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Interval formulas and atoms for a Lebesgue-Stieltjes measure

Statement

Assume the Axiom of Countable Choice. Let $F : R \to R$ be nondecreasing and right-continuous, and let $μ_{F}$ be its Lebesgue-Stieltjes measure. Then for every $a < b$ ,

$μ_{F} ((a, b)) = F (b^{-}) - F (a),$

$μ_{F} ([a, b]) = F (b) - F (a^{-}),$

$μ_{F} ([a, b)) = F (b^{-}) - F (a^{-}),$

and

$μ_{F} ({a}) = F (a) - F (a^{-}) .$

Consequently $a$ is an atom of $μ_{F}$ in the sense of An atom of a measure on $R$ if and only if $F (a) > F (a^{-})$ , and the set of atoms of $μ_{F}$ is at most countable.

Facts & Assumptions

Given: The Axiom of Countable Choice, a nondecreasing right-continuous function $F : R \to R$ , its Lebesgue-Stieltjes measure $μ_{F}$ , and real numbers $a < b$ .

[L1]

Assuming Countable Choice, the measure $μ_{F}$ satisfies $μ_{F} ((u, v]) = F (v) - F (u)$ for all $u < v$ . (Assuming countable choice, finite-on-compacts Borel measures on $R$ correspond to nondecreasing right-continuous functions modulo constants)

[L2]

Measures are continuous from below along increasing set limits; they are continuous from above along decreasing set limits when one set has finite measure (Continuity from below for measures, Continuity from above when one set has finite measure).

[L3]

For a bounded-variation function on a compact interval, every well-posed one-sided limit exists, every discontinuity is of the first kind, and there are at most countably many discontinuities (A bounded-variation function has at most countably many discontinuities, all of the first kind).

[L4]

Assuming Countable Choice, a countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L5]

A point $x$ is an atom of a Borel measure $μ$ exactly when $μ ({x}) > 0$ (An atom of a measure on $R$ ).

[L6]

If $A \subseteq B$ are measurable and $μ (B) < + \infty$ , then $μ (B ∖ A) = μ (B) - μ (A)$ (Measure of a set difference when the smaller set has finite measure).

Proof

technique · direct

1.1L1L2L3algebra

Put $t_{n} : = b - (b - a) / (n + 2)$ . Then $a < t_{n} < b$ for every $n$ and $t_{n} ↑ b$ .

Because $(a, b) = ⋃_{n} (a, t_{n}]$ , continuity from below and [L1] give

$μ_{F} ((a, b)) = \lim_{n} μ_{F} ((a, t_{n}]) = \lim_{n} (F (t_{n}) - F (a)) = F (b^{-}) - F (a) .$

Here $F ∣_{[a, b]}$ has bounded variation because it is nondecreasing, so [L3] ensures that the displayed left limit exists. [L1, L2, L3, algebra]

2.1step 1.1L1L2L3

Put $s_{n} : = a - 1 / (n + 1)$ . Then $s_{n} < a$ and $s_{n} ↑ a$ , while the intervals $(s_{n}, b]$ decrease to $[a, b]$ .

The first interval has finite measure because $F$ is real-valued, so continuity from above and [L1] give

$μ_{F} ([a, b]) = \lim_{n} μ_{F} ((s_{n}, b]) = \lim_{n} (F (b) - F (s_{n})) = F (b) - F (a^{-}) .$

The restriction $F ∣_{[a - 1, a]}$ has bounded variation because it is nondecreasing, so [L3] ensures that the displayed left limit exists. [L1, L2, L3, algebra]

3.1step 2.1L1L2L3

The intervals $(s_{n}, a]$ decrease to ${a}$ , and continuity from above gives $μ_{F} ({a}) = F (a) - F (a^{-})$ .

Indeed, $(s_{0}, a]$ has finite measure and $μ_{F} ({a}) = \lim_{n} μ_{F} ((s_{n}, a]) = \lim_{n} (F (a) - F (s_{n})) = F (a) - F (a^{-}) .$ [step 2.1, L1, L2, L3]

The same argument gives $μ_{F} ({b}) = F (b) - F (b^{-})$ . Therefore

$μ_{F} ([a, b)) = μ_{F} ([a, b]) - μ_{F} ({b}) = F (b^{-}) - F (a^{-}) .$

Together with step 1.1, this proves all four interval formulas. [step 1.1, step 2.1, step 3.1, L1, L6, algebra]

4.1step 3.1L5

By step 3.1 and [L5], the point $a$ is an atom of $μ_{F}$ exactly when $μ_{F} ({a}) = F (a) - F (a^{-}) > 0$ , which is exactly the jump condition $F (a) > F (a^{-})$ .

5.1step 4.1L3

Fix $m \geq 1$ . The restriction $F ∣_{[- m, m]}$ is nondecreasing, hence of bounded variation on $[- m, m]$ .

So [L3] makes its discontinuity set at most countable. Every atom of $μ_{F}$ in $(- m, m)$ is an interior jump point by step 4.1, hence lies in that countable discontinuity set. Therefore the atoms in $(- m, m)$ are at most countable. By [L4], their union over $m \geq 1$ is at most countable, and this union contains every atom of $μ_{F}$ . [given, step 4.1, L3, L4]

6.1step 1.1step 2.1step 3.1step 4.1step 5.1

Steps 1.1 through 5.1 prove the claimed interval formulas, the atom criterion, and the countability of the atom set.

[step 1.1, step 2.1, step 3.1, step 4.1, step 5.1] ∎

Depends on

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Section 1.5 (standard reference, not scraped)
John K. Hunter, Measure Theory, Section 2.9 (standard reference, not scraped)