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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 !

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    $\begingroup$ 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 $x^p$. $\endgroup$ – Rhys Steele Jan 18 at 17:30
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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 $\alpha\to\infty.$

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  • $\begingroup$ 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 $\endgroup$ – rapidracim Jan 18 at 17:42
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    $\begingroup$ @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 $n$. $\endgroup$ – Rhys Steele Jan 18 at 17:45
  • $\begingroup$ @RhysSteele I see it now, thanks ! $\endgroup$ – rapidracim Jan 18 at 17:47

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