# Show that if $p > 1$ and $\sup_{n \geq 1} |X_n| \in L^p$ then $\{X_n, n \geq 1\}$ is uniformly integrable.

Let $$(\Omega,\mathcal{\Sigma,\mathbb{P}})$$ be a complete probability space.

I have to show that $$\lim_{\alpha \to \infty} \sup_{n \geq 1}\int_{\{|X_n| \geq \alpha\}}|X_n|\, d\mathbb{P} = 0$$

assuming title's assumptions and let $$\alpha > 0$$

my attempt :

$$| X_n| \leq \sup_{n\geq 1}|X_n| \leq [\sup_{n\geq 1}|X_n|]^p$$

therefore

$$\int_{\{|X_n| \geq \alpha\}}|X_n|\, d\mathbb{P} \leq \int_{\{|X_n| \geq \alpha\}} [\sup_{n\geq 1}|X_n|]^p\, d\mathbb{P} \leq \mathbb{E}[ [\sup_{n\geq 1}|X_n|]^p] < \infty$$

so for every $$n \geq 1$$

$$\int_{\{|X_n| \geq \alpha\}}|X_n|\, d\mathbb{P} < \infty$$

however I'm not sure this completes the proof because although it's finite it could possibly depend on $$n$$ and taking the supremum with $$n$$'s in an expression can cause a blow up.

any hints or help will be greatly appreciated, thanks !

• I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$. – Rhys Steele Jan 18 at 17:30

We can use the fact that $$\int_{|X_n|\ge \alpha}|X_n|d\Bbb P\le\frac{1}{\alpha^{p-1}}\int_{|X_n|\ge \alpha}|X_n|^pd\Bbb P.$$ If $$\sup\limits_{n\in\Bbb N}\|X_n\|_p=M<\infty$$ (our case is a special case of this), then we have $$\int_{|X_n|\ge \alpha}|X_n|d\Bbb P\le\frac{1}{\alpha^{p-1}}M^p,\quad\forall n\in\Bbb N.$$ This gives $$\sup_{n\in\Bbb N}\int_{|X_n|\ge \alpha}|X_n|d\Bbb P\le\frac{1}{\alpha^{p-1}}M^p$$ and we get the desired conclusion by taking $$\alpha\to\infty.$$
• thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $\|\sup_{n \geq 1} X_n\|_p < \infty \implies \sup_{n \geq 1}\|X_n\|_p < \infty$, I'm gonna think about this – rapidracim Jan 18 at 17:42
• @rapidracim Note that $\int |X_n|^p \leq \int (\sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$. – Rhys Steele Jan 18 at 17:45