1 72 Chapter 3 • Topology of the Real Number System(μ5) If A and Bare measurable, then
μ(A - B) =μ(A) - μ(An B), and
μ(AU B) =μ(A)+ μ(B) - μ(An B).
(μ6) If A, Bare measurable and B ~A, then μ(B) ::::; μ(A).
(μ7) If {An : n E N} is a (countable) collection of measurable sets such
that μ(A1) < oo and A1 2 A2 2 · · · 2 An 2 · · ·, then
μ Ca
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An) =}~ ... ~μ(An)·(μ8) Every interval I is measurable, and μ(I) = the length of I.
(μ9) A set A has "measure zero" by Definition 3.4.19 iff μ(A) = 0.These properties are somewhat redundant; see Exercises 15 and 16.
Assuming the existence of such a "measure function," we are able to con-
struct intriguing Cantor-like sets that do not have measure zero. As a warm-up,
we use these ideas to give an alternative proof of Theorem 3.4.21.
Theorem 3.4.21 (Again) The Cantor set has measure zero.
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Alternate Proof. From the definition of the Cantor set, C = n Cn, with
n=l
μ(Cn+I) = ~μ(Cn), and C1 2 C2;;? · · · 2 Cn 2 · · · .Thus, by (μ7),
μ(C) = n-+oo lim μ(Cn) = n-+oo lim (~r = o. •
Corollary 3.4.22 The complement of the Cantor set in [O, l] has measure 1.
Proof. Exercise 17. •Definition 3.4.23 ("Fat" Cantor-like Sets) Pick any number 0 <a < 1.
We construct a Cantor-like set of measure a.
We shall do so by removing intervals from [O, l], analogous to the "middle
thirds" intervals removed in Definition 3.4.1, leaving a set whose complement
in [O, l] has measure 1 - a.
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We begin with a geometric series L rn with r < 1 whose sum is 1 - a.
n=l
(We want f: rn = 1 - a; i.e., r = 1 - a, so taker=
1- a.) We shall
n=l 1 - r 2 - a
describe the sequence {Rn} of sets of intervals removed at each step , as we did
in Definition 3.4.1. First, note that 0 < r < 1·
Let R 1 = the open interval of length r centered in [O, 1 J. Then the set
C 1 = [O, l] -R 1 consists of 2 disjoint closed intervals , each of length^1 ;r.