# 7.6: Measure Spaces. More on Outer Measures

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I. In §5, we considered premeasure spaces, stressing mainly the idea of $$\sigma$$-subadditivity (Note 5 in §5). Now we shall emphasize $$\sigma$$-additivity.

## Definition 1

A premeasure

$m : \mathcal{M} \rightarrow[0, \infty]$

is called a measure (in $$S$$) iff $$\mathcal{M}$$ is a $$\sigma$$-ring (in $$S$$), and $$m$$ is $$\sigma$$-additive on $$\mathcal{M}.$$

If so, the system

$(S, \mathcal{M}, m)$

is called a measure space; $$m X$$ is called the measure of $$X \in \mathcal{M}$$; $$\mathcal{M}$$-sets are called $$m$$-measurable sets.

Note that $$m$$ is nonnegative and $$m \emptyset=0,$$ as $$m$$ is a premeasure (Definition 2 in §5).

## Corollary $$\PageIndex{1}$$

Measures are $$\sigma$$-additive, $$\sigma$$-subadditive, monotone, and continuous.

Proof

Use Corollary 2 in §5 and Theorem 2 in §4, noting that $$\mathcal{M}$$ is a $$\sigma$$-ring.$$\quad \square$$

## Corollary $$\PageIndex{2}$$

In any measure space $$(S, \mathcal{M}, m),$$ the union and intersection of any sequence of $$m$$-measurable sets is $$m$$-measurable itself. So also is $$X-Y$$ if $$X, Y \in \mathcal{M}.$$

This is obvious since $$\mathcal{M}$$ is a $$\sigma$$-ring.

As measures and other premeasures are understood to be $$\geq 0,$$ we often write

$m : \mathcal{M} \rightarrow E^{*}$

for

$m : \mathcal{M} \rightarrow[0, \infty].$

We also briefly say "measurable" for "$$m$$-measurable."

Note that $$\emptyset \in \mathcal{M},$$ but not always $$S \in \mathcal{M}$$.

## Examples

(a) The volume of intervals in $$E^{n}$$ is a $$\sigma$$-additive premeasure, but not a measure since its domain (the intervals) is not a $$\sigma$$-ring.

(b) Let $$\mathcal{M}=2^{S}.$$ Define

$(\forall X \subseteq S) \quad m X=0.$

Then $$m$$ is trivially a measure (the zero-measure). Here each set $$X \subseteq S$$ is measurable, with $$m X=0$$.

(c) Let again $$\mathcal{M}=2^{S}.$$ Let $$m X$$ be the number of elements in $$X,$$ if finite, and $$m X=\infty$$ otherwise.

Then $$m$$ is a measure ("counting measure"). Verify!

(d) Let $$\mathcal{M}=2^{S}.$$ Fix some $$p \in S.$$ Let

$m X=\left\{\begin{array}{ll}{1} & {\text { if } p \in X}, \\ {0} & {\text { otherwise }}.\end{array}\right.$

Then $$m$$ is a measure (it describes a "unit mass" concentrated at $$p$$).

(e) A probability space is a measure space $$(S, \mathcal{M}, m$$), with

$S \in \mathcal{M} \text { and } m S=1.$

In probability theory, measurable sets are called events; $$m X$$ is called the probability of $$X,$$ often denoted by $$p X$$ or similar symbols.

In Examples (b), (c), and (d),

$\mathcal{M}=2^{S} \text { (all subsets of } S \text{).}$

More often, however,

$\mathcal{M} \neq 2^{S},$

i.e., there are nonmeasurable sets $$X \subseteq S$$ for which $$m X$$ is not defined.

Of special interest are sets $$X \in \mathcal{M},$$ with $$m X=0,$$ and their subsets. We call them $$m$$-null or null sets. One would like them to be measurable, but this is not always the case for subsets of $$X.$$

This leads us to the following definition.

## Definition 2

A measure $$m : \mathcal{M} \rightarrow E^{*}$$ is called complete iff all null sets (subsets of sets of measure zero) are measurable.

We now develop a general method for constructing complete measures.

II. From §5 (Note 5) recall that an outer measure in $$S$$ is a $$\sigma$$-subadditive premeasure defined on all of $$2^{S}$$ (even if it is not derived via Definition 3 in §5). In Examples (b), (c), and (d), $$m$$ is both a measure and an outer measure. (Why?)

An outer measure

$m^{*} : 2^{S} \rightarrow E^{*}$

need not be additive; but consider this fact:

$\text { Any set } A \subseteq S \text { splits } S \text { into two parts: } A \text { itself and }-A.$

It also splits any other set $$X$$ into $$X \cap A$$ and $$X-A;$$ indeed,

$X=(X \cap A) \cup(X-A) \text { (disjoint).}$

We want to single out those sets $$A$$ for which $$m^{*}$$ behaves "additively," i.e., so that

$m^{*} X=m^{*}(X \cap A)+m^{*}(X-A).$

This motivates our next definition.

## Definition 3

Given an outer measure $$m^{*} : 2^{S} \rightarrow E^{*}$$ and a set $$A \subseteq S,$$ we say that $$A$$ is $$m^{*}$$-measurable iff all sets $$X \subseteq S$$ are split "additively" by $$A;$$ that is,

$(\forall X \subseteq S) \quad m^{*} X=m^{*}(X \cap A)+m^{*}(X-A).$

As is easily seen (see Problem 1), this is equivalent to

$(\forall X \subseteq A)(\forall Y \subseteq-A) \quad m^{*}(X \cup Y)=m^{*} X+m^{*} Y.$

The family of all $$m^{*}$$-measurable sets is usually denoted by $$\mathcal{M}^{*}.$$ The system $$\left(S, \mathcal{M}^{*}, m^{*}\right)$$ is called an outer measure space.

Note 1. Definition 3 applies to outer measures only. For measures, "$$m$$-measurable" means simply "member of the domain of $$m$$" (Definition 1).

Note 2. In (1) and (2), we may equivalently replace the equality sign $$(=)$$ by $$(\geq).$$ Indeed, $$X$$ is covered by

$\{X \cap A, X-A\},$

and $$X \cup Y$$ is covered by $$\{X, Y\};$$ so the reverse inequality $$(\leq)$$ anyway holds, by subadditivity.

Our main objective is to prove the following fundamental theorem.

## Theorem $$\PageIndex{1}$$

In any outer measure space

$\left(S, \mathcal{M}^{*}, m^{*}\right),$

the family $$\mathcal{M}^{*}$$ of all $$m^{*}$$-measurable sets is a $$\sigma$$-field in $$S,$$ and $$m^{*},$$ when restricted to $$\mathcal{M}^{*},$$ is a complete measure (denoted by $$m$$ and called the $$m^{*}$$-induced measure; so $$m^{*}=m$$ on $$\mathcal{M}^{*}$$).

Proof

We split the proof into several steps (lemmas).

## lemma 1

$$\mathcal{M}^{*}$$ is closed under complementation:

$\left(\forall A \in \mathcal{M}^{*}\right) \quad-A \in \mathcal{M}^{*}.$

Indeed, the measurability criterion (2) is same for $$A$$ and $$-A$$ alike.

## lemma 2

$$\emptyset$$ and $$S$$ are $$\mathcal{M}^{*}$$ sets. So are all sets of outer measure 0.

Proof

Let $$m^{*} A=0.$$ To prove $$A \in \mathcal{M}^{*},$$ use (2) and Note 2.

Thus take any $$X \subseteq A$$ and $$Y \subseteq-A.$$ Then by monotonicity,

$m^{*} X \leq m^{*} A=0$

and

$m^{*} Y \leq m^{*}(X \cup Y).$

Thus

$m^{*} X+m^{*} Y=0+m^{*} Y \leq m^{*}(X \cup Y),$

as required.

In particular, as $$m^{*} \emptyset=0, \emptyset$$ is $$m^{*}$$-measurable $$\left(\emptyset \in \mathcal{M}^{*}\right)$$.

So is $$S$$ (the complement of $$\emptyset)$$ by Lemma 1.$$\quad \square$$

## lemma 3

$$\mathcal{M}^{*}$$ is closed under finite unions:

$\left(\forall A, B \in \mathcal{M}^{*}\right) \quad A \cup B \in \mathcal{M}^{*}.$

Proof

This time we shall use formula (1). By Note 2, it suffices to show that

$(\forall X \subseteq S) \quad m^{*} X \geq m^{*}(X \cap(A \cup B))+m^{*}(X-(A \cup B)).$

Fix any $$X \subseteq S;$$ as $$A \in \mathcal{M}^{*},$$ we have

$m^{*} X=m^{*}(X \cap A)+m^{*}(X-A).$

Similarly, as $$B \in \mathcal{M}^{*},$$ we have (replacing $$X$$ by $$X-A$$ in (1))

\begin{aligned} m^{*}(X-A) &=m^{*}((X-A) \cap B)+m^{*}(X-A-B) \\ &=m^{*}(X \cap-A \cap B)+m^{*}(X-(A \cup B)), \end{aligned}

since

$X-A=X \cap-A$

and

$X-A-B=X-(A \cup B).$

Combining (4) with (3), we get

$m^{*} X=m^{*}(X \cap A)+m^{*}(X \cap-A \cap B)+m^{*}(X-(A \cup B)).$

Now verify that

$(X \cap A) \cup(X \cap-A \cap B) \supseteq X \cap(A \cup B).$

As $$m$$ is subadditive, this yields

$m^{*}(X \cap A)+m^{*}(X \cap-A \cap B) \geq m^{*}(X \cap(A \cup B)).$

Combining with (5), we get

$m^{*} X \geq m^{*}(X \cap(A \cup B))+m^{*}(X-(A \cup B)),$

so that $$A \cup B \in \mathcal{M}^{*},$$ indeed.$$\quad \square$$

Induction extends Lemma 3 to all finite unions of $$\mathcal{M}^{*}$$-sets.

Note that by Problem 3 in §3, $$\mathcal{M}^{*}$$ is a set field, hence surely a ring. Thus Corollary 1 in §1 applies to it. (We use it below.)

## lemma 4

Let

$X_{k} \subseteq A_{k} \subseteq S, \quad k=0,1,2, \ldots,$

with all $$A_{k}$$ pairwise disjoint.

Let $$A_{k} \in \mathcal{M}^{*}$$ for $$k \geq 1.$$ ($$A_{0}$$ and the $$X_{k}$$ need not be $$\mathcal{M}^{*}$$-sets.) Then

$m^{*}\left(\bigcup_{k=0}^{\infty} X_{k}\right)=\sum_{k=0}^{\infty} m^{*} X_{k}.$

Proof

We start with two sets, $$A_{0}$$ and $$A_{1};$$ so

$A_{1} \in \mathcal{M}^{*}, A_{0} \cap A_{1}=\emptyset, X_{0} \subseteq A_{0}, \text { and } X_{1} \subseteq A_{1}.$

As $$A_{0} \cap A_{1}=\emptyset,$$ we have $$A_{0} \subseteq-A_{1};$$ hence also $$X_{0} \subseteq-A_{1}$$.

since $$A_{1} \in \mathcal{M}^{*},$$ we use formula (2), with

$X=X_{1} \subseteq A_{1} \text { and } Y=X_{0} \subseteq-A,$

to obtain

$m^{*}\left(X_{0} \cup X_{1}\right)=m^{*} X_{0}+m^{*} X_{1}.$

Thus (6) holds for two sets.

Induction now easily yields

$(\forall n) \sum_{k=0}^{n} m^{*} X_{k}=m^{*}\left(\bigcup_{k=0}^{n} X_{k}\right) \leq m^{*}\left(\bigcup_{k=0}^{\infty} X_{k}\right)$

by monotonicity of $$m^{*}.$$ Now let $$n \rightarrow \infty$$ and pass to the limit to get

$\sum_{k=0}^{\infty} m^{*} X_{k} \leq m^{*}\left(\bigcup_{k=0}^{\infty} X_{k}\right).$

As $$\bigcup X_{k}$$ is covered by the $$X_{k},$$ the $$\sigma$$-subadditivity of $$m^{*}$$ yields the reverse inequality as well. Thus (6) is proved.$$\quad \square$$

Proof of Theorem 1. As we noted, $$\mathcal{M}^{*}$$ is a field. To show that it is also closed under countable unions (a $$\sigma$$-field), let

$U=\bigcup_{k=1}^{\infty} A_{k}, \quad A_{k} \in \mathcal{M}^{*}.$

We have to prove that $$U \in \mathcal{M}^{*};$$ or by (2) and Note 2,

$(\forall X \subseteq U)(\forall Y \subseteq-U) \quad m^{*}(X \cup Y) \geq m^{*} X+m^{*} Y.$

We may safely assume that the $$A_{k}$$ are disjoint. (If not, replace them by disjoint sets $$B_{k} \in \mathcal{M}^{*},$$ as in Corollary 1 §1.)

To prove (7), fix any $$X \subseteq U$$ and $$Y \subseteq-U,$$ and let

$X_{k}=X \cap A_{k} \subseteq A_{k},$

$$A_{0}=-U,$$ and $$X_{0}=Y,$$ satisfying all assumptions of Lemma 4. Thus by (6), writing the first term separately, we have

$m^{*}\left(Y \cup \bigcup_{k=1}^{\infty} X_{k}\right)=m^{*} Y+\sum_{k=1}^{\infty} m^{*} X_{k}.$

But

$\bigcup_{k=1}^{\infty} X_{k}=\bigcup_{k=1}^{\infty}\left(X \cap A_{k}\right)=X \cap \bigcup_{k=1}^{\infty} A_{k}=X \cap U=X$

(as $$X \subseteq U).$$ Also, by $$\sigma$$-subadditivity,

$\sum m^{*} X_{k} \geq m^{*} \bigcup X_{k}=m^{*} X.$

Therefore, (8) implies (7); so $$\mathcal{M}^{*}$$ is a $$\sigma$$-field.

Moreover, $$m^{*}$$ is $$\sigma$$-additive on $$\mathcal{M}^{*},$$ as follows from Lemma 4 by taking

$X_{k}=A_{k} \in \mathcal{M}^{*}, A_{0}=\emptyset.$

Thus $$m^{*}$$ acts as a measure on $$\mathcal{M}^{*}$$.

By Lemma 2, $$m^{*}$$ is complete; for if $$X$$ is "null" ($$X \subseteq A$$ and $$m^{*} A=0$$), then $$m^{*} X=0;$$ so $$X \in \mathcal{M}^{*},$$ as required.

Thus all is proved.$$\quad \square$$

We thus have a standard method for constructing measures: From a premeasure

$\mu : \mathcal{C} \rightarrow E^{*}$

in $$S,$$ we obtain the $$\mu$$-induced outer measure

$m^{*} : 2^{S} \rightarrow E^{*} \text{ (§5);}$

this, in turn, induces a complete measure

$m : \mathcal{M}^{*} \rightarrow E^{*}.$

But we need more: We want $$m$$ to be an extension of $$\mu,$$ i.e.,

$m=\mu \text { on } \mathcal{C},$

with $$\mathcal{C} \subseteq \mathcal{M}^{*}$$ (meaning that all $$\mathcal{C}$$-sets are $$m^{*}$$-measurable). We now explore this question.

## lemma 5

Let $$(S, \mathcal{C}, \mu)$$ and $$m^{*}$$ be as in Definition 3 of §5. Then for a set $$A \subseteq S$$ to be $$m^{*}$$-measurable, it suffices that

$m^{*} X \geq m^{*}(X \cap A)+m^{*}(x-A) \quad \text {for all } X \in \mathcal{C}.$

Proof

We must show that (9) holds for any $$X \subseteq S,$$ even not a $$\mathcal{C}$$-set.

This is trivial if $$m^{*} X=\infty.$$ Thus assume $$m^{*} X<\infty$$ and fix any $$\varepsilon>0$$.

By Note 3 in §5, $$X$$ must have a basic covering $$\left\{B_{n}\right\} \subseteq \mathcal{C}$$ so that

$X \subseteq \bigcup_{n} B_{n}$

and

$m^{*} X+\varepsilon>\sum \mu B_{n} \geq \sum m^{*} B_{n}.$

(Explain!)

Now, as $$X \subseteq \cup B_{n},$$ we have

$X \cap A \subseteq \bigcup B_{n} \cap A=\bigcup\left(B_{n} \cap A\right).$

Similarly,

$X-A=X \cap-A \subseteq \bigcup\left(B_{n}-A\right).$

Hence, as $$m^{*}$$ is $$\sigma$$-subadditive and monotone, we get

\begin{aligned} m^{*}(X \cap A)+m^{*}(X-A) & \leq m^{*}\left(\bigcup\left(B_{n} \cap A\right)\right)+m^{*}\left(\bigcup\left(B_{n}-A\right)\right) \\ & \leq \sum\left[m^{*}\left(B_{n} \cap A\right)+m^{*}\left(B_{n}-A\right)\right]. \end{aligned}

But by assumption, (9) holds for any $$\mathcal{C}$$-set, hence for each $$B_{n}.$$ Thus

$m^{*}\left(B_{n} \cap A\right)+m^{*}\left(B_{n}-A\right) \leq m^{*} B_{n},$

and (11) yields

$m^{*}(X \cap A)+m^{*}(X-A) \leq \sum\left[m^{*}\left(B_{n} \cap A\right)+m^{*}\left(B_{n}-A\right)\right] \leq \sum m^{*} B_{n}.$

Therefore, by (10),

$m^{*}(X \cap A)+m^{*}(X-A) \leq m^{*} X+\varepsilon.$

Making $$\varepsilon \rightarrow 0,$$ we prove (10) for any $$X \subseteq S,$$ so that $$A \in \mathcal{M}^{*},$$ as required.$$\quad \square$$

## Theorem $$\PageIndex{2}$$

Let the premeasure

$\mu : \mathcal{C} \rightarrow E^{*}$

be $$\sigma$$-additive on $$\mathcal{C}, a$$ semiring in $$S.$$ Let $$m^{*}$$ be the $$\mu$$-induced outer measure, and

$m : \mathcal{M}^{*} \rightarrow E^{*}$

be the $$m^{*}$$-induced measure. Then

(i) $$\mathcal{C} \subseteq \mathcal{M}^{*}$$ and

(ii) $$\mu=m^{*}=m$$ on $$\mathcal{C}$$.

Thus $$m$$ is a $$\sigma$$-additive extension of $$\mu$$ (called its Lebesgue extension) to $$\mathcal{M}^{*}$$.

Proof

By Corollary 2 in §5, $$\mu$$ is also $$\sigma$$-subadditive on the semiring $$\mathcal{C}.$$ Thus by Theorem 2 in §5, $$\mu=m^{*}$$ on $$\mathcal{C}.$$

To prove that $$\mathcal{C} \subseteq \mathcal{M}^{*},$$ we fix $$A \in \mathcal{C}$$ and show that $$A$$ satisfies (9), so that $$A \in \mathcal{M}^{*}.$$

Thus take any $$X \in \mathcal{C}.$$ As $$\mathcal{C}$$ is a semiring, $$X \cap A \in \mathcal{C}$$ and

$X-A=\bigcup_{k=1}^{n} A_{k} \text { (disjoint)}$

for some sets $$A_{k} \in \mathcal{C}.$$ Hence

\begin{aligned} m^{*}(X \cap A)+m^{*}(X-A) &=m^{*}(X \cap A)+m^{*} \bigcup_{k=1}^{n} A_{k} \\ & \leq m^{*}(X \cap A)+\sum_{k=1}^{n} m^{*} A_{k}. \end{aligned}

As

$X=(X \cap A) \cup(X-A)=(X \cap A) \cup \bigcup A_{k} \text { (disjoint),}$

the additivity of $$\mu$$ and the equality $$\mu=m^{*}$$ on $$\mathcal{C}$$ yield

$m^{*} X=m^{*}(X \cap A)+\sum_{k=1}^{n} m^{*} A_{k}.$

Hence by (12),

$m^{*} X \geq m^{*}(X \cap A)+m^{*}(X-A);$

so by Lemma 5, $$A \in \mathcal{M}^{*},$$ as required.

Also, by definition, $$m=m^{*}$$ on $$\mathcal{M}^{*},$$ hence on $$\mathcal{C}.$$ Thus

$\mu=m^{*}=m \text { on } \mathcal{C},$

as claimed.$$\quad \square$$

Note 3. In particular, Theorem 2 applies if

$\mu : \mathcal{M} \rightarrow E^{*}$

is a measure (so that $$\mathcal{C}=\mathcal{M}$$ is even a $$\sigma$$-ring).

Thus any such $$\mu$$ can be extended to a complete measure $$m$$ (its Lebesgue extension) on a $$\sigma$$-field

$\mathcal{M}^{*} \supseteq \mathcal{M}$

via the $$\mu$$-induced outer measure (call it $$\mu^{*}$$ this time), with

$\mu^{*}=m=\mu \text { on } \mathcal{M}.$

Moreover,

$\mathcal{M}^{*} \supseteq \mathcal{M} \supseteq \mathcal{M}_{\sigma}$

(see Note 2 in §3); so $$\mu^{*}$$ is $$\mathcal{M}$$-regular and $$\mathcal{M}^{*}$$-regular (Theorem 3 of §5).

Note 4. A reapplication of this process to $$m$$ does not change $$m$$ (Problem 16).

This page titled 7.6: Measure Spaces. More on Outer Measures is shared under a CC BY license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) .