In a proof I found the following conversion
$$E\left[|X|\mathbf{1}_{[a,b]}(Y)\right] = E\left[|X|P(a \le Y \le b)\right]$$
I understand, why $E\left[\mathbf{1}_{[a,b]}(Y)\right] = P(a \le Y \le b)$, but I do not understand the above equation. Why is it true?
Note: In my case I have $X=Y_i$ and $Y=V_n^i+\Theta \frac{Y_i}{\sqrt{n}}$. The random variables $Y_i$, $V_n^i$ and $\Theta$ are independent.
There was just a typographical error in the textbook. Instead of $E[|X|\cdot \mathbf{1}_{[a,b]}(Y)]$ it has to be $E[|X| \cdot E[\mathbf{1}_{[a,b]}(Y)]]$. Then it is obvious, that
$$E[|X| \cdot E[\mathbf{1}_{[a,b]}(Y)]] = E[|X| \cdot P(a \le Y \le b)]$$