TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Dominated convergence in $L^{p}$

Statement

Let $1 \leq p < \infty$ . If $X_{n} \to X$ almost surely and $∣ X_{n} ∣ \leq Y$ almost surely for every $n$ , where $Y \in L^{p} (P)$ , then $X_{n} \to X$ in $L^{p}$ .

Facts & Assumptions

Given: $1 \leq p < \infty$ , $X_{n} \to X$ almost surely, and $∣ X_{n} ∣ \leq Y$ almost surely with $Y \in L^{p}$ .

[L1]

Dominated convergence gives convergence of integrals under one integrable majorant (Dominated convergence).

[L2]

$L^{p}$ convergence is convergence of the $p$ th absolute moments of the difference ( $L^{p}$ convergence for random variables).

Proof

technique · direct

1.1

Intersect the countably many full-measure events on which $∣ X_{n} ∣ \leq Y$ with the full-measure convergence event. Outside the resulting null set, $X_{n} \to X$ and $∣ X_{n} ∣ \leq Y$ for every $n$ , so $∣ X ∣ \leq Y$ there. Thus $∣ X_{n} - X ∣^{p} \to 0$ almost everywhere and $∣ X_{n} - X ∣^{p} \leq (2 ∣ Y ∣)^{p}$ .

given

2.1

Since $(2 ∣ Y ∣)^{p}$ is integrable, [L1] applied to step 1.1 yields [step 1.1, L1, L2] $E ∣ X_{n} - X ∣^{p} \to 0$ . By [L2], this is $X_{n} \to X$ in $L^{p}$ .

step 1.1L1L2∎

Depends on

Used by

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Sources

Rick Durrett, Probability: Theory and Examples, 5th ed., Theorem 1.5.8 (standard reference, not scraped)

Dominated convergence in LpLp