Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Almost-sure convergence does not imply convergence of expectations

Statement refuted

Almost-sure convergence alone need not imply convergence of expectations.

Facts & Assumptions

Given: Lebesgue probability space (0,1) and Xn=(n+1)1(0,1/(n+1)) for n∈N.

[L1]

Almost-sure convergence is pointwise convergence off a null set (Almost-sure convergence of real random variables).

[L2]

Expectation is integration against the probability measure (Expectation of a nonnegative or integrable random variable).

Counterexample

technique · direct
1.1

For every x∈(0,1), eventually x>1/(n+1), so Xn(x)=0. [L1] Hence Xn→0 almost surely by [L1].

L1
2.1

Yet [L2] gives [step 1.1, L2] EXn=∫01/(n+1)(n+1) dx=1 for every n, whereas E0=0. Thus the expectations do not converge.

step 1.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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