ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A piecewise-quadratic distribution function recovers its density

Example

Assume the Axiom of Countable Choice. Define $F (x) : = {\begin{cases} 0, & x \leq 0, \\ x^{2}, & 0 \leq x \leq 1, \\ 1, & x \geq 1. \end{cases}$ Let $μ_{F}$ be the Lebesgue-Stieltjes measure of $F$ . Then $μ_{F} (E) = \int_{E} 2 x χ_{[0, 1]} (x) d λ (x) (E \in B (R)),$ so the density recovered from $F$ is $2 x χ_{[0, 1]} (x)$ .

Facts & Assumptions

Given: The piecewise-quadratic distribution function $F$ above.

[L1]

A nondecreasing right-continuous function on $R$ defines a Lebesgue--Stieltjes measure. Two Borel measures finite on compact sets and agreeing on all half-open intervals are equal (Assuming countable choice, a nondecreasing right-continuous function defines a Borel measure on $R$ , The interval data on $(a, b]$ determines the Borel measure uniquely).

[L3]

A nonnegative measurable density defines a measure (The measure with density $f$ relative to $μ$ ).

[L2]

For an absolutely continuous signed measure and a sigma-finite positive base satisfying a common finite exhaustion, a Radon--Nikodym derivative is represented by a measurable function whose measurable-set integrals recover the measure (The Radon-Nikodym derivative as an almost-everywhere equivalence class).

[L4]

Integrals over null sets vanish (A nonnegative integral over a null set vanishes).

Verification

technique · direct

1.1L1givenalgebra

The function $F$ is nondecreasing and right-continuous, so [L1] gives a Borel measure $μ_{F}$ . For every $a < b$ , direct integration shows $\int_{(a, b]} 2 x χ_{[0, 1]} (x) d λ (x) = F (b) - F (a),$ because both sides are $0$ off $[0, 1]$ , and on $[0, 1]$ they equal $b^{2} - a^{2}$ or the corresponding truncated interval increment.

2.1step 1.1L1L2L3L4∎

By [L3], $E \mapsto \int_{E} 2 x χ_{[0, 1]} d λ$ is a finite Borel measure and hence is finite on compact sets. The Lebesgue--Stieltjes measure $μ_{F}$ is also finite, because $F$ has total increment $1$ . The two measures agree on every half-open interval by step 1.1, so [L1] makes them equal on all Borel sets. Thus [L4] makes $μ_{F} ≪ λ$ ; $λ$ is sigma-finite and $[- n, n]$ is a common finite exhaustion. Therefore [L2] identifies $2 x χ_{[0, 1]}$ as a representative of $d μ_{F} / d λ$ .

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