ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-27

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The arctangent distribution function generates a Borel probability measure

Example

Assume the Axiom of Countable Choice and let

$F (x) : = \frac{\arctan (x)}{π} + \frac{1}{2} .$

Then $F$ is increasing and continuous, so it defines a Lebesgue-Stieltjes measure $μ_{F}$ . Its half-open interval values are

$μ_{F} ((a, b]) = \frac{\arctan (b) - \arctan (a)}{π},$

and $μ_{F}$ is a probability measure.

Facts & Assumptions

Given: The Axiom of Countable Choice, the function $F (x) = \arctan (x) / π + 1 / 2$ , and its Lebesgue-Stieltjes measure $μ_{F}$ .

[L1]

Assuming Countable Choice, every increasing right-continuous function defines a Lebesgue-Stieltjes measure, with half-open interval values given by increments. (Assuming countable choice, a nondecreasing right-continuous function defines a Borel measure on $R$ )

[L2]

Measures are continuous from below along increasing sets. (Continuity from below for measures)

Verification

technique · direct

1.1givenL1

The function $F$ is increasing and continuous, hence right-continuous, so [L1] gives a measure $μ_{F}$ with the following interval values.

$μ_{F} ((a, b]) = F (b) - F (a) = \frac{\arctan (b) - \arctan (a)}{π} .$

2.1step 1.1L2

The intervals $(- n - 1, n + 1]$ increase to $R$ .

$μ_{F} ((- n - 1, n + 1]) = F (n + 1) - F (- n - 1) = \frac{2 \arctan (n + 1)}{π} .$

Because $\arctan (n + 1) \to π / 2$ , [L2] gives $μ_{F} (R) = \lim_{n} 2 \arctan (n + 1) / π = 1$ . So $μ_{F}$ is a probability measure. [step 1.1, L2] ∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Section 1.5 (standard reference, not scraped)