# 7.6: Measure Spaces. More on Outer Measures

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- Elias Zakon
- University of Windsor via The Trilla Group (support by Saylor Foundation)

**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).