from previously selected balls.

Since $m(E) < \infty$ , the process terminates after finitely many steps, and the remaining set has small measure. The full Vitali covering lemma (which we'll see in later courses) provides more precise control. ∎

∎

Theorem 1.9: Approximation by Simple Sets

For any measurable set $E$ and any $\epsilon > 0$ , there exist:

A finite union of disjoint closed rectangles $F$ such that $m(E \triangle F) < \epsilon$ , where $\triangle$ denotes symmetric difference.
A finite union of open rectangles $G$ such that $E \subseteq G$ and $m(G \setminus E) < \epsilon$ .

Proof of Theorem 1.9:

Proof:

(1) By regularity, there exists a closed set $F_1$ with $F_1 \subseteq E$ and $m(E \setminus F_1) < \epsilon/2$ . Since $F_1$ is closed and bounded (we can restrict to a large ball), it is compact. Cover $F_1$ by finitely many open rectangles, then take the closure of their union and use the structure of rectangles to approximate.

(2) By outer regularity, there exists an open set $G$ with $E \subseteq G$ and $m(G \setminus E) < \epsilon$ . Since $G$ is open, it is a countable union of open rectangles. By continuity from below, we can approximate by a finite union. ∎

∎

Example 1.10: Constructing a Non-Measurable Set

Problem: Describe the construction of a Vitali set (a non-measurable set).

Solution:

Define an equivalence relation on $[0,1]$ by $x \sim y$ if $x - y \in \mathbb{Q}$ . Using the Axiom of Choice, select one element from each equivalence class to form a set $V$ .

For each rational $q \in [-1, 1]$ , consider the translate $V_q = V + q$ . These sets are disjoint and their union contains $[0,1]$ and is contained in $[-1, 2]$ .

If $V$ were measurable, then by translation invariance, all $V_q$ would have the same measure. But then

1 \leq m\left(\bigcup_q V_q\right) \leq 3

and by countable additivity (if measurable), this would give either 0 or $\infty$ , a contradiction. Therefore, $V$ is not measurable.

Note: This construction requires the Axiom of Choice. In models of set theory without AC, all sets might be measurable.

Corollary 1.4: Measurable Sets are Dense

For any set $E$ with $m^*(E) < \infty$ and any $\epsilon > 0$ , there exists a measurable set $F$ such that $E \subseteq F$ and $m^*(F \setminus E) < \epsilon$ .

In other words, any set can be approximated from outside by a measurable set with arbitrarily small error.

Remark 1.3: Historical Context

The development of Lebesgue measure was a major breakthrough in analysis. Before Lebesgue, mathematicians struggled with the concept of "size" for irregular sets.

Henri Lebesgue's 1902 thesis introduced the outer measure approach, which was later refined by Carathéodory's criterion. This construction provides a measure that:

Extends the intuitive notion of volume to a large class of sets
Satisfies countable additivity (crucial for integration theory)
Is translation invariant (natural for Euclidean space)
Allows approximation by simple sets (enabling computations)

The discovery of non-measurable sets (requiring the Axiom of Choice) showed that we cannot measure all sets, but the class of measurable sets is large enough for all practical purposes in analysis.

Remark 1.4: Connection to Integration

The theory of measurable sets is the foundation for Lebesgue integration, which we will study in later courses. The key insight is that we can integrate functions over measurable sets, and the measure of a set tells us "how much" of the domain contributes to the integral.

For example, if a function is zero almost everywhere (except on a null set), its integral will be zero, regardless of what the function does on that null set. This is why null sets and "almost everywhere" properties are so important.

Example 1.11: Outer Measure of a Line Segment in ℝ²

Problem: Compute the outer measure of the line segment $L = \{(x, 0) : 0 \leq x \leq 1\}$ in $\mathbb{R}^2$ .

Solution:

For any $\epsilon > 0$ , we can cover $L$ by rectangles of the form $[0, 1] \times [-\delta, \delta]$ for small $\delta > 0$ .

The volume of such a rectangle is $2\delta$ . By choosing $\delta < \epsilon/2$ , we get $m^*(L) \leq \epsilon$ .

Since $\epsilon > 0$ was arbitrary, $m^*(L) = 0$ .

This shows that "lower-dimensional" sets (like curves in the plane) have measure zero in higher dimensions. This is a general phenomenon: a $k$ -dimensional manifold in $\mathbb{R}^n$ for $k < n$ has $n$ -dimensional Lebesgue measure zero.

Example 1.12: Outer Measure of a Disk

Problem: Show that for a disk $D = \{(x,y) : x^2 + y^2 \leq r^2\}$ in $\mathbb{R}^2$ , we have $m^*(D) = \pi r^2$ .

Solution:

The disk $D$ is contained in the square $[-r, r] \times [-r, r]$ , which has area $4r^2$ , so $m^*(D) \leq 4r^2$ .

For a better estimate, we can cover $D$ by a grid of small squares. As the grid becomes finer, the sum of areas of squares covering $D$ approaches $\pi r^2$ .

More rigorously, we can inscribe and circumscribe $D$ by polygons. The inscribed polygon has area approaching $\pi r^2$ from below, and the circumscribed polygon from above.

Since $D$ is closed (hence measurable), we have $m(D) = m^*(D) = \pi r^2$ , which matches our geometric intuition.

Theorem 1.10: Product of Measurable Sets

If $E_1 \subseteq \mathbb{R}^{n_1}$ and $E_2 \subseteq \mathbb{R}^{n_2}$ are measurable, then $E_1 \times E_2 \subseteq \mathbb{R}^{n_1 + n_2}$ is measurable, and

m_{n_1+n_2}(E_1 \times E_2) = m_{n_1}(E_1) \cdot m_{n_2}(E_2)

where $m_k$ denotes $k$ -dimensional Lebesgue measure.

Proof of Theorem 1.10:

Proof:

First, if $E_1$ and $E_2$ are rectangles, the result is immediate from the definition of volume.

For general measurable sets, use approximation by rectangles and the fact that the product of σ-algebras is contained in the product σ-algebra. The measure equality follows from Fubini's theorem (which we'll study later) or by direct approximation. ∎

∎

Example 1.13: Measure of a Cylinder

Problem: Compute the 3-dimensional Lebesgue measure of the cylinder $C = \{(x,y,z) : x^2 + y^2 \leq r^2, 0 \leq z \leq h\}$ .

Solution:

The cylinder is the product of a disk $D = \{(x,y) : x^2 + y^2 \leq r^2\}$ in $\mathbb{R}^2$ and the interval $[0, h]$ in $\mathbb{R}$ .

By Theorem 1.10,

m_3(C) = m_2(D) \cdot m_1([0,h]) = \pi r^2 \cdot h = \pi r^2 h

This matches the familiar formula for the volume of a cylinder from elementary geometry.

Corollary 1.5: Measure of Cartesian Products

If $E_j \subseteq \mathbb{R}^{n_j}$ are measurable for $j = 1, \ldots, k$ , then

m_{n_1+\cdots+n_k}(E_1 \times \cdots \times E_k) = \prod_{j=1}^k m_{n_j}(E_j)

Remark 1.5: Computational Techniques

When computing outer measures or Lebesgue measures, several techniques are useful:

Covering arguments: Find coverings that give upper bounds, then show these bounds are tight
Regularity: Approximate by open or closed sets, which are easier to work with
Product structure: Use Theorem 1.10 to compute measures of product sets
Symmetry: Use translation, rotation, or reflection invariance
Limiting arguments: Use continuity from above/below to compute limits of measures

These techniques will be essential when we move to integration theory, where we need to compute measures of level sets and other geometric objects.

Theorem 1.11: Operations on Measurable Sets

Let $E$ and $F$ be measurable sets. Then:

$E \cup F$ is measurable
$E \cap F$ is measurable
$E \setminus F$ is measurable
If $E_1, E_2, \ldots$ are measurable, then $\bigcap_{j=1}^{\infty} E_j$ is measurable

Moreover, for disjoint measurable sets $E_1, E_2, \ldots$ , we have:

m\left(\bigcup_{j=1}^{\infty} E_j\right) = \sum_{j=1}^{\infty} m(E_j)

Proof of Theorem 1.11:

Proof:

(1) Since measurable sets form a σ-algebra (Theorem 1.2), finite unions are measurable.

(2) By De Morgan's law, $E \cap F = (E^c \cup F^c)^c$ . Since complements and unions of measurable sets are measurable, intersections are measurable.

(3) Note that $E \setminus F = E \cap F^c$ . Since $F$ is measurable, $F^c$ is measurable, and by (2), the intersection is measurable.

(4) By De Morgan's law, $\bigcap_{j=1}^{\infty} E_j = \left(\bigcup_{j=1}^{\infty} E_j^c\right)^c$ . Since countable unions and complements preserve measurability, countable intersections are measurable.

The countable additivity property follows from Theorem 1.3. ∎

∎

Example 1.14: Computing Outer Measure of a Complex Set

Problem: Let $E = \bigcup_{n=1}^{\infty} [n, n + \frac{1}{n^2}]$ be a union of disjoint intervals. Compute $m^*(E)$ .

Solution:

Since the intervals $[n, n + \frac{1}{n^2}]$ are disjoint and measurable, by countable additivity (Theorem 1.3),

m^*(E) = m(E) = \sum_{n=1}^{\infty} m\left([n, n + \frac{1}{n^2}]\right) = \sum_{n=1}^{\infty} \frac{1}{n^2}

We recognize this as the sum of the reciprocals of squares, which converges to $\frac{\pi^2}{6}$ (the Basel problem).

Therefore, $m^*(E) = \frac{\pi^2}{6}$ .

Verification: Each interval has measure $\frac{1}{n^2}$ , and since they are disjoint, the total measure is the sum. The convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2}$ ensures the result is finite.

Example 1.15: Computing Outer Measure Using Symmetry and Decomposition

Problem: Compute the outer measure of the set $E = \{(x, y) \in \mathbb{R}^2 : |x| + |y| \leq 1\}$ (a diamond shape).

Solution:

By symmetry, we can compute the measure of $E$ by considering the first quadrant and multiplying by 4.

In the first quadrant, $E$ is the triangle with vertices $(0, 0)$ , $(1, 0)$ , and $(0, 1)$ , bounded by $x + y = 1$ .

The area of this triangle is $\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$ .

By symmetry, the total area is:

m^*(E) = m(E) = 4 \cdot \frac{1}{2} = 2

Alternative approach: We can also compute this using Fubini's theorem (to be covered later) or by approximating with rectangles. Since $E$ is closed and bounded, it is compact and hence measurable.

The boundary of $E$ is a null set (a curve in $\mathbb{R}^2$ ), so the measure equals the area computed geometrically.

Remark 1.7: Computational Techniques and Common Pitfalls

When computing outer measures, several techniques and common pitfalls should be noted:

Symmetry: Use geometric symmetry to reduce computations. For example, symmetric sets can often be computed by considering one symmetric component and multiplying.
Decomposition: Break complex sets into simpler measurable pieces. If the pieces are disjoint, use countable additivity.
Approximation: For sets that are not obviously measurable, use regularity to approximate by open or closed sets.
Null sets: Remember that boundaries of "nice" sets (like polygons, smooth curves) have measure zero, so they don't affect the measure.

Common pitfalls:

Assuming countable additivity for all sets: Outer measure is only countably subadditive. Countable additivity holds only for measurable sets.
Ignoring overlap: When computing unions, be careful not to double-count overlapping regions. Use the inclusion-exclusion principle or decompose into disjoint sets.
Boundary issues: The measure of a set and its closure may differ if the boundary has positive measure (though this is rare for "nice" sets).
Infinite sums: When summing measures of disjoint sets, ensure the series converges. If it diverges, the measure is infinite.

These techniques become especially important when computing integrals, where we need to measure level sets and other geometric objects.

Remark 1.6: Comparison with Other Measures

Lebesgue measure is not the only way to assign "size" to sets. Other important measures include:

Counting measure: For finite sets, the number of elements
Hausdorff measure: For fractals and lower-dimensional sets
Probability measures: In probability theory, measures with total mass 1
Dirac measures: Concentrated at a single point

However, Lebesgue measure is the most natural choice for $\mathbb{R}^n$ because it:

Agrees with geometric volume for simple sets
Is translation invariant
Is countably additive
Has good approximation properties

Example 1.16: Outer Measure of a Fractal-Like Set

Problem: Consider the set $E = \bigcap_{n=1}^{\infty} E_n$ where $E_1 = [0,1]$ and $E_{n+1}$ is obtained by removing the middle third of each interval in $E_n$ (the Cantor set construction). Compute $m^*(E)$ .

Solution:

This is the Cantor set construction. At step $n$ , we have $2^n$ intervals, each of length $\frac{1}{3^n}$ , so $m(E_n) = \left(\frac{2}{3}\right)^n$ .

Since $E_1 \supseteq E_2 \supseteq \cdots$ and $m(E_1) = 1 < \infty$ , by continuity from above (Theorem 1.8),

m^*(E) = m(E) = \lim_{n \to \infty} m(E_n) = \lim_{n \to \infty} \left(\frac{2}{3}\right)^n = 0

Therefore, the Cantor set has measure zero, as shown in Example 1.4. This demonstrates that uncountable sets can have measure zero.

Example 1.17: Measure of Intersection of Nested Sets

Problem: Let $E_n = (0, 1/n)$ for $n \in \mathbb{N}$ . Compute $m\left(\bigcap_{n=1}^{\infty} E_n\right)$ .

Solution:

Note that $E_1 \supseteq E_2 \supseteq \cdots$ and $\bigcap_{n=1}^{\infty} E_n = \bigcap_{n=1}^{\infty} (0, 1/n) = \emptyset$ .

However, we cannot directly apply continuity from above because $m(E_1) = 1$ is finite, but the intersection is empty.

Actually, we can apply Theorem 1.8: since $m(E_1) = 1 < \infty$ and $E_n \downarrow \emptyset$ , we have

m\left(\bigcap_{n=1}^{\infty} E_n\right) = \lim_{n \to \infty} m(E_n) = \lim_{n \to \infty} \frac{1}{n} = 0

This confirms that the intersection has measure zero, which is consistent with it being empty.

Corollary 1.6: Measure of Symmetric Differences

For measurable sets $E$ and $F$ , the symmetric difference $E \triangle F = (E \setminus F) \cup (F \setminus E)$ is measurable, and

m(E \triangle F) = m(E) + m(F) - 2m(E \cap F)

This follows from Theorem 1.11 and the fact that $E \setminus F$ and $F \setminus E$ are disjoint measurable sets.

Remark 1.8: Advanced Topics and Extensions

The theory of Lebesgue measure extends in several important directions:

Hausdorff measure: For sets of fractional dimension (fractals), Hausdorff measure provides a more appropriate notion of "size". The Hausdorff dimension of the Cantor set is $\log_3 2$ .
Signed measures: Measures that can take negative values, important for the Radon-Nikodym theorem and Lebesgue decomposition.
Complex measures: Measures taking complex values, used in harmonic analysis and Fourier theory.
Abstract measure spaces: The theory extends to general measure spaces, not just $\mathbb{R}^n$ , enabling applications to probability, ergodic theory, and functional analysis.
Product measures: Measures on product spaces, leading to Fubini's theorem for multiple integrals.

These extensions demonstrate the power and generality of measure theory as a foundation for modern analysis.

Corollary 1.7: Approximation by Elementary Sets

For any measurable set $E$ with $m(E) < \infty$ and any $\epsilon > 0$ , there exists a finite union of disjoint closed rectangles $R$ such that $m(E \triangle R) < \epsilon$ .

This follows from regularity (Theorem 1.4) and the fact that open sets can be approximated by finite unions of rectangles.

Remark 1.2: Key Insights

Key takeaways from this course:

Outer measure provides a systematic way to assign "size" to any subset of $\mathbb{R}^n$
The Carathéodory criterion identifies exactly which sets should be considered measurable
Measurable sets form a σ-algebra, ensuring closure under countable operations
On measurable sets, outer measure becomes countably additive (Lebesgue measure)
Regularity allows approximation of measurable sets by simpler sets (open/closed)
Null sets play a crucial role in "almost everywhere" properties
Borel sets are measurable, but there exist non-Borel measurable sets
Continuity properties (from above/below) are essential for limit operations
Non-measurable sets exist (with AC), but are pathological and not encountered in practice

Practice Quiz

Measure Foundations & Outer Measure

Questions

Correct

Accuracy

What is the definition of outer measure

m^*(E)

for a set

E \subseteq \mathbb{R}^n

Easy

Not attempted

A set

E

is measurable (in the sense of Carathéodory) if and only if:

Medium

Not attempted

Which of the following is NOT a property of outer measure?

Medium

Not attempted

A null set is defined as:

Easy

Not attempted

The regularity property states that for a measurable set

E

Hard

Not attempted

E

is measurable, then

E^c

is:

Easy

Not attempted

The outer measure of a countable set in

\mathbb{R}

is:

Medium

Not attempted

Which property distinguishes Lebesgue measure from outer measure?

Hard

Not attempted

E_1, E_2, \ldots

are measurable sets, then

\bigcup_{i=1}^{\infty} E_i

is:

Medium

Not attempted

The Carathéodory extension theorem shows that:

Hard

Not attempted

Frequently Asked Questions

Why do we need outer measure instead of just using volume for all sets?

Not all sets have a well-defined 'volume' in the geometric sense. Outer measure provides a systematic way to assign a 'size' to any subset of ℝⁿ, even pathological ones. It generalizes the intuitive notion of volume to sets that may be very irregular.

What is the relationship between outer measure and Lebesgue measure?

Lebesgue measure is the restriction of outer measure to the σ-algebra of measurable sets. On measurable sets, outer measure becomes countably additive and is called Lebesgue measure. Outer measure itself is only subadditive.

Why is the Carathéodory criterion important?

The Carathéodory criterion identifies exactly which sets should be considered 'measurable'. It ensures that on measurable sets, outer measure has the crucial property of countable additivity, which is essential for integration theory.

Are all open sets measurable?

Yes, all open sets in ℝⁿ are measurable. In fact, the Borel σ-algebra (generated by open sets) is contained in the σ-algebra of measurable sets. This means Borel sets are measurable, but there are also non-Borel measurable sets.

What is a null set and why are they important?

A null set is a set with outer measure zero. Null sets are important because properties that hold 'almost everywhere' (i.e., except on a null set) are sufficient for many purposes in analysis. For example, two functions that differ only on a null set have the same integral.

How does regularity help in practice?

Regularity allows us to approximate measurable sets by simpler sets (open or closed). This is crucial for proofs and computations, as we can often reduce problems about arbitrary measurable sets to problems about open or closed sets, which are easier to work with.

Can a set have positive outer measure but zero Lebesgue measure?

No. If a set has positive outer measure, it must be measurable (or we wouldn't be able to define its Lebesgue measure properly). If it's measurable and has positive outer measure, then its Lebesgue measure equals its outer measure and is positive.