Outer Measures

From Measures to Outer Measures

A measure assigns sizes to sets in a specified sigma-algebra. But when we begin with an arbitrary subset of the real line, we do not yet know whether it belongs to the Borel sigma-algebra or to a larger measurable collection. Outer measure provides a way to assign an extended nonnegative size to every subset, whether or not that subset is measurable. The construction uses covers: a set is small if it can be covered by intervals whose total length is small.

This distinction will matter in the next tutorial, where we will use outer measure to identify which sets are measurable. For now, the central task is to define outer measure carefully and verify that the covering construction has the properties required of it. We use $[0,\infty]$ for the extended nonnegative real numbers, so the value $+\infty$ is allowed.

Definition: Let $X$ be a set. A function $\mu^*:\mathcal{P}(X)\to[0,\infty]$ is an outer measure if:

$\mu^*(\varnothing)=0$;
if $A\subseteq B\subseteq X$, then $\mu^*(A)\leq\mu^*(B)$;
for every sequence of subsets $A_1,A_2,\ldots$ of $X$, $\mu^*(\bigcup_{n=1}^{\infty}A_n)\leq\sum_{n=1}^{\infty}\mu^*(A_n)$.

The last property is countable subadditivity. An outer measure is defined on the entire power set $\mathcal{P}(X)$, unlike a measure, whose domain is a sigma-algebra. The axioms look similar to familiar properties of measures, but countable additivity is not required: outer measure is designed to give an upper bound on the size of a union, even when the sets involved are not yet known to be measurable.

Lebesgue Outer Measure on the Real Line

For an interval $I$ in $\mathbb{R}$, its length, denoted $|I|$, is the difference between its right and left endpoints. Define the Lebesgue outer measure of $E\subseteq\mathbb{R}$ by considering all finite or countable collections of open intervals that cover $E$. The covering cost is the sum of the intervals’ lengths, and the outer measure is the least possible cost in the sense of an infimum.

Definition: For $E\subseteq\mathbb{R}$, its Lebesgue outer measure is

m^*(E)=\inf\left\{\sum_{n}|I_n|: E\subseteq\bigcup_n I_n,\ \text{each }I_n\text{ is an open interval, and the collection is finite or countable}\right\}.

The empty set has the empty cover, whose total cost is zero.

The infimum need not be achieved by any particular cover. It is enough that there are covers with costs arbitrarily close to the infimum. Also, an open interval can itself be used as a covering interval. A closed interval $[a,b]$, however, cannot itself be used as a covering interval under this definition, which allows only open intervals. It can be covered by slightly larger open intervals.

Why the Covering Construction Is an Outer Measure

Theorem (Lebesgue Outer Measure Is an Outer Measure): The function $m^*$ defined by countable open-interval covers satisfies the three outer-measure axioms on $\mathbb{R}$.

Proof. The empty set has a cover of cost zero, so $m^*(\varnothing)=0$. Now suppose $A\subseteq B$. Every open-interval cover of $B$ is also a cover of $A$. Thus the infimum of the costs of covers of $A$ is no greater than the infimum of the costs of covers of $B$, giving $m^*(A)\leq m^*(B)$.

It remains to prove countable subadditivity. Let $E_1,E_2,\ldots\subseteq\mathbb{R}$. If $\sum_{n=1}^{\infty}m^*(E_n)=\infty$, the required inequality holds because its right-hand side is $+\infty$. Suppose instead that this sum is finite. Then every $m^*(E_n)$ is finite. Fix $\varepsilon>0$. By the definition of infimum, for each $n$ there is a finite or countable open-interval cover of $E_n$ whose total length is less than $m^*(E_n)+\varepsilon/2^n$.

Collect all the intervals in all these covers. This is a countable collection: it is a countable union of finite or countable collections. It covers $\bigcup_{n=1}^{\infty}E_n$, and its total length is at most the sum of the individual covering costs. Therefore

m^*\left(\bigcup_{n=1}^{\infty}E_n\right) \leq \sum_{n=1}^{\infty}\left(m^*(E_n)+\frac{\varepsilon}{2^n}\right) =\sum_{n=1}^{\infty}m^*(E_n)+\varepsilon.

This holds for every $\varepsilon>0$, so $m^*(\bigcup_n E_n)\leq\sum_n m^*(E_n)$. All three axioms are proved. $\square$

The argument uses the infimum in an important way: a cover need not attain the outer measure, but for any positive error we can choose a cover whose cost is within that error. Assigning errors with a summable total lets us combine infinitely many approximate covers without accumulating an uncontrolled amount of extra length.

Outer Measure of an Interval

To calculate outer measure exactly for a closed interval, the upper bound comes from a slightly larger open interval. For the lower bound, we need the elementary fact that a finite collection of intervals covering a closed interval must have total length at least that interval’s length.

Here is a direct justification of that fact. Given finitely many open intervals covering $[a,b]$, list $a$, $b$, and every endpoint of those intervals that lies between $a$ and $b$, in increasing order, omitting repetitions. These points divide $[a,b]$ into finitely many adjacent subintervals. Each subinterval’s interior lies in at least one of the covering intervals: membership cannot change in its interior, since no covering-interval endpoint lies there. Assign each subinterval to one such covering interval. The sum of the lengths assigned to any one interval is at most its length, because the assigned subintervals have disjoint interiors and lie within it. Summing over the covering intervals shows that their total length is at least $b-a$.

Theorem (Outer Measure of a Closed Interval): For real numbers $a\leq b$,

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

Proof. For any $\varepsilon>0$, the open interval $(a-\varepsilon/2,b+\varepsilon/2)$ covers $[a,b]$ and has length $b-a+\varepsilon$. Hence $m^*([a,b])\leq b-a+\varepsilon$ for every $\varepsilon>0$, and therefore $m^*([a,b])\leq b-a$.

For the reverse inequality, consider any finite or countable open-interval cover of $[a,b]$. The Heine–Borel Theorem says that $[a,b]$ is compact, so this open cover has a finite subcover. By the finite interval-cover argument above, the total length of that finite subcover is at least $b-a$. Since all lengths are nonnegative, the cost of the original cover is at least the cost of the finite subcover, and thus is also at least $b-a$. This holds for every cover, so taking the infimum gives $m^*([a,b])\geq b-a$. The two bounds prove the formula. When $a=b$, the same argument gives lower bound zero, and the upper-bound argument still applies. $\square$

Worked Example: The Outer Measure of a Closed Interval

Consider $E=[2,5]$. The theorem gives $m^*(E)=5-2=3$. To see the upper-bound cover concretely, for any $\varepsilon>0$ use the open interval $(2-\varepsilon/2,5+\varepsilon/2)$, whose length is $$ (5+\varepsilon/2)-(2-\varepsilon/2)=3+\varepsilon. $$ So the infimum of cover costs is at most $3$. No open interval cover can have total cost below $3$: compactness gives a finite subcover, and the finite interval-cover argument forces its total length to be at least $3$. The value $3$ is an infimum; the closed interval is not itself an allowed open covering interval.

Worked Example: The Outer Measure of an Open Interval

Let $E=(1,4)$. The open interval $(1,4)$ itself is an allowed cover and has length $3$, so $m^*(E)\leq3$. For a lower bound, take any $\delta$ with $0<\delta<3/2$. The closed interval $[1+\delta,4-\delta]$ is contained in $E$. By monotonicity and the closed-interval theorem, $$ m^*(E)\geq m^*([1+\delta,4-\delta])=(4-\delta)-(1+\delta)=3-2\delta. $$ This is true for every such $\delta$. If $m^*(E)<3$, choosing $\delta$ small enough that $3-2\delta>m^*(E)$ would contradict the inequality. Hence $m^*(E)\geq3$, and $m^*((1,4))=3$. The open interval and its closed counterpart have the same outer measure, although the covers used to establish their upper bounds are different.

Countable Sets and Infinite Outer Measure

A countable set can be covered by intervals with arbitrarily small total length. This fact is useful even before deciding whether the set is measurable: outer measure is defined for every subset of $\mathbb{R}$.

Theorem (Every Countable Set Has Outer Measure Zero): If $E\subseteq\mathbb{R}$ is countable, then $m^*(E)=0$.

Proof. If $E$ is empty, the result follows from $m^*(\varnothing)=0$. Otherwise, list its elements as $x_1,x_2,\ldots$, allowing repetitions if $E$ is finite. Given $\varepsilon>0$, cover $x_n$ by the open interval centered at $x_n$ of radius $\varepsilon/2^{n+1}$. This interval has length $\varepsilon/2^n$, and the intervals cover $E$. Their total length is

\sum_{n=1}^{\infty}\frac{\varepsilon}{2^n}=\varepsilon.

Thus $m^*(E)\leq\varepsilon$ for every $\varepsilon>0$. Since outer measure is nonnegative, $m^*(E)=0$. $\square$

Worked Example: Covering the Reciprocal Integers

Let $E=\{1/n:n\in\mathbb{N}\}$. For a fixed $\varepsilon>0$, cover $1/n$ by $$ I_n=\left(\frac{1}{n}-\frac{\varepsilon}{2^{n+1}},\frac{1}{n}+\frac{\varepsilon}{2^{n+1}}\right). $$ The point $1/n$ is the midpoint of $I_n$, and the interval has length $\varepsilon/2^n$. Since every element of $E$ appears in one of these intervals, they cover $E$, with total length $\sum_{n=1}^{\infty}\varepsilon/2^n=\varepsilon$. As this is possible for every positive $\varepsilon$, $m^*(E)=0$. This conclusion is about outer measure; the earlier result that countable subsets of the real line are Lebesgue null additionally concerns their measurability and measure.

Worked Example: The Outer Measure of the Real Line

The outer measure of $\mathbb{R}$ is infinite. For every positive integer $k$, $[-k,k]\subseteq\mathbb{R}$, so monotonicity and the closed-interval theorem give $$ m^*(\mathbb{R})\geq m^*([-k,k])=2k. $$ No finite number is at least $2k$ for every positive integer $k$. Therefore $m^*(\mathbb{R})=+\infty$. This argument uses only monotonicity and the interval calculation; it does not require choosing a particular cover of the whole line.

What Outer Measure Does—and Does Not—Tell Us

Outer measure gives every subset a size, but its value alone does not establish that a set is measurable. In particular, the outer-measure axioms do not assert that outer measure is additive on disjoint sets. Countable subadditivity provides an inequality in one direction; additivity requires a further condition on the set being measured. The next tutorial introduces that condition through Carathéodory measurability.

The covering definition is also sensitive to the allowed shapes of covers. Here the covers are open intervals, and their costs are lengths. A proof using a closed interval itself as an allowed covering interval would not follow this definition. Instead, the upper bound for a closed interval uses an open interval slightly larger than it; the lower bound comes from compactness and the finite-cover length argument.

Key takeaway: Lebesgue outer measure assigns every subset of $\mathbb{R}$ the infimum of the total lengths of its countable open-interval covers. It is an outer measure, gives $[a,b]$ outer measure $b-a$, and assigns outer measure zero to every countable set.

Check Your Understanding

Use the definitions and proofs in this tutorial to answer the following questions.

Which three axioms must a function satisfy to be an outer measure?
Why does the proof of countable subadditivity choose covers within $\varepsilon/2^n$ of each individual outer measure?
Why is compactness useful for proving the lower bound $m^*([a,b])\geq b-a$?
How can a countable set be covered with intervals whose total length is less than any prescribed positive number?
Why does knowing $m^*(E)$ not, by itself, prove that $E$ is measurable?

What You'll Learn