Measure, Integration & Real Analysis, 2021a

Section 2B Measurable Spaces and Functions 25
2B
Measurable Spaces and Functions
The last result in the previous section showed that outer measure is not additive.
Could this disappointing result perhaps be fixed by using some notion other than
outer measure for the size of a subset of R? The next result answers this question by
showing that there does not exist a notion of size, called the Greek letter mu (μ)in
the result below, that has all the desirable properties.
Property (c) in the result below is called countable additivity. Countable additivity
is a highly desirable property because we want to be able to prove theorems about
limits (the heart of analysis!), which requires countable additivity.
2.22 nonexistence of extension of length to all subsets of R
There does not exist a function μ with all the following properties:
(a) μ is a function from the set of subsets of R to [0, ∞].
(b) μ(I) =l(I) for every open interval I of R.
( ⋃ ∞
(c) μ
k=1
of R.
A k
)
=
∞
∑ μ(A k ) for every disjoint sequence A 1 , A 2 ,...of subsets
k=1
(d) μ(t + A) =μ(A) for every A ⊂ R and every t ∈ R.
Proof Suppose there exists a function μ
with all the properties listed in the statement
of this result.
Observe that μ(∅) = 0, as follows
from (b) because the empty set is an open interval with length 0.
We will show that μ has all the
properties of outer measure that
were used in the proof of 2.18.
If A ⊂ B ⊂ R, then μ(A) ≤ μ(B), as follows from (c) because we can write B
as the union of the disjoint sequence A, B \ A, ∅, ∅,...; thus
μ(B) =μ(A)+μ(B \ A)+0 + 0 + ···= μ(A)+μ(B \ A) ≥ μ(A).
If a, b ∈ R with a < b, then (a, b) ⊂ [a, b] ⊂ (a − ε, b + ε) for every ε > 0.
Thus b − a ≤ μ([a, b]) ≤ b − a + 2ε for every ε > 0. Hence μ([a, b]) = b − a.
If A 1 , A 2 ,...is a sequence of subsets of R, then A 1 , A 2 \ A 1 , A 3 \ (A 1 ∪ A 2 ),...
is a disjoint sequence of subsets of R whose union is ⋃ ∞
k=1
A k . Thus
( ⋃ ∞
μ
k=1
) (
A k = μ A 1 ∪ (A 2 \ A 1 ) ∪ ( A 3 \ (A 1 ∪ A 2 ) ) )
∪···
= μ(A 1 )+μ(A 2 \ A 1 )+μ ( A 3 \ (A 1 ∪ A 2 ) ) + ···
∞
≤ ∑ μ(A k ),
k=1
where the second equality follows from the countable additivity of μ.
Measure, Integration & Real Analysis, by Sheldon Axler