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Uniform integrability characterizes $L^{1}$ convergence under probability convergence

Statement

Suppose $X_{n} \to X$ in probability. Then $X_{n} \to X$ in $L^{1}$ if and only if ${X_{n} : n \in N}$ is uniformly integrable.

Facts & Assumptions

Given: $X_{n} \to X$ in probability and each $X_{n}$ is integrable.

[L1]

$L^{1}$ convergence makes the sequence together with its limit uniformly integrable ( $L^{1}$ convergence implies uniform integrability).

[L2]

Uniform integrability plus probability convergence gives $L^{1}$ convergence (Uniform integrability plus convergence in probability implies $L^{1}$ convergence).

Proof

technique · direct

1.1

If $X_{n} \to X$ in $L^{1}$ , [L1] makes the larger family uniformly integrable. [L1] Thus its subfamily ${X_{n} : n \in N}$ is uniformly integrable.

2.1

Conversely, if $(X_{n})$ is uniformly integrable, [L2] applies to the given probability convergence. [L2] It yields $X \in L^{1}$ and $X_{n} \to X$ in $L^{1}$ .

L2∎

Depends on

Used by

Uniform integrability repairs the expectation limit Example

Dependency tree · two levels

9 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 characterizes L1L1 convergence under probability convergence