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). <http://en.wikipedia.org/wiki/Uniform_integrability>


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 $\mathbb EX\_i\to\mathbb EX$?



Is an L_1 bounded sequence of random variables with uniformly converging CDFs uniformly integrable?
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.


