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I asked a slightly similar question here: Does Convergence in probability implies convergence of the mean?, but now I wish to examine a stricter scenario: Let $\{X_n\}_{n=1}^\infty$ be a sequence of random variables converging a.s to a const $c$. Is it required for the sequence to be uniformly integrable in order to imply $\lim_{n\to \infty} EX_n = c$?

And what about $\lim_{n\to \infty} EYh(X_n) = E[Y]h(c)$ for some random variable $Y$ and a continuous function $h$? Under which regularity conditions does the last equality holds?

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  • $\begingroup$ Think again here $~X_n~$ not converges to $~0~$ almost surely. When you are taking $~X_n=2^n~$ then probability approaches towards $~0~$ but not zero. Definition is $~P[\lim X_n=0]=1~$, but here when $~X_n=0~$ probability is not exactly $~1~$, lies near $~1~$; hence contradicts the definition. $\endgroup$
    – Anonymous
    Commented Sep 10, 2019 at 17:17

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For the first question, try $P[X_n=0]=1-1/n^2$, $P[X_n=2^n]=1/n^2$, then $X_n\to0$ almost surely but $E[X_n]$ does not converge to $0$. (The second question is unclear.)

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  • $\begingroup$ Thanks. But, in order for the mean value to converge, do I have to enforce uniform integrability, or because of the a.s. convergence, I can use a weaker condition? $\endgroup$
    – SBM
    Commented Oct 20, 2013 at 15:14
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    $\begingroup$ A classical way to deduce $L^1$ convergence from almost sure convergence is to assume uniform integrability, yes. $\endgroup$
    – Did
    Commented Oct 20, 2013 at 15:36

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