1

Given $A$ is an integrable random variable and $P(A_n)\rightarrow 0$ as $n\rightarrow \infty$. Prove that $\int_{A_n} A\ dP\rightarrow 0$ as $n\rightarrow \infty$.

My Solution By using SLNN result, which gives $P(A_n)\rightarrow 0$ as $n\rightarrow \infty$ implies $1_{A_n}\rightarrow 0$ as $n\rightarrow \infty$ almost surely. Since $A$ is a random variable, $A\cdot 1_{A_n}\rightarrow 0.A= 0$ almost surely. Since $A$ is integrable, $\int_{\omega} A\cdot 1_{A_n} dP\rightarrow \int_{\omega} 0\ dP$ as $\ n\rightarrow \infty$, or $\int_{A_n} A\ dP = \int_{\omega} A\cdot 1_{A_n} dP\rightarrow 0$ as $n\rightarrow \infty$.

My Question Is the solution above correct? If not, please help me correct it.

ghjk
  • 2,937
  • Which SLLN result are you using? – aduh Sep 28 '16 at 03:18
  • @aduh: the one I stated out.... – ghjk Sep 28 '16 at 03:19
  • Sorry, I'm finding this kind of hard to follow. Why do you state $A = 0$ almost surely? If that were true, then $\int A dP = 0$ and you're done. – aduh Sep 28 '16 at 03:25
  • @aduh: I thought that's a property of random variable and convergence? Like, if $x\rightarrow y$ almost surely, and $z$ is a random variable, then $zx\rightarrow zy$ almost surely. That's also the step I'm skeptical about. You have another solution? – ghjk Sep 28 '16 at 03:26

1 Answers1

0

Your argument doesn't look right, as pointed out in the comments. Here's an alternative proof.

Suppose the conclusion is false. Then there exists an $\epsilon > 0$ and a sequence $(A_n)$ of events such that $P(A_n) < 2^{-n}$ and $\int_{A_{n}} A dP > \epsilon '$. By the Borel-Cantelli Lemma, $P(\limsup A_n) = 0$. But, letting $F = \limsup A_{n}$, (Reverse) Fatou's Lemma implies that $\int_F A dP \geq \limsup \int_{A_n} A dP > \epsilon$. This is a contradiction.

aduh
  • 8,894
  • Thank you very much for your help. I check your Reverse Fatou's Lemma, and that does not look right: https://en.wikipedia.org/wiki/Fatou%27s_lemma#Reverse_Fatou_lemma. Why do you have $\int_{F} AdP$? – ghjk Sep 28 '16 at 03:57
  • $\int \limsup \mathbf{1}{A_n} A \ dP \geq \limsup \int \mathbf{1}{A_n} A \ dP$. – aduh Sep 28 '16 at 04:00
  • But $F = \lim \sup A_{n}$, not $\lim \sup[1_{A_n}]$. – ghjk Sep 28 '16 at 04:04
  • 1
    Maybe this will help you http://math.stackexchange.com/questions/72690/liminf-and-limsup-with-characteristic-indicator-function – aduh Sep 28 '16 at 04:09
  • I'm so dumb. Thank you for your useful link! I got it now. Basically, you could move the limit to the subscript. – ghjk Sep 28 '16 at 04:35