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For a compact subset of R, measure zero and content zero coincide

Statement

Let K⊆R be compact (Open cover, subcover, compact subset of R (every open cover has a finite subcover), and sequentially compact subset), equivalently closed and bounded (A subset of R is compact if and only if it is closed and bounded). Then

K has measure zero⟺K has content zero

(Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover)).

The implication from content zero to measure zero is A set of content zero has measure zero and needs no hypothesis on K. The other direction is the one that uses compactness, and it uses it exactly as A sequence of intervals covering [a,b] has total length at least b−a, so no interval of positive length has measure zero does: a countable cover is enlarged to an open cover at an arbitrarily small cost in total length, and compactness reduces the open cover to a finite one.

Facts & Assumptions

Given: A compact set K⊆R and a real ε>0. Throughout, θ:=2−1.

[L1]

A is null when for every real η>0 there are sequences (ak), (bk) with ak≤bk, A⊆⋃k[ak,bk] and ∑k<i(bk−ak)≤η for every i; A has content zero when the same holds with a finite list (Measure zero (a countable cover by intervals of total length below every ε) and content zero (a finite such cover)).

[L2]

A set of content zero is null (A set of content zero has measure zero).

[L3]

[c,d] has length d−c≥0 for c≤d; (c,d) is the open interval with the same endpoints and is contained in [c,d] (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L5]

K is compact: from every family of open sets whose union contains K, either K=∅ and the empty subfamily covers it, or there are m∈N and members U0,…,Um of the family whose union contains K; compactness is equivalent to being closed and bounded (Open cover, subcover, compact subset of R (every open cover has a finite subcover), and sequentially compact subset, A subset of R is compact if and only if it is closed and bounded).

[L6]

Powers and the geometric series: θ0=1, θk+1=θkθ, θk>0, and ∑k=0∞θk=2 for θ=2−1; a series of nonnegative terms has all its partial sums at most its sum (Integer powers am, For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L7]

Finite sums: additivity, scaling, splitting and monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L8]

Every finite list of naturals has an upper bound in N, by induction on its length and the totality of the order of N (The principle of mathematical induction, Trichotomy of the order on N, Order on the natural numbers).

[L9]

Ordered-field arithmetic: 0<1, so 2>0, 4>0, 8>0 and t⋅8−1>0 for t>0; adding a constant and multiplying by a positive preserve an inequality (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Proof

technique · direct
1.1L2suffices: only the forward direction remains

One direction is immediate: if K has content zero then K is null by [L2], with no hypothesis on K used. It remains to prove the converse for compact K.

1.2L1cases

If K=∅, then for every real ε>0 the single interval [0,0] covers K and has total length 0≤ε, so K has content zero by [L1]. Hence suppose K≠∅ for the rest of the proof.

2.1step 1.1givenL1L9choose

Assume K is null and let the real ε>0 be given. By [L1] applied with η:=ε⋅2−1>0 fix sequences (ak), (bk) with ak≤bk, K⊆⋃k[ak,bk] and ∑k<i(bk−ak)≤ε⋅2−1 for every i∈N.

3.1step 2.1L3L4L6L9

Put δk:=ε⋅8−1⋅θk, a positive real by [L6] and [L9], and Jk:=(ak−δk, bk+δk), an open set by [L4] containing [ak,bk] by [L3] and [L9]. Hence { Jk:k∈N } is a family of open sets whose union contains K, and the closed interval [ak−δk, bk+δk] has length (bk−ak)+2δk=(bk−ak)+ε⋅4−1⋅θk by [L3] and [L9].

4.1step 1.2step 3.1L3L5L8choose

By [L5] there are m∈N and members Jk0,…,Jkm of that family covering K, and by [L8] there is N∈N with kt≤N for every t≤m; then K⊆⋃k≤NJk⊆⋃k≤N[ak−δk, bk+δk] by [L3].

5.1step 2.1step 3.1step 4.1L6L7L9

The total length of that finite list is ∑k≤N((bk−ak)+ε⋅4−1θk)=∑k<N+1(bk−ak)+ε⋅4−1∑k<N+1θk≤ε⋅2−1+ε⋅4−1⋅2=ε, by [L7], step 2.1, [L6] and [L9].

6.1step 1.1step 1.2step 4.1step 5.1L1∎

So for every real ε>0 the finite list of step 4.1 covers K with total length at most ε, which by [L1] is exactly the statement that K has content zero; together with step 1.1 the two notions coincide on compact sets.

Remarks

Depends on

Dependency tree · 2 levels

Sources