CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A distribution function need not have a density

Statement refuted

A distribution function need not arise from a Lebesgue density.

Facts & Assumptions

Given: A Bernoulli random variable $X$ with $P (X = 1) = p$ and $P (X = 0) = 1 - p$ , where $0 < p < 1$ .

[L1]

The cumulative distribution function is $F_{X} (x) = P (X \leq x)$ (Cumulative distribution function of a real random variable).

[L2]

Atoms of a law are positive point masses (Atoms and continuity points of a law).

Counterexample

technique · direct

1.1

The CDF of $X$ is $F_{X} (x) = {\begin{cases} 0, & x < 0, \\ 1 - p, & 0 \leq x < 1, \\ 1, & x \geq 1. \end{cases}$ So the law has atoms at $0$ and $1$ by [L2].

L1L2given

2.1

If this law were given by a Lebesgue density $f$ , then every singleton would have probability $\int_{{a}} f (x) d x = 0.$ In particular $P (X = 0) = 0$ , contradicting $P (X = 0) = 1 - p > 0$ .

step 1.1givenalgebra

3.1

Therefore this distribution function has no Lebesgue density.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

J. R. Norris, Probability and Measure, Section 2.3 (standard reference, not scraped)