$\lim_{n\to\infty}E[XI_{A_n}]=0.$

My school has ended and this caught my attention when I was reviewing.

Given a probability space $(\Omega, \mathcal{A}, P)$. Let $X \in L^1$ on $(\Omega, \mathcal{A}, P)$ and let $\left\{A_n\in\mathcal{A}:n\geq1\right\}$ be a sequence of sets such that $\lim_{n\to\infty}P(A_n)=0$. Show that
$\lim_{n\to\infty}E[XI_{A_n}]=0.$ (Not assuming that $\lim_{n\to\infty}XI_{A_n}=0$ a.s.).

My approach at that time is to use the linearity of $E$ and indicator method. Where goes
$$\begin{gather} \lim_{n\to\infty}E[XI_{A_n}] = E[X]P[\lim_{n\to\infty}A_n]\to 0 \end{gather}$$ Which I thought is quite neat. However, the answer provided by our TA was.

Is my approach not rigorous, and may anyone explain a bit on the TA's solution? Thanks.

Thanks again sincerely for all the replies, they are all really instructive!

Solution 1:

This is equivalent to the following elementary measure-theory result: if $f \in L^1(X, \mathcal{M}, \mu)$, then for any $\epsilon>0$ there is a $\delta>0$ such that $E \in \mathcal{M}$, $\mu(E)<\delta$ implies $\int_{E} |f| < \epsilon$.

See here for a proof.

Solution 2:

Your TA's solution needs correction. I would write it like this [with as few changes as possible]:

$|XI_{\{|X|>n\}}| \leq |X|, X \in L^1$

By the Dominated Convergence Theorem, $$\lim_{n\to\infty}E\left[|XI_{\{|X|>n\}}|\right] = E\left[\lim_{n\to\infty}|XI_{\{|X|>n\}}|\right] = E[0]=0$$ Note that, $\forall M > 0$, $$\begin{align*}0 \leq \left|E[XI_{A_n}]\right| \leq E\left[|X|I_{A_n}\right] &= E\left[|X|I_{A_n}I_{\{|X|>M\}}\right] + E\left[|X|I_{A_n}I_{\{|X|\leq M\}}\right] \\ &\leq E\left[|X|I_{\{|X|>M\}}\right] + MP(A_n)\end{align*}$$ $$\implies 0 \leq \lim_{n\to\infty} |E[XI_{A_n}]| \leq \lim_{n\to\infty}\left(E\left[|X|I_{\{|X|>M\}}\right] + MP(A_n)\right) = E\left[|X|I_{\{|X|>M\}}\right]$$ $$\implies 0 \leq \lim_{n\to\infty} |E[XI_{A_n}]| = \lim_{M\to\infty}\lim_{n\to\infty} |E[XI_{A_n}]| \leq \lim_{M\to\infty} E\left[|X|I_{\{|X|>M\}}\right] = 0$$ Thus, $\lim_{n\to\infty} E[XI_{A_n}] = 0$

Solution 3:

Integrable random variable $X$ has a property that for any $\varepsilon>0$, there exists $\delta>0$ such that $\int_{B}|X|dP<\varepsilon$ whenever $B\in\mathcal{A}$ and $P(B)<\delta$.

Your assertion follows from this fact immediately.

To prove this fact: Firstly, note that this is true if $X$ is bounded. For if $|X|\leq M$ for some $M>0$, then given $\varepsilon>0$ we simply take $\delta=\frac{1}{2M}\varepsilon$.

For the general case, let $Y_{n}=\min(|X|,n)$. By Monotone Convergence Theorem Theorem, $\int Y_{n}dP\rightarrow\int|X|dP$. In particular, given $\varepsilon>0$, we choose $n$ such that $0\leq\int\left(|X|-Y_{n}\right)dP<\frac{\varepsilon}{2}.$ Since $Y_{n}$ is bounded, we can further choose $\delta>0$ such that $\int_{B}Y_{n}dP<\frac{\varepsilon}{2}$ whenever $P(B)<\delta$. Now, let $B\in\mathcal{A}$ with $P(B)<\delta$. We have that \begin{eqnarray*} & & \int_{B}|X|dP\\ & = & \int_{B}\left(|X|-Y_{n}\right)dP+\int_{B}Y_{n}dP\\ & \leq & \int\left(|X|-Y_{n}\right)dP+\int_{B}Y_{n}dP\\ & < & \frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ & = & \varepsilon. \end{eqnarray*}