52 2 Abstract Measure Theory

2.5 Outer Measures

In Chapter 1, our goal was to extend the notion of measure from volumes of
boxes to arbitrary subsets of Rd . We could not do this in a “good” way for
all sets, so we constructed exterior Lebesgue measure on the measure space
(Rd , P(Rd )), and then restricted to the class of Lebesgue measurable subsets
in order to obtain Lebesgue measure. In contrast, this chapter deals with
abstract measures which, by definition, satisfy countable additivity on the σ-
algebra with the measure. But this leaves us with an important question: How
can we construct an abstract measure? The goal of this section and the next
is generalize the idea of deriving Lebesgue measure from exterior Lebesgue
measure to an abstract setting. Specifically, given a class of subsets of X that
we know how we want to measure, we will construct an exterior or outer
measure µ∗ that is defined on all subsets of X but is not a true measure,
and then construct a actual measure µ by restricting µ∗ to an appropriate
σ-algebra Σ of “measurable sets.”
It will be easier if we break this process into two parts:
(i) the construction of an outer measure µ∗ , and
(ii) the construction of a measure µ from the outer measure µ∗ .

It will also be conceptually more clear if we tackle the second item first. So,
in this section we show how to obtain a measure µ from an outer measure µ∗ ,
while in the next section we will consider the issue of constructing an outer
measure µ∗ from scratch.
Our first task is to precisely define outer measures. Considering the exam-
ple of exterior Lebesgue measure, it seems that the most important require-
ments are that an outer measure µ∗ should be defined on all subsets of X
and that it should satisfy countable subadditivity. However, there is another
important but hidden property: Monotonicity. Although countable additivity
implies monotonicity, countable subadditivity does not. Therefore we need to
include monotonicity as part of the definition of an outer measure.

Definition 2.28 (Outer Measure). Let X be a nonempty set. An outer

measure or exterior measure on X is a function µ∗ : P(X) → [0, ∞] that
satisfies the following conditions.
(a) µ∗ (∅) = 0.
(b) Monotonicity: If A ⊆ B, then µ∗ (A) ≤ µ∗ (B).
(c) Countable subadditivity: If E1 , E2 , . . . ⊆ X, then
S  X
µ∗ Ek ≤ µ∗ (Ek ). ♦
k k

c 2011 by Christopher Heil

 2.5 Outer Measures 53

Given an outer measure µ∗ , our goal is to create a σ-algebra Σ on X

such that µ∗ restricted to Σ will be countably additive. The elements of Σ
will be our “good sets,” the sets that are measurable with respect to µ∗ . But
how do we define measurability for an arbitrary outer measure? We might
not be given a topology on X, so we cannot define measurability in terms
of surrounding open sets, as we did for Lebesgue measure (Definition 1.19).
However, the formulation of Lebesgue measurability given by Carathéodory’s
Criterion (Theorem 1.45) does not involve topology, and as such it is the
appropriate motivation for the following definition.

Definition 2.29 (Measurable Set). Let µ∗ be an outer measure on a set X.

Then a set E ⊆ X is µ∗ -measurable, or simply measurable for short, if

∀ A ⊆ X, µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A\E). ♦

Remark 2.30. (a) It is often helpful to recall that µ∗ (A\E) = µ∗ (A ∩ E C ), so

we can equivalently write the condition for measurability as

∀ A ⊆ X, µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ E C ).

(b) The empty set is µ∗ -measurable by virtue of the fact that µ∗ (∅) = 0.
(c) By subadditivity, we always have the inequality

µ∗ (A) ≤ µ∗ (A ∩ E) + µ∗ (A\E).

Hence, to establish that a set E is measurable, we just have to prove that the
opposite inequality holds for every set A ⊆ X. ♦

The next exercise, which follows by combining subadditivity with mono-

tonicity, asks for a proof that every subset of X that has outer measure zero
is measurable.

Exercise 2.31. Let µ∗ be an outer measure on X. Show that if E ⊆ X and

µ∗ (E) = 0 then E is µ∗ -measurable. ♦

Now we will prove that the collection of µ-measurable sets forms a σ-

algebra on X, and µ∗ restricted to this σ-algebra is a complete measure on X.

Theorem 2.32 (Carathéodory’s Theorem). If µ∗ is an outer measure on

a set X, then the following statements hold.
(a) The family 
Σ = E ⊆ X : E is µ∗ -measurable
is a σ-algebra on X.
(b) µ = µ∗ |Σ is a measure on (X, Σ).
(c) µ is a complete measure. In fact, every set E ⊆ X that satisfies µ∗ (E) = 0
is µ∗ -measurable.
54 2 Abstract Measure Theory

Proof. (a) Σ is not empty since the empty set is µ∗ -measurable.

To show that Σ is closed under complements, fix any set E ∈ Σ and
let A be an arbitrary subset of X. Using the fact that E is µ∗ -measurable, we
compute that

µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A\E)
= µ∗ (A ∩ E) + µ∗ (A ∩ E C )
= µ∗ (A ∩ E C ) + µ∗ A ∩ (E C )C

= µ∗ (A ∩ E C ) + µ∗ (A\E C ).

Hence E C is µ∗ -measurable, so E C ∈ Σ.
Ultimately we want to show that Σ is closed under countable unions, but
to begin we will show that it is closed under finite unions. By induction, to
prove this it suffices to show that if E, F ∈ Σ then E ∪ F ∈ Σ. Choose any
set A ⊆ X. By subadditivity,

µ∗ (A) ≤ µ∗ (A ∩ (E ∪ F )) + µ∗ (A ∩ (E ∩ F )C ).

To prove the opposite inequality, note that since F is µ∗ -measurable and A∩E
is a subset of X, we have

µ∗ (A ∩ E ∩ F ) + µ∗ (A ∩ E ∩ F C ) = µ∗ (A ∩ E).

Similarly,

µ∗ (A ∩ E C ∩ F ) + µ∗ (A ∩ E C ∩ F C ) = µ∗ (A ∩ E C ).

Applying these equalities and using the fact that µ∗ is finitely subadditive,
we see that

µ∗ (A ∩ (E ∪ F )) + µ∗ (A ∩ (E ∪ F )C ).
= µ∗ (A ∩ E ∩ F ) ∪ (A ∩ E ∩ F C ) ∪ (A ∩ E C ∩ F ) + µ∗ (A ∩ (E ∪ F )).

≤ µ∗ (A ∩ E ∩ F ) + µ∗ (A ∩ E ∩ F C ) + µ∗ (A ∩ E C ∩ F ) + µ∗ (A ∩ E C ∩ F C )
= µ∗ (A ∩ E) + µ∗ (A ∩ E C )
= µ∗ (A),

where at the final equality we have used the fact that E is µ∗ -measurable.
Therefore E ∪ F ∈ Σ.
Before proceeding, we make an observation. Suppose that E, F ∈ Σ are
disjoint. Since E is measurable, we have

µ∗ (E ∪ F ) = µ∗ (E ∪ F ) ∩ E + µ∗ (E ∪ F ) ∩ E C = µ∗ (E) + µ∗ (F ).



 2.5 Outer Measures 55

Thus, µ∗ is finitely additive on Σ. However, µ∗ need not be finitely additive

on arbitrary subsets of X!
Now we will show that Σ is closed under countable unions. By Problem 2.4,
it suffices to show that Σ is closed under countable disjoint unions. So, assume
that E1 , E2 , . . . ∈ Σ are disjoint sets and define
∞
S n
S
F = Ek and Fn = Ek , n ∈ N.
k=1 k=1

Note that Fn ∈ Σ since Σ is closed under finite unions.

Choose any set A ⊆ X. We claim that
n
X
µ∗ (A ∩ Fn ) = µ∗ (A ∩ Ek ). (2.4)
k=1

Now, if we knew that µ∗ was finitely additive on all subsets of X then equation
(2.4) would be immediate. However, we only know that µ∗ is finitely additive
on the µ∗ -measurable sets. Since A is an arbitrary set, this does not help us.
Instead, we prove equation (2.4) by induction.
Since F1 = E1 , equation (2.4) is trivial when n = 1. Suppose that equation
(2.4) holds for some integer n ≥ 1. Then, since En+1 is µ∗ -measurable,
 n+1
S 
µ∗ (A ∩ Fn+1 ) = µ∗ A ∩ Ek
k=1
 n+1   n+1 
S S C
= µ∗ A ∩ Ek ∩ En+1 + µ∗ A ∩ Ek ∩ En+1
k=1 k=1
n
S
= µ∗ (A ∩ En+1 ) + µ∗ (A ∩ En ) (by disjointness)
k=1
n
X
= µ∗ (A ∩ En+1 ) + µ∗ (A ∩ Ek ).
k=1

It follows that equation (2.4) holds for all n by induction.

Next we compute that
n
X
µ∗ (A ∩ Ek ) + µ∗ (A ∩ F C )
k=1
n
X
≤ µ∗ (A ∩ Ek ) + µ∗ (A ∩ FnC ) (since F C ⊆ FnC )
k=1

= µ∗ (A ∩ Fn ) + µ∗ (A ∩ FnC ) (by equation (2.4))

= µ∗ (A) (since Fn ∈ Σ). (2.5)

Applying subadditivity and taking the limit as n → ∞,

56 2 Abstract Measure Theory

µ∗ (A) ≤ µ∗ (A ∩ F ) + µ∗ (A ∩ F C ) (subadditivity)
∞
X
≤ µ∗ (A ∩ Ek ) + µ∗ (A ∩ F C ) (subadditivity)
k=1

≤ µ∗ (A) (by equation (2.5)).

Therefore equality holds in the preceding lines:

∞
X
µ∗ (A) = µ∗ (A ∩ Ek ) + µ∗ (A ∩ F C ) (2.6)
k=1

= µ∗ (A ∩ F ) + µ∗ (A ∩ F C ),

so F ∈ Σ.
(b) To show that µ∗ restricted to Σ is countably additive, let E1 , E2 , . . .
be disjoint sets in Σ, and set F = ∪Ek . Then since F is µ∗ -measurable, by
applying equation (2.6) with A = F we see that
∞
X ∞
X
µ∗ (F ) = µ∗ (F ∩ Ek ) + µ∗ (F ∩ F C ) = µ∗ (Ek ).
k=1 k=1

Hence µ∗ is countably additive on Σ, and therefore µ = µ∗ |Σ is a measure.

(c) This follows from Exercise 2.31. ⊓
⊔

Additional Problems

2.21. Let µ∗ be an outer measure on X, and let A, B ⊆ X be µ∗ -measurable.

Show that µ∗ (A ∪ B) + µ∗ (A ∩ B) = µ∗ (A) + µ∗ (B).

2.22. Let µ∗ be an outer measure on X. Show that if A, B are disjoint subsets

of X and A is µ∗ -measurable, then µ∗ (A ∪ B) = µ∗ (A) + µ∗ (B).

2.23. Given an uncountable set X, define µ∗ (E) = 0 if E ⊆ X is countable,

and µ∗ (E) = 1 if E ⊆ X is uncountable, Show that µ∗ is an outer measure
on X, and identify the µ∗ -measurable subsets of X.