We say that a family of random variables $X_n, n \geq 1$ is uniformly integrable if

$$\lim_{M \rightarrow \infty} \sup_{n} E[|X_n| 1_{|X_n|>m}]=0.$$

I am struggling with some proofs and could need some help. are my ideas correct? how does one conclude correctly? Is there an easier proof, ...

  • We want to show that $\sup_n ||X_n||_p < \infty$ for some $p> 1$ implies uniform integrability.

$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} E[|X_n|^p],$$ using Jensen.

Since $\sup_n ||X_n||_p < \infty$, we also have $\sup_n E[|X_n|^p] < \infty$, and the claim follows by letting $M \rightarrow \infty$.

  • Now we want to show that a finite family of random variables in $L^1$ is always uniformly integrable.

Let $n \in N$ for some finite set $N$. Define $$M_0:=\max_{n \in N} |X_n|.$$ Then we have $$E[|X_n| 1_{|X_n|>M_0}]= E[|X_n|\cdot 0 ] = 0,$$ and hence we can take the $sup$ to get for all $M \geq M_0$ that

$$\sup_n E[|X_n| 1_{|X_n|> M}] = 0.$$

The result follows by taking the limit $M \rightarrow \infty$. Do we need here something like monotonce or dominated convergence? Is this proof valid? If not, how would one prove it? is there a more elegant way of proving it?

  • When $E[\sup_n |X_n|] < \infty$, then the sequence is uniformly integrable.

How can one interchange order of $\sup$ and expectation? I have no idea!


$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} E[|X_n|^p],$$ using Jensen.

You didn't apply Jensen's inequality correctly; it should read

$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} \left( E[|X_n|^p] \right)^{\color{red}{\frac{1}{p}}}.$$

[...] and the claim follows by letting $M \rightarrow \infty$.

No, it's not that simple. Letting $M \to \infty$ you get

$$\lim_{M \to \infty} \sup_n \mathbb{E}(|X_n| 1_{|X_n|>M}) \leq \sup_{n \in \mathbb{N}} \|X_n\|_p,$$

but that's not good enough; you have to show that the limit equals $0$. Hint for this problem: Use Markov's inequality, i.e.

$$\mathbb{E}(|X_n| 1_{\{|X_n|>M}) \leq \frac{1}{M^{p-1}} \mathbb{E}(|X_n|^p 1_{|X_n|>M}) \leq \frac{1}{M^{p-1}} \mathbb{E}(|X_n|^p).$$

Define $$M_0:=\max_{n \in N} |X_n|.$$ Then we have $$E[|X_n| 1_{|X_n|>M_0}]= E[|X_n|\cdot 0 ] = 0,$$

No this doesn't work, because $M_0$ depends on $\omega$. Unfortunately, this means that your approach fails. Hint for this one: Using e.g. the dominated convergence theorem check first that the set $\{f\}$ is uniformly integrable. Extend the approach to finitely many integrable random variables.

When $E[\sup_n |X_n|] < \infty$, then the sequence is uniformly integrable.

Hint: By assumption, $Y := \sup_n |X_n|$ is integrable and $|X_n| \leq Y$ for all $n \in \mathbb{N}$. Consequently,

$$\mathbb{E}(|X_n| 1_{|X_n|>M}) \leq \mathbb{E}(|Y| 1_{|Y|>M}) \qquad \text{for all $M>0$ and $n \in \mathbb{N}$.}$$

Now use the fact that $\{Y\}$ is uniformly integrable (see question nr. 2).

  • $\begingroup$ Thanks for your answer. I have never attended a course in measure theory and also only attended a basic course in analysis. So it is a bit hard for me. Can you maybe give me some more hints? I really tried to do something with it, but all tries are quite hopeless. $\endgroup$ – user136457 Jan 9 '16 at 15:47
  • 3
    $\begingroup$ @user136457 Giving more hints means, basically, giving full solutions. There are a lot of similar questions on stackexchange; you just have to search for them. Have e.g. a look at these questions: math.stackexchange.com/q/729217, math.stackexchange.com/q/1191626 (And, honestly, there are 3 questions, I gave you three hints and after ~45 minutes you are telling me that you "really" tried it... do you think that 15 minutes per question is enough; in particular if you are not familiar with measure theory?) $\endgroup$ – saz Jan 9 '16 at 16:08
  • $\begingroup$ Note that there are several ways to showing these properties. One very helpful way of proving uniform integrability is using the de la Vallée-Poussin criterion which you can find in several textbooks (Cinlar page 73, and probably somewhere in Kallenberg). $\endgroup$ – Olorun Jan 11 '16 at 7:11

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