0.3Outer Measure on the Reals

Updated 2 Sep 2026Chapter (PDF)

This is not a section in Folland and content is drawn from a variety of places, such as Chapters 1-2 of MIRA.

[0.3.1]Definition(Length of Open Interval)#

We denote the length of an open interval by

\ell(I) = \begin{cases} b - a & I = (a, b), \\ 0 & I = \emptyset, \\ \infty & I = (-\infty, a), (a, \infty), \text{ or } (-\infty, \infty). \end{cases}

If we fully cover a set $A$ with open intervals in the most conservative way possible, the length of all intervals in the cover would seem to be a good notion of “size” for $A$ . And so it is.

[0.3.2]Definition(Outer Measure)#

The outer measure of a set $A \subseteq \bR$ is defined by

\mu^*(A) = \inf\Big\{ \sum_{k = 1}^\infty \ell(I_k): I_1, I_2, \dots \text{ are open intervals such that } A \subseteq \bigcup_{k = 1}^\infty I_k \Big\}.

The measure takes $\infty$ as a value—that is, $\mu^*: \bR \to [0, \infty]$ . Thus, for example, $\mu^*(\bR) = \infty$ . The outer measure has many desirable properties.

[0.3.3]Theorem(Countable Sets Are Measure Zero)#

Suppose $A \subseteq \bR$ is countable. Then $\mu^*(A) = 0$ .

Proof.

Suppose $A = \{a_1, a_2, \dots\}$ is a countable subset of $\bR$ . Fix some $\epsilon > 0$ . For each $k \in \bZ_+$ , define $I_k = (a_k - \epsilon/2^k, a_k + \epsilon/2^k)$ . Then note

\mu^*(A) \le \sum_{k = 1}^\infty \ell(I_k) = 2\epsilon \sum_{k = 1}^\infty \frac{1}{2^k} = 2\epsilon.

Since $\epsilon$ is arbitrary, we have that $\mu^*(A) = 0$ .

[0.3.4]Theorem(Outer Measure Preserves Set Inclusion Order)#

Suppose $A, B$ are subsets of $\bR$ such that $A \subseteq B$ . Then $\mu^*(A) \le \mu^*(B)$ .

[0.3.5]Theorem(Outer Measure is Translation Invariant)#

Let $A$ be any subset of $\bR$ and $t$ any arbitrary number in $\bR$ . Then $\mu^*(A) = \mu^*(t + A)$ .

Let $(1, 4)$ and $(3, 5)$ be open intervals, whose union is $(1, 5)$ . Then note that

\mu^*((1, 4) \cup (3, 5)) < \mu^*((1, 4)) + \mu^*((3, 5)),

because the LHS is $4$ and the RHS is $5$ . The intuition behind this is that the RHS counts $(3, 4)$ twice, whereas LHS counts it only once. We will generalize this to show the outer measure on $\bR$ is countably subadditive.

[0.3.6]Theorem(Outer Measure is Countably Subadditive)#

Suppose $A_1, A_2, \dots$ is a sequence of subsets of $\bR$ . Then

\mu^*\left( \bigcup_{k = 1}^\infty A_k \right) \le \sum_{k = 1}^\infty \mu^*(A_k).

Proof.

If $\mu^*(A_k) = \infty$ for any $k \in \bZ_+$ , the result would be immediate, so suppose $\mu^*(A_k) < \infty$ for every $k \in \bZ_+$ . Fix $\epsilon > 0$ . For each $k \in \bZ_+$ , define a sequence of open intervals $A_{1, k}, A_{2, k}, \dots$ that cover $A_k$ in such a way that

\sum_{j = 1}^\infty \ell(A_{j, k}) \le \frac{\epsilon}{2^k} + \mu^*(A_k).

Simply consider the collection $\{A_{j, k}\}_{j, k \in \bZ_+}$ . Note the sum over these indices yields

\sum_{k = 1}^\infty \sum_{j = 1}^\infty \ell(A_{j, k}) \le \epsilon + \sum_{k = 1}^\infty \mu^*(A_k).

Note that this collection also covers $\bigcup_{k = 1}^\infty A_k$ by construction, and so

\mu^*\left( \bigcup_{k = 1}^\infty A_k \right) \le \sum_{k = 1}^\infty \sum_{j = 1}^\infty \ell(A_{j, k}) \le \epsilon + \sum_{k = 1}^\infty \mu^*(A_k).

Since $\epsilon > 0$ is arbitrary, we have achieved that

\mu^*\left( \bigcup_{k = 1}^\infty A_k \right) \le \sum_{k = 1}^\infty \mu^*(A_k).

Countable subadditivity implies finite subadditivity, since we can let $A_k = \emptyset$ for every $k > n$ , then we have

\mu^*\left( \sum_{k = 1}^n A_k \right) \le \sum_{k = 1}^n \mu^*(A_k).

[0.3.7]Theorem(Outer Measure of Closed Intervals)#

$\mu^*([a, b]) = b - a$ .

Proof.

Note that, for any $\epsilon > 0$ , the sequence of open sets $(a - \epsilon, b + \epsilon), \emptyset, \emptyset, \dots$ forms an open cover for $[a, b]$ , and so

\mu^*([a, b]) \le (b - a) + 2\epsilon \le b - a.

We simply show that $\mu^*([a, b]) \ge b - a$ . Let $\{I_n\}_{n \in \bZ_+}$ be any countable open cover for $[a, b]$ . By compactness of $[a, b]$ , there exists a finite subcover $\{I_k\}_{k = 1}^n$ . We prove by induction that

\sum_{k = 1}^n \mu^*(I_k) \ge b - a,

and so $\sum_{k = 1}^\infty \mu^*(I_k) \ge \sum_{k = 1}^n \mu^*(I_k) \ge b - a$ , meaning $\mu^*([a, b]) = b - a$ .

If $[a, b] \subseteq I_1$ , then $\mu^*(I_1) \ge b - a$ , of course. Now suppose we know that, if $[a, b] = I_1 \cup I_2 \cup \cdots \cup I_n$ , then

\sum_{k = 1}^n \mu^*(I_k) \ge b - a.

Suppose $[a, b] \subseteq I_1 \cup \cdots \cup I_{n + 1}$ . Write $I_{n + 1} = (c, d)$ , and without loss of generality suppose $b \in I_{n + 1}$ . Note that if $c \le a$ , then $\mu^*(I_{n + 1}) \ge b + a$ , so we are done. Thus suppose $a < c < b < d$ , and so $[a, c] \subseteq I_1 \cup \cdots \cup I_n$ . By inductive hypothesis, note that

\sum_{k = 1}^{n + 1} \mu^*(I_k) \ge (c - a) + \mu^*(I_{n + 1}) \ge (c - a) + (d - c) = d - a \ge b - a.

Thus $\mu^*([a, b]) = b - a$ .

[0.3.8]Corollary(Intervals in

\bR

are Uncountable)#

Every interval in $\bR$ that contains at least two distinct elements is uncountable.

Proof.

Suppose $I$ is an interval containing $a, b \in \bR$ with $a < b$ . Then

\mu^*(I) \ge \mu^*([a, b]) = b - a \ne 0.

Countable sets are measure zero, and so $I$ is not countable.

The outer measure does have some unpleasant properties, however. We would expect that, for any disjoint sets $A, B$ , the outer measure of $A \cup B$ is equal to the sum of the outer measure of $A$ and $B$ respectively. Unfortunately, this is not true in general.

[0.3.9]Theorem(Nonadditivity of Outer Measure)#

There exist disjoint subsets $A, B \subseteq \bR$ such that $\mu^*(A \cup B) \ne \mu^*(A) + \mu^*(B)$ .

Proof.

Consider the set $[-1, 1]$ . For any $a, b \in [-1, 1]$ , define the equivalence relation

a \sim b \iff a - b \in \bQ.

Then the equivalence relation $\sim$ partitions $[-1, 1]$ in such a way that

[-1, 1] = \bigcup_{a \in [-1, 1]} [a]_\sim.

For each coset, choose exactly one point, then let $V$ be the set of all these points. That is, $V \cap [a]_\sim$ has a single, unique element for each $a \in [-1, 1]$ .

Let $r_1, r_2, \dots$ be a sequence of points such that $[-2, 2] \cap \bQ = \{r_1, r_2, \dots\}$ . Then note

[-1, 1] \subseteq \bigcup_{k = 1}^\infty \, (r_k + V).

Indeed, for any $a \in [-1, 1]$ , note there is some $v \in V \cap [a]_\sim$ . Then $a - v \in \bQ$ , meaning there is some $k \in \bZ_+$ such that $a = r_k + v$ , meaning $a \in r_k + V$ as desired. By order properties and countable subadditivity of the outer measure, note

2 = \mu^*([-1, 1]) \le \mu^*\left( \bigcup_{k = 1}^\infty r_k + V \right) \le \sum_{k = 1}^\infty \mu^*(r_k + V) = \sum_{k = 1}^\infty \mu^*(V).

Thus $\mu^*(V) > 0$ .

Let $n \in \bZ_+$ . Note that

\bigcup_{k = 1}^n \, (r_k + V) \subseteq [-3, 3],

since $V \subseteq [-1, 1]$ and $r_k \in [-2, 2]$ for each $k$ . Again, by order properties of the outer measure, note

\mu^*\left( \bigcup_{k = 1}^n \, (r_k + V) \right) \le \sum_{k = 1}^n \mu^*(V) \le 6.

But also note that

\sum_{k = 1}^n \mu^*(V) = n \mu^*(V).

Simply choose $n \in \bZ_+$ such that $n \mu^*(V) > 6$ , and so

\mu^*\left( \bigcup_{k = 1}^n \, (r_k + V) \right) \le \sum_{k = 1}^n \mu^*(r_k + V).

Note each $r_k + V$ is disjoint. Since $\mu^*(A_1 \cup \cdots \cup A_n) \ne \mu^*(A_1) + \dots + \mu^*(A_n)$ in general for disjoint $A_1, \dots, A_n$ , by induction we have $\mu^*(A \cup B) \ne \mu^*(A) + \mu^*(B)$ in general.

[0.3.10]Remark(Intuition behind Disproof of Additivity of Outer Measure)#

We start by taking $[-1, 1]$ and forming equivalence classes that group numbers that differ by a rational together. From each equivalence class, we choose one point, and we take the set of these points to be $V$ . Crucially, note that no elements of $V$ differ by a rational, otherwise two elements of $V$ would be in the same equivalence class, contrary to construction. We then make copies of the set, shifting $V$ by all the rationals in $[-2, 2]$ , written as a sequence $r_1, r_2, \dots$ . The translates $r_1 + V, r_2 + V, \dots$ are all disjoint from eachother, since if they overlapped, there would be two numbers in $V$ that differed by a rational. We will refer to $r_k + V$ as a “copy” of $V$ , since it is still isomorphic to $V$ .

If we take infinitely many of these copies and union them together, we note it forms a cover for $[-1, 1]$ . Thus the outer measure of this union is greater than $2$ , establishing the outer measure of $V$ is nonzero.

Now, we look at finitely many copies of $V$ , say $n$ of them. The union of $n$ many copies of $V$ lie in $[-3, 3]$ , and so the outer measure of this union is bounded above by $6$ . But the sum of $n$ many outer measures of $V$ is given by $n$ multiplied by the outer measure of $V$ , which is nonzero. This sum grows without bound, so there exists some $n$ such that $n\mu^*(V) > 6$ , completing the disproof.