Measure, Integration & Real Analysis, 2021a

140 Chapter 5 Product Measures
Volume of Unit Ball in R n
The proof of the next result provides good experience in working with the Lebesgue
measure λ n . Recall that tE = {tx : x ∈ E}.
5.41 measure of a dilation
Suppose t > 0. IfE ∈B n , then tE ∈B n and λ n (tE) =t n λ n (E).
Proof
Let
E = {E ∈B n : tE ∈B n }.
Then E contains every open subset of R n (because if E is open in R n then tE is open
in R n ). Also, E is closed under complementation and countable unions because
(
t(R n \ E) =R n ⋃ ∞ ) ∞⋃
\ (tE) and t E k = (tE k ).
Hence E is a σ-algebra on R n containing the open subsets of R n . Thus E = B n .In
other words, tE ∈B n for all E ∈B n .
To prove λ n (tE) =t n λ n (E), first consider the case n = 1. Lebesgue measure on
R is a restriction of outer measure. The outer measure of a set is determined by the
sum of the lengths of countable collections of intervals whose union contains the set.
Multiplying the set by t corresponds to multiplying each such interval by t, which
multiplies the length of each such interval by t. In other words, λ 1 (tE) =tλ 1 (E).
Now assume n > 1. We will use induction on n and assume that the desired result
holds for n − 1. IfA ∈B n−1 and B ∈B 1 , then
λ n
(
t(A × B)
) = λn
(
(tA) × (tB)
)
5.42
k=1
k=1
= λ n−1 (tA) · λ 1 (tB)
= t n−1 λ n−1 (A) · tλ 1 (B)
= t n λ n (A × B),
giving the desired result for A × B.
For m ∈ Z + , let C m be the open cube in R n centered at the origin and with side
length m. Let
E m = {E ∈B n : E ⊂ C m and λ n (tE) =t n λ n (E)}.
From 5.42 and using 5.13(b), we see that finite unions of measurable rectangles
contained in C m are in E m . You should verify that E m is closed under countable
increasing unions (use 2.59) and countable decreasing intersections (use 2.60, whose
finite measure condition holds because we are working inside C m ). From 5.13 and
the Monotone Class Theorem (5.17), we conclude that E m is the σ-algebra on C m
consisting of Borel subsets of C m . Thus λ n (tE) =t n λ n (E) for all E ∈B n such that
E ⊂ C m .
Now suppose E ∈B n . Then 2.59 implies that
as desired.
λ n (tE) = lim
m→∞ λ n
(
t(E ∩ Cm ) ) = t n lim
m→∞ λ n(E ∩ C m )=t n λ n (E),
Measure, Integration & Real Analysis, by Sheldon Axler