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Because of this question, we know that weak convergence plus uniform integrability implies convergence in expectation. But can one prove that if we have weak convergence and convergence in expectation of non-negative RVs, then we must have uniform integrability?

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On $(0,1)$ with Lebesgue measure let $X_n=n1_{(0,\frac 1 n)} -n1_{(-\frac 1 n,0)} $ and $0$. Then, $X_n \to X$ a.s., hence in distribution. Also, $EX_n=0=EX$ for all $n$. But $(X_n)$ is not uniformly integrable since $\int_{X_n \geq n} X_ndP \geq 1$.

For nonnegaive r.v.'s the answer is in the affirmative. This can be proved using two theorems: 1) Skhorohod Theorem 2) Sheffe's Lemma. By the first theorem we may replave convergence in distrbution by a.s. convergence. Then second theorem implies that $E|X_n-X| \to 0$ which implies uniform integrability.

Kavi Rama Murthy
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  • Good counterexample, but see my edit for a missing condition, the $X_i$ must be $\geq 0$ –  Dec 16 '22 at 09:08
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    It is not good to change questions after answers have appeared but I have anyway answered your new question also. @Eulerian – Kavi Rama Murthy Dec 16 '22 at 09:19