TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1

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

L1L2L3algebra

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]

2.1

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

step 1.1L1L2L3

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]

3.1

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

step 2.1L1L2L3

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]

4.1

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 $F (a) > F (a^{-})$ .

step 3.1L5

5.1

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

step 4.1L3

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]

6.1

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

step 1.1step 2.1step 3.1step 4.1step 5.1

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

Depends on

Used by

Dependency tree · two levels

33 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)