Measure, Integration & Real Analysis, 2021a

16 Chapter 2 Measures
The next result shows that outer measure does the right thing with respect to set
inclusion.
2.5 outer measure preserves order
Suppose A and B are subsets of R with A ⊂ B. Then |A| ≤|B|.
Proof Suppose I 1 , I 2 ,...is a sequence of open intervals whose union contains B.
Then the union of this sequence of open intervals also contains A. Hence
|A| ≤
∞
∑ l(I k ).
k=1
Taking the infimum over all sequences of open intervals whose union contains B, we
have |A| ≤|B|.
We expect that the size of a subset of R should not change if the set is shifted to
the right or to the left. The next definition allows us to be more precise.
2.6 Definition translation; t + A
If t ∈ R and A ⊂ R, then the translation t + A is defined by
t + A = {t + a : a ∈ A}.
If t > 0, then t + A is obtained by moving the set A to the right t units on the real
line; if t < 0, then t + A is obtained by moving the set A to the left |t| units.
Translation does not change the length of an open interval. Specifically, if t ∈ R
and a, b ∈ [−∞, ∞], then t +(a, b) =(t + a, t + b) and thus l ( t +(a, b) ) =
l ( (a, b) ) . Here we are using the standard convention that t +(−∞) =−∞ and
t + ∞ = ∞.
The next result states that translation invariance carries over to outer measure.
2.7 outer measure is translation invariant
Suppose t ∈ R and A ⊂ R. Then |t + A| = |A|.
Proof Suppose I 1 , I 2 ,...is a sequence of open intervals whose union contains A.
Then t + I 1 , t + I 2 ,...is a sequence of open intervals whose union contains t + A.
Thus
|t + A| ≤
∞
∑ l(t + I k )=
k=1
∞
∑ l(I k ).
k=1
Taking the infimum of the last term over all sequences I 1 , I 2 ,...of open intervals
whose union contains A,wehave|t + A| ≤|A|.
To get the inequality in the other direction, note that A = −t +(t + A). Thus
applying the inequality from the previous paragraph, with A replaced by t + A and t
replaced by −t,wehave|A| = |−t +(t + A)| ≤|t + A|. Hence |t + A| = |A|.
Measure, Integration & Real Analysis, by Sheldon Axler