Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let X be integrable and $A_n$ a sequence of subsets such that $ \lim_{n\to \infty} {P(A_n)} =0$. Show that $E X 1_{A_n} \to 0$.

share|improve this question
1  
Dear Kamini, your question is not clear, probably there are some typos. Is $F=X$? Is $1A_n$ the indicator function of $A_n$? in this case I think it is better to write $1_{A_n}$. –  Hans Oct 17 '12 at 17:40
    
What is $X$???? –  copper.hat Oct 17 '12 at 17:40
    
Yes 1An the indicator function of An. Sorry I am not used to TeX commands. –  Kamini Oct 17 '12 at 17:49
    
The $X$ has disappeared from the expectation??? –  copper.hat Oct 17 '12 at 18:14

2 Answers 2

Presumably you mean $X$ is integrable?

Since $X$ is integrable, the measure $\mu(A) = E (|X| 1_A)$ is absolutely continuous (AC) with respect to $P$. Since $\mu$ is AC we have $\forall \epsilon>0$, there exists $\delta >0$ such that if $P(A)< \delta$ then $\mu(A) < \epsilon$.

Since $P(A_n) \to 0$, it follows that $\mu(A_n) \to 0$. Since $|E(X 1_{A})| \leq E (|X| 1_A) = \mu(A)$, it follows that $E(X 1_{A_n}) \to 0$.

An alternative approach: (This just explicitly demonstrates that the measure $\mu$ above is absolutely continuous with respect to $P$.)

Let $\epsilon >0$. Let $L_M = \{t| |X(t)| \leq M\}$. Then $|X| 1_{L_M} \leq |X|$, and $\lim_{M\to\infty} |X(t)| 1_{L_M}(t) = |X(t)|$, hence the dominated convergence theorem gives $\lim_{M\to\infty} \int_{L_M} |X| = \int |X|$. It follows that there exists $M$ such that $\int_{L_M^C} |X| < \frac{\epsilon}{2}$. Note that if $t \in L_M$, then $|X(t)|\leq M$.

Now suppose that $P(A) < \frac{\epsilon}{2M}$. Then $|\int_A X| \leq \int_A |X| = \int_{A\cap L_M} |X| + \int_{A \setminus L_M} |X|$. Since $A \setminus L_M \subset L_M^C$, we have $\int_{A \setminus L_M} |X| < \frac{\epsilon}{2}$, and $\int_{A\cap L_M} |X| \leq M P(A) < \frac{\epsilon}{2}$. Consequently, $|\int_A X| < \epsilon$.

Since $P(A_n) \to 0$, it follows that $\int_{A_n} X \to 0$. Since $E X 1_{A_n} = \int_{A_n} X$, the result follows.

share|improve this answer
    
Is Dominated Convergence Theorem related to this? –  Kamini Oct 19 '12 at 5:37
    
Copper Hat, Don't we have to do any integration at all? I am sorry I still can't understand the stated logic. Could you pls explain more? –  Kamini Oct 19 '12 at 5:45
    
I added a little more detail... –  copper.hat Oct 19 '12 at 6:32
    
Thank you very much. It's clear to me now. –  Kamini Oct 19 '12 at 10:54
    
Then you could accept this answer. –  Did Nov 3 '12 at 10:47

Actually, $E 1_{A_{n}}$ is $\mathrm{Pr}(A_{n})$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.