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Uniform integrability plus convergence in probability implies $L^{1}$ convergence

Statement

If $X_{n} \to X$ in probability and ${X_{n} : n \in N}$ is uniformly integrable, then $X \in L^{1} (P)$ and $X_{n} \to X$ in $L^{1}$ .

Facts & Assumptions

Given: Integrable real random variables $X_{n}$ , a real random variable $X$ , probability convergence, and uniform integrability of $(X_{n})$ .

[L1]

Probability convergence is convergence in measure for the probability measure (Convergence in probability).

[L2]

On a finite measure space, convergence in measure plus uniform integrability is equivalent to $L^{1}$ convergence (Vitali convergence theorem on finite and sigma-finite measure spaces).

Proof

technique · direct

1.1

The underlying measure has total mass one, hence is finite; [L1] converts the hypothesis to convergence in measure.

2.1

Apply the finite-measure reverse implication of [L2] to $(X_{n})$ . [step 1.1, L2] It supplies $X \in L^{1}$ and $E ∣ X_{n} - X ∣ \to 0$ , namely $L^{1}$ convergence.

step 1.1L2∎

Depends on

Used by

Uniform integrability characterizes L¹ convergence under probability convergence Corollary

Dependency tree · two levels

17 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

Rick Durrett, Probability: Theory and Examples, 5th ed., Theorem 4.6.3 (standard reference, not scraped)

Uniform integrability plus convergence in probability implies L1L1 convergence