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Assuming countable choice, a nondecreasing right-continuous function defines a Borel measure on R

Statement

Assume the Axiom of Countable Choice. Let F:RR be nondecreasing and right-continuous. Then there is a Borel measure μF on R, finite on compact sets in the sense of A Borel measure on R that is finite on compact sets, such that

μF((a,b])=F(b)F(a)for all a<b.

Facts & Assumptions

Given: Countable choice, a nondecreasing right-continuous function F:RR, and its interval set function μ0,F on the half-open interval algebra.

[L1]

The interval set function μ0,F is a premeasure on the half-open interval algebra. (The Stieltjes interval set function is a premeasure)

[L2]

Assuming countable choice, a premeasure extends to a measure on the sigma-algebra it generates. (Assuming countable choice, a premeasure extends through its induced outer measure)

[L3]

The family of half-open intervals (a,b] with a<b generates the Borel sigma-algebra B(R). (Seven generating families for the Borel sigma-algebra on the real line)

[L4]

A compact subset of R is bounded. (A compact subset of R is closed and bounded)

Proof

technique · direct
1.1

By [L1], μ0,F is a premeasure, so [L2] gives a measure μ on the generated sigma-algebra σ(H) extending μ0,F.

L1L2L3

By [L3], that sigma-algebra is B(R), and therefore

μ((a,b])=μ0,F((a,b])=F(b)F(a)

for every a<b. [L1, L2, L3]

2.1

Let KR be compact. By [L4] there is R>0 with K[R,R](R1,R].

step 1.1L4algebra

So monotonicity and step 1.1 give

μ(K)μ((R1,R])=F(R)F(R1)<+.

Thus μ is finite on compact sets. [step 1.1, L4, algebra]

3.1

The measure μ of steps 1.1 and 2.1 is therefore a Borel measure on R finite on compact sets and having the prescribed half-open interval values.

step 1.1step 2.1

This is the required μF. [step 1.1, step 2.1] ∎

Depends on

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