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Changing my question in light of Dan's answer. Thanks, Dan.

Consider a sequence of real random variables $X_i$ bounded in $L_1$, that is $\mathbb E\left|X_i\right|\leq M$ for all $i$. Suppose that they converge in distribution to $X$ (which by Fatou's lemma will also be in the $M$-ball of $L_1$). This is not enough to say that $\mathbb EX_i\to\mathbb EX$. We need uniform integrability (it is necessary for example if the RVs are nonnegative).

Boundedness in $L_1$ is not enough for uniform integrability. For example, the nonnegative RVs $X_i$ with CDF $(1-1/i)$ on $[0,i)$ and $1$ on $[i,\infty)$ are all in the 1-ball of $L_1$ but are not uniformly integrable. But their CDFs also do not converge uniformly.

So suppose we have $X_i$ bounded in $L_1$ converging in distribution to $X$ but also the CDFs converge uniformly $$\left|\left|F_i-F\right|\right|_\infty\to0.$$ Is it the case that $X_i$ are uniformly integrable and/or $\mathbb EX_i\to\mathbb EX$?

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You need uniform integrability. To change notation a bit, suppose $X_i$ and $X$ have distributions $F_i$ and $F$, respectively. Suppose for now only that $F_i(x) \rightarrow F(x)$ at each continuity point $x$ of $F$. The following are from Billingsley's Convergence of Probability Measures, Theorems 3.5 and 3.6: (1) If $g(X_n)$ are uniformly integrable, then $E[g(X_n)] \rightarrow E[g(X)]$. (2) If $g \ge 0$, then the converse holds.

I doubt the uniform convergence or uniform continuity can gain you anything. For example, if $X$ has a Cauchy distribution and $X_n = (X \wedge n) \vee (-n)$, then the CDFs of $X_n$ converge uniformly to that of $X$ and $E[X_n] = 0$, but $E[X]$ fails to exist.

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Thanks for your answer, Dan. Indeed I see from the converse (Thm 3.6) that I will definitely need uniform integrability for sure. But maybe uniform convergence will help get that. Suppose $g(X_i)$, $g(X)$ are $L_1$ bounded. I.e. $\mathbb E |g(X_i)|\leq M$, unlike your example. That's not enough for uniform integrability, true, but if the CDFs converge uniformly maybe it is? It feels right but I can't seem to show it. The converse of Vitali's convergence theorem seems maybe relevant ( – user30746 Jan 29 '13 at 20:08

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