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$L^{1}$ convergence implies uniform integrability

Statement

If $X_{n} \to X$ in $L^{1}$ on a probability space, then ${X, X_{0}, X_{1}, \dots}$ is uniformly integrable.

Facts & Assumptions

Given: Integrable $X_{n}, X$ with $E ∣ X_{n} - X ∣ \to 0$ .

[L1]

Uniform integrability is vanishing uniformly of the large-value tail integrals (A uniformly integrable family).

[L2]

$L^{1}$ convergence means $E ∣ X_{n} - X ∣ \to 0$ (Convergence in L^1(mu)).

Proof

technique · direct

1.1L2algebra

Given $ε > 0$ , choose $N$ from [L2] so that the following tail estimate holds for $n \geq N$ . [L2, algebra] $E ∣ X_{n} - X ∣ < ε / 4$ for $n \geq N$ . For $M > 0$ , $E [∣ X_{n} ∣ 1_{∣ X_{n} ∣ > M}] \leq 2 E ∣ X_{n} - X ∣ + 2 E [∣ X ∣ 1_{∣ X ∣ > M / 2}]$ for $n \geq N$ , by splitting at $∣ X ∣ > M / 2$ .

2.1step 1.1L1choose∎

Choose $M$ so the tail of $X$ in step 1.1 is below $ε / 4$ and control the finite initial family separately. [step 1.1, L1, choose] The finitely many functions $X, X_{0}, \dots, X_{N - 1}$ each have tail below $ε$ . Then step 1.1 gives the same bound for all later $X_{n}$ . By [L1] the whole family is uniformly integrable.

Depends on

Used by

Dependency tree · two levels

11 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)

L1L1 convergence implies uniform integrability