Almost sure convergence implies uniform integrability (Submartingales)

Question

Question: Assume $M_n \rightarrow M_{\infty}$ almost surely where the extended sequence $\{M_0,M_1,...,M_{\infty}\}$ is a submartingale. Show that $(M_n)_{n \ge 0}$ is uniformly integrable. That is $\lim_{a \to \infty}\sup_nE[|M_n|\cdot \mathbb{I}_{\{|M_n|> a\}}]=0$

Proof: $\sup_n P(M_n^+ > \epsilon) \le \sup_n \frac{E[M_n^+]}{\epsilon} \le \frac{E[M_\infty]}{\epsilon} \rightarrow 0$ as $\epsilon \to \infty$ (by Markov's inequality and the assumption above). This implies that $\sup_nE[M_n^+\cdot \mathbb{I}_{\{M_n^+ > \epsilon\}}] \le \sup_nE[M_\infty^+\cdot \mathbb{I}_{\{M_n^+ > \epsilon\}}] \rightarrow 0$ as $\epsilon \rightarrow \infty$. We now have that $M^+$ is uniformly integrable and $\lim_{n \to \infty}E[M_n^+] = E[M_\infty^+].$ Since, $\lim_{n \to \infty}E[M_n] = E[M_\infty]$ we also have $\lim_{n \to \infty}E[M_n^-] = E[M_\infty^-]$ which implies $M^-$ is uniformly integrable. Therefore, $M = M^+ - M^-$ is uniformly integrable.

Could someone please explain how $\sup_n P(M_n^+ > \epsilon) \rightarrow 0$ as $\epsilon \to \infty$ implies $\sup_nE[M_n^+\cdot \mathbb{I}_{\{M_n^+ > \epsilon\}}] \rightarrow 0$ as $\epsilon \to \infty$? I believe this equation is saying that if the probability of an event goes to zero then the expected value of a random variable restricted to this event also has to go to zero. Why is this always true?

Also, how are we able to conclude just from $\lim_{n \to \infty}E[M_n^-] = E[M_\infty^-]$ that $M^-$ is uniformly integrable?

Answer 1

For the first question, the implication follows directly assuming integrability of $M_n^+$ (i.e. $\mathbb{P}(M_n^+>\epsilon) = \mathbb{E}\mathbb{I}_{\{M_n^+>\epsilon\}}\to 0$ and adding the integrable $M_n^+$ doesn't change anything). Since $(M_n)$ is a (sub)martingale, this condition is met.

As for the second question, we have that $M_n\overset{a.s.}{\to} M_\infty$ and hence $M^-_n\overset{a.s.}{\to} M^-_\infty$. Crucially, $M^-_n \geq 0$ for all $n$, and hence $\mathbb{E}[M^-_n] = \mathbb{E}[|M^-_n|]$. This (a sequence $X_n\overset{a.s.}{\to}X$ is u.i. if $\mathbb{E}[|X_n|] = \mathbb{E}[|X|]$) is a classical and very useful condition for uniform integrability. Without spoiling the fun, proving it relies on DCT to get to $\mathbb{E}[X_n\mathbb{I}_{\{X_n>c\}}] \to \mathbb{E}[X\mathbb{I}_{\{X>c\}}]$ for $c>0$. In fact, proving this statement is a homework question here (exercise 6, file includes solutions as well).

Answer 2

Are you confident that this proof is correct? I'm also confused.

Note that you didn't copy $\sup_n E[M_\infty^+ 1_{\{M_n^+>\varepsilon\}}]$ correctly down to the bottom half of your question. But I don't understand how this expression is obtained anyway; we know that $M_n^+ \leq E[M_\infty^+|{\cal F}_n]$, but I don't know that we know $M_n^+ \leq M_\infty^+$. Even if what you wrote is correct, I don't understand how the conclusion is drawn.

If I were to prove this, I would use the following fact which I think is standard: If $X \in L^1(\Omega,{\cal F},P)$, then the collection $\{E[X|{\cal G}]: {\cal G} \subseteq {\cal F}\}$ is uniformly integrable. You can apply this fact to say that $E[M_\infty^+|{\cal F}_n]$ is uniformly integrable. Since $$ \sup_n E[M_n^+ 1_{\{M_n^+>\varepsilon\}}] \leq \sup_n E[E[M_\infty^+|{\cal F}_n] 1_{\{E[M_\infty^+|{\cal F}_n]>\varepsilon\}}], $$ you can use uniform integrability of $E[M_\infty^+|{\cal F}_n]$ to draw the conclusion.

I think for the second question, the answer already given is correct.