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

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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

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.

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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 $\mathrm{Pr}(A_{n})$.

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