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Suppose $\{\xi_n\}$ is a sequence of random variables on a probability space such that $E \xi_n^2 \le c $ for some constant $c.$ Assume that $\xi_n \to \xi $ almost surely as $n \to \infty$. Prove that $E \xi $ is finite and $E \xi_n \to E \xi.$

This question has been asked before, but I don't want to take the path outlined there. I'm trying to use just the theorems from my class notes: Egorov, Lebesgue Dominated Convergence, Levi Monotone Convergence, Fatou Lemma.

Attempt: By Fatou's Lemma, $E \liminf\limits_{n \to \infty} \xi_n^2 \le \liminf\limits_{n \to \infty} E \xi_n^2 = c.$ Since $\xi^2 = \lim\limits_{n \to \infty} \xi_n^2 = \liminf\limits_{n \to \infty} \xi_n^2$ almost surely, I'm tempted to conclude that $E \xi^2 \le c \, (*),$ which would solve the problem.

Assuming $(*)$ is true for the moment, we get $E |\xi| \le \sqrt{c}$ by Cauchy-Schwarz, so $E \xi$ is finite and all that remains to show is $E|\xi_n - \xi| \to 0.$ Fix $\epsilon > 0.$ By Egorov's Theorem, there exists $A$ such that $\xi_n \to \xi$ uniformly on $A$ and $P(A) \ge 1 - \frac{\epsilon}{4\sqrt{c}}.$ Then $E|\xi_n - \xi| = \int_A |\xi_n - \xi| dP + \int_{\Omega \setminus A} |\xi_n - \xi| dP.$ Using uniform convergence, there exists $N$ such that the first integral is $\le \sup\limits_{\omega \in A} |\xi_n - \xi| dP \le \epsilon/2$ for all $n \ge N.$ The second integral is at most $\frac{\epsilon}{4\sqrt{c}} (E|\xi_n|+E|\xi|) \le \epsilon/2.$ Thus, $E|\xi_n - \xi| \le \epsilon$ for all $n \ge N,$ which implies $E|\xi_n - \xi| \to 0.$

Now how do I show $(*)$? Let $A = \{w : \xi_n(\omega) \not\to \xi(\omega)\}$ be the set of measure zero on which convergence fails. What we showed is that $\int_{\Omega \setminus A} \xi^2 dP \le c.$ If we show that $\int_A \xi^2 dP = 0,$ that would finish things off. I believe this follows from the convention that $\infty \cdot 0 = 0,$ and should still apply even though we only know a priori that $\xi$ is measurable, not integrable.

Anyways, it seems I ended up answering my own question, so I just want to know if my proof is correct.

Edit: Inspired by this question, I found a fix. Let $B = \Omega \setminus A.$ Cauchy-Schwarz on the indicator function (earlier I was using $E1 =1,$ which is too weak) gives $\int_B |\xi_n - \xi| dP = E(|\xi_n - \xi| I_B) \le (E I_B E (\xi_n-\xi)^2)^{1/2} \le (P(B) \cdot 2c)^{1/2},$ which really is bounded now.

Display name
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1 Answers1

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There is a problem with your proof: You said $\int_{\Omega \setminus A} |\xi_n-\xi|dP \leq \frac {\epsilon} {4\sqrt c }(E|\xi_n|+E|\xi_i|)$. This is not correct. (It is not true that $\int_E |f| d\mu \leq P(E)\int |f|d\mu$). It is hard to get any control over $E|\xi_n I_{\Omega \setminus A}|$? I think this where the idea of uniform integrbility comes in. I doubt if this approach can be fixed.

If $A$ has measure $0$ then $\int |f| d\mu=0$ for any measurable function $f$. This is an easy consequence of the way integrals are defined: first verify it for simple functions and then take limits. Thus any measurable function is integrable over a set of measure $0$ and its integral is $0$.

Kavi Rama Murthy
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  • I didn't expect the mistake to be there. It seems like using the fact that convergence almost surely implies convergence also won't help show that $E|\xi_n - \xi| \to 0.$ For any $\epsilon > 0,$ you'll have the integral over almost all of $\Omega$ is $\le \epsilon(1-\delta),$ but that last part will be the integral of $|\xi_n - \xi|$ over a set of measure $\delta > 0.$ And no matter how much you shrink $\delta,$ the integral will not be guaranteed to shrink. To top it off, no one in my class complained about what would happen to the small part when you split up the integral. – Display name Nov 06 '20 at 12:49
  • But I have an idea. The measure $\mu(A) = \int_A f dP$ where $f = |\xi_n| + |\xi|$ is absolutely continuous with respect to $P$ since $P(A) = 0 \Rightarrow \mu(A) = 0.$ An alternate but equivalent (this was a different problem which I solved more rigorously) definition of absolute continuity is the existence of, for any $\epsilon > 0,$ a $\delta>0$ such that $P(A) < \delta \Rightarrow \mu(A) < \epsilon.$ Thus, we may fix such a $\delta$ for $\mu$ and $\epsilon/2$ at the start, then replace $1- \epsilon/(4 \sqrt{c})$ with $1-\delta.$ What do you think? – Display name Nov 06 '20 at 12:55
  • Your $\delta$ depends on $n$. That is a serious issue. – Kavi Rama Murthy Nov 06 '20 at 13:50
  • In that case, I've exhausted all the tools my class uses. Before resorting to the notion of uniform integrability, I'll ask my professor what he intended and see if any students found an approach that avoids all the problems we pointed out collectively. – Display name Nov 06 '20 at 14:01
  • It turns out that all you had to do was use Cauchy-Schwarz again, but on the indicator function. Check my edit because at this point I'm starting to get suspicious of any easy fixes. – Display name Nov 06 '20 at 16:06
  • @Displayname Your proof is correct now. – Kavi Rama Murthy Nov 06 '20 at 23:22