TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

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Almost-sure convergence implies convergence in probability

Statement

If $X_{n} \to X$ almost surely, then $X_{n} \to X$ in probability.

Facts & Assumptions

Given: $X_{n} \to X$ almost surely.

[L1]

Real random variables are measurable, so differences, absolute values, threshold events, and their indicators are measurable (Random elements and real random variables).

[L2]

Dominated convergence sends an almost-everywhere convergent integrable sequence with one integrable majorant to convergence of integrals (Dominated convergence).

Proof

technique · direct

1.1

Fix $ε > 0$ and set $I_{n} = 1_{{∣ X_{n} - X ∣ > ε}}$ . [given, L1] By [L1] these are measurable; the hypothesis gives $I_{n} \to 0$ almost surely, and $0 \leq I_{n} \leq 1$ .

givenL1

2.1

Apply [L2] to the indicators from step 1.1 with majorant $1$ . [step 1.1, L2] $P (∣ X_{n} - X ∣ > ε) = E I_{n} \to 0$ , which is the required probability convergence.

step 1.1L2∎

Depends on

Used by

Dependency tree · two levels

18 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

S. Roch, Lecture 3: Modes of convergence, Theorem 3.12 (standard reference, not scraped)