I have got a statement which seems really trivial to me, but I was not able to prove it, so maybe someone here can help me out.
We consider a Family of random variables $(X_n)_{n\in\mathbb{N}}$ such that $\lim_{K\to\infty}\sup_{n\in\mathbb{N}}\mathbb{P}(|X_n|>K)=0$. Now we take $Y\in L^1(\mathbb{P})$. I wonder if there was a simple way to show $$\lim_{K\to \infty}\sup_{n\in\mathbb{N}}\mathbb{E}[\mathbb{I}_{\left\lbrace|X_n|>K \right\rbrace }Y]=0.$$
I am grateful for every help!
Following up on my comment.
Let $\epsilon > 0$. Assume that for all $n$, tere is an event $A_n$ with probability less than $2^{-n}$ such that $|Y|1_{A_n}$ has an expected value at least $\epsilon$.
Consider $B_n=\cup_{p \geq n}{A_p}$. Then the sequence $B_n$ decreases to $C=\lim\,\sup A_n$ which (by Borel-Cantelli) has null measure. So $|Y|1_{B_n}$ is decreasing to $0$, is dominated by the integrable function $|Y|$ but has an expected value at least $\epsilon$.
So there is a $\delta > 0$ such that if $P(A) < \delta$, then $|Y|1_A$ has expected value at most $\epsilon$.
Edit: requested clarification.
Let $\epsilon > 0$.
There is $\delta > 0$ such that if any event $A$ has probability at most $\delta$ then $\mathbb{E}[1_A|Y|] \leq \epsilon$.
There is $K > 0$ such that for every $k \geq K$, $n \geq 1$, $X_n > k$ has probability at most $\delta$.
Therefore, for every $k \geq K$, $n \geq 1$, $\mathbb{E}[|Y|1(X_n > k)] \leq \epsilon$. In other words, for all $k \geq K$, $\sup_n\,\mathbb{E}[|Y|1(X_n > k)] \leq \epsilon$.