2.5 Outer Measure and Measurable sets.

Note The results of this section concern any given outer measure λ.
If an outer measure λ on a set X were a measure then it would be additive.
In particular, given any two sets A, B ⊆ X we have that A ∩ B and A ∩ B c
are disjoint with (A ∩ B) ∪ (A ∩ B c ) = A and so we would have

λ(A) = λ(A ∩ B) + λ(A ∩ B c ).

We will see later that this does not necessarily hold for all A and B but
it does lead to the following definition.
Definition Let λ be an outer measure on a set X. Then E ⊆ X is said to
be measurable with respect to λ (or λ-measurable) if

λ(A) = λ(A ∩ E) + λ(A ∩ E c ) for all A ⊆ X. (7)

(This can be read as saying that we take each and every possible “test
set”, A, look at the measures of the parts of A that fall within and without
E, and check whether these measures add up to that of A.)
Since λ is subadditive we have λ(A) ≤ λ(A∩E)+λ(A∩E c ) so, in checking
measurability, we need only verify that

λ(A) ≥ λ(A ∩ E) + λ(A ∩ E c ) for all A ⊆ X. (8)

Let M = M(λ) denote the collection of λ-measurable sets.
Theorem 2.6
M is a field.
Proof
Trivially φ and X are in M.
Take any E1 , E2 ∈ M and any test set A ⊆ X. Then

λ(A) = λ(A ∩ E1 ) + λ(A ∩ E1c ).

Now apply the definition of measurability for E2 with the test set A ∩ E1c to
get

λ(A ∩ E1c ) = λ((A ∩ E1c ) ∩ E2 ) + λ((A ∩ E1c ) ∩ E2c )

= λ(A ∩ E1c ∩ E2 ) + λ(A ∩ (E1 ∪ E2 )c ).

Combining

λ(A) = λ(A ∩ E1 ) + λ(A ∩ E1c ∩ E2 ) + λ(A ∩ (E1 ∪ E2 )c ). (9)

1
We hope to use the subadditivity of λ on the first two term on the right hand
side of (9). For the sets there we have

(A ∩ E1 ) ∪ (A ∩ E1c ∩ E2 ) = A ∩ (E1 ∪ (E1c ∩ E2 ))

= A ∩ ((E1 ∪ E1c ) ∩ (E1 ∪ E2 ))
= A ∩ (X ∩ (E1 ∪ E2 ))
= A ∩ (E1 ∪ E2 ).

So

λ(A ∩ E1 ) + λ(A ∩ E1c ∩ E2 ) ≥ λ(A ∩ (E1 ∪ E2 )).

Substituting in (9) gives

λ(A) ≥ λ(A ∩ (E1 ∪ E2 )) + λ(A ∩ (E1 ∪ E2 )c ).

So we have verified (8) for E1 ∪ E2 , that is, E1 ∪ E2 ∈ M.
Observe that the definition of λ-measurable sets is symmetric in that
E ∈ M if, and only if, E c ∈ M. Thus

E1 \ E2 = E1 ∩ E2c = (E1c ∪ E2 )c ∈ M.
Hence M is a field. ¥
Proposition 2.7 If G, F ∈ M(λ) are disjoint then

λ(A ∩ (G ∪ F )) = λ(A ∩ G) + λ(A ∩ F )

for all A ⊆ X.
Proof
Let A ⊆ X be given. Apply the definition of λ-measurability to G with
the test set A ∩ (G ∪ F ). Then

λ(A ∩ (G ∪ F )) = λ((A ∩ (G ∪ F )) ∩ G)
+λ((A ∩ (G ∪ F )) ∩ Gc ). (10)

Yet

(G ∪ F ) ∩ G = (G ∩ G) ∪ (F ∩ G)
= G ∪ (F ∩ G) = G

2
since F ∩ G ⊆ G. Also

(G ∪ F ) ∩ Gc = (G ∩ Gc ) ∪ (F ∩ Gc )
= φ ∪ F = F,

because F and G disjoint means F ⊆ Gc and so F ∩ Gc = F . Thus (9)

becomes

λ(A ∩ (G ∪ F )) = λ(A ∩ G) + λ(A ∩ F ).

¥
Using induction it is possible to prove the following.
Corollary 2.8 For all n ≥ 1 if {Fi }1≤i≤n is a finite collection of disjoint
sets from M(λ) then
à n
! n
[ X
λ A∩ Fi = λ (A ∩ Fi ) for all A ⊆ X.
i=1 i=1

Proof Left to students. ¥

Corollary 2.9 If {Fi }i≥1 is a countable collection of disjoint sets from M(λ)
then
à ∞
! ∞
[ X
λ A∩ Fi = λ (A ∩ Fi ) for all A ⊆ X.
i=1 i=1

Proof S Sn
For any n ≥ 1 we have A ∩ ∞i=1 Fi ⊇ A ∩ i=1 Fi and so by monotonicity
we have

à ∞
! Ã n
!
[ [
λ A∩ Fi ≥ λ A∩ Fi
i=1 i=1
n
X
= λ (A ∩ Fi )
i=1

by Corollary 2.8. Let n → ∞ to get

à ∞
! ∞
[ X
λ A∩ Fi ≥ λ (A ∩ Fi ) .
i=1 i=1

The reverse inequality follows from subadditivity. ¥

3
Theorem 2.10
M(λ) is a σ-field and λ restricted to M(λ) is a measure.
Proof
Let {Ei }i≥1 be a countable collection from M. They might not be disjoint
so define F1 = E1 and
i−1
[
Fi = Ei \ Ej
j=1

for all S
i > 1. The Fi are
S∞disjoint S
and Fi ∈ M since M is a field. Let
Gm = j=1 Fj and G = j=1 Fj = ∞
m
i=1 Ei . Then for any A ⊆ X and for
any n ≥ 1 we have

λ(A) = λ(A ∩ Gn ) + λ(A ∩ Gcn ) since Gn ∈ M

Xn
= λ(A ∩ Fi ) + λ(A ∩ Gcn ) by Corollary 2.8
i=1
n
X
≥ λ(A ∩ Fi ) + λ(A ∩ Gc ) since Gc ⊆ Gcn .
i=1

True for all n means that

∞
X
λ(A) ≥ λ(A ∩ Fi ) + λ(A ∩ Gc )
i=1
à ∞
!
[
= λ A∩ Fi + λ(A ∩ Gc )
i=1
by Corollary 2.8, since the Fi are disjoint,
= λ (A ∩ G) + λ(A ∩ Gc ).

True for all A ⊆ X means that G ∈ M(λ).

Choosing A = X in Corollary 2.9 shows that λ is σ-additive on M(λ).
Hence λ is a measure on M(λ). ¥
∗
Example † With the Lebesgue-Stieltjes outer measure µF of example 8 we
can now form the σ-field M(µ∗F ). This is known as the collection of Lebesgue-
Stieltjes measurable sets and is denoted by LF . If F (x) = x, it is simply
known as the collection of Lebesgue measurable sets and is denoted by L.
We now specialise to those outer measures constructed, as in (4), from
measures defined in a ring.

4
 S
Theorem 2.11 Let R be a ring of sets in X such that X = ∞ i=1 Ei for
some Ei ∈ R. Let µ be a measure on R and let µ∗ be the outer measure on
X constructed from µ as in (4). Then
(i) the elements of R are µ∗ -measurable sets,
(ii) µ∗ = µ on R.
Proof
(i) Let E ∈ R and a test set A ⊆ X be given. If µ∗ (A) = +∞ then (7)
is trivially satisfied so assume that µ∗ (A) < +∞.
Let ε > 0 be given. By theSdefinition (4) there exists a countable collection
{Ei }i≥1 ⊆ R such that A ⊆ i≥1 Ei and
∞
X
µ∗ (A) ≤ µ(Ei ) < µ∗ (A) + ε.
i=1

Yet µ is a measure on R and Ei , E ∈ R so

µ(Ei ) = µ(Ei ∩ E) + µ(Ei ∩ E c ).

Combining we see

∞
X
µ∗ (A) + ε > (µ(Ei ∩ E) + µ(Ei ∩ E c ))
i=1
≥ µ∗ (A ∩ E) + µ∗ (A ∩ E c ),

since {Ei ∩ E}i≥1 and {Ei ∩ E c }i≥1 are covers for A ∩ E and A ∩ E c respec-
tively. True for all ε > 0 means that (7) is satisfied and so E ∈ M and thus
R ⊆ M.
(ii) Let E ∈ R be given. Then since E is a cover from R for E we have
that
X
µ∗ (E) = inf µ(Ei ) ≤ µ(E).
all covers

Take any other cover {Ei }i≥1 of E. AsSin TheoremS∞2.10 replace the Ei
by disjoint sets Fi ⊆ Ei , Fi ∈ R and where ∞ F
i=1 i = i=1 Ei . Then

5
 Ã ∞
! ∞
[ [
µ(E) = µ E ∩ Fi since E ⊆ Fi ,
à ∞ i=1 ! i=1
[
=µ (E ∩ Fi )
i=1
∞
X
= µ (E ∩ Fi ) since µ is additive on R,
i=1
X∞ ∞
X
≤ µ (Fi ) ≤ µ (Ei )
i=1 i=1

since E ∩ Fi ⊆ Fi ⊆ Ei and µ is monotonic on R. So µ(E) is a lower

bound for the sums for which µ∗ (E) is the greatest lower bound and thus
µ(E) ≤ µ∗ (E).
Hence µ(E) = µ∗ (E). ¥
∗
We now see that it is reasonable to say that µ extends µ.
Further, let A be a collection of subsets of X.
Definition We say that the extended real-valued function φ : A → R∗ is
σ-finite if Sfor all A ∈ A there exists a countable collection {An }n≥1 ⊆ A such
that A ⊆ n≥1 An and |φ(An )| < ∞ for all n ≥ 1.
Then it can be shown that
Theorem 2.12
If, in addition to the conditions of Theorem 2.11, µ is σ-finite on R then
the extension to M is unique and is also σ-finite.
Proof not given here. ¥
Example 8† Recall that µF was first defined on P and then extended to E
in Corollary 2.3. In the last example the σ-field M (µ∗F ), known as LF , was
constructed. Now Theorem 2.11 implies that E ⊆ LF and µ∗F = µF on E. So
it is reasonable to write µ∗F simply as µF on LF , called the Lebesgue-Stieltjes
measure. If F (x) = x, we write µF simply as µ, called the Lebesgue measure
on R.
Note† LF ⊇ E ⊇ P so LF is a σ-field containing P. But B, the Borel
sets of R, is the smallest σ-field containing P. Hence LF ⊇ B, true for all
distribution functions F .