A fixed proof that m∗ extends m
February 2, 2009
Let A be a ring of subsets of a set X and let m : A → [0, ∞] be a measure
on A. We know:
1. m is increasing: if A, B ∈ A with B ⊆ A then m(B) ≤ m(A);
2. A is closed under finite intersections; and
S∞
3. m is countably
if B1 , B2 , · · · ∈ A with A =
Psubadditive:
∞
then m(A) ≤ i=1 m(Bi ).
i=1
Bi ∈ A
The outer-measure of m is defined as the function m∗ : P(X) → [0, ∞],
∗
m (B) = inf
∞
nX
m(Ai ) : A1 , A2 , · · · ∈ A with B ⊆
i=1
∞
[
o
Ai .
i=1
Theorem 6. Let A be a ring of subsets of a set X and let m : A → [0, ∞]
be a measure. Then m∗ extends m; that is, m∗ (A) = m(A) for every A ∈ A.
Proof. Let A ∈ A. Consider
S the sets A1 = A and Ai = ∅ for i ≥ 2. Since
A1 , A2 , · · · ∈ A and A = ∞
i=1 Ai ,
m∗ (A) ≤
∞
X
m(Ai ) = m(A) + 0 + 0 + · · · = m(A).
i=1
∗
To establish
S∞ the inequality m(A) ≤ m (A), suppose that A1 , A2 , · · · ∈ A
with A ⊆ i=1 Ai . Let Bi = A ∩ Ai ; then
S∞ Bi ⊆ ASi ,∞so m(Bi ) ≤ m(A
S∞ i ) by (1).
Moreover, Bi ∈ A by (2) and A = A∩ i=1 Ai = i=1 (A∩Ai ) = i=1 Bi ∈ A,
so
(3) P
P∞
m(A) ≤ ∞
i=1 m(Bi ) ≤
i=1 m(Ai ).
P∞
S∞
So m(A) ≤
i=1 m(Ai ) whenever A1 , A2 , · · · ∈ A with A ⊆
i=1 Ai .
∗
Hence m(A) ≤ m (A).
IS
tried to do the latter part of this proof in class by arguing as follows:
A⊆ ∞
i=1 Ai , so
(1)
(3) P
S
∞
m(A) ≤ m( ∞
A
)
i=1 i ≤
i=1 m(Ai ).
S
S∞
The problem is that ∞
i=1 Ai need not be in A, in which case m( i=1 Ai )
doesn’t make sense! Introducing the sets Bi = A ∩ Ai fixes this.
1