How to show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$? Let $X$ be a positive countinuous random variable. How do I show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$?
It seems that I have missed something important.
$\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\int\limits_{-\infty}^{\infty}x\cdot f_{X\mathbb{1}_{\lbrace X\le a \rbrace}}(x)dx$ and i want to justify it equals to $\int\limits_{-\infty}^{a}x\cdot f_X(x)dx$.
 A: The equation is false because $E(X\mathbb{I}_{\{X \le a \}})  \ne P(X \le a )$. In fact, we have rather $P(X \le a ) = E(\mathbb{I}_{\{X \le a \}})$.
For your second question: why we have $E[X \mathbb{1}_{\{X>a\}}]\le \sqrt{E[X^2]}\sqrt{P\{X>a\}}$ ?
In fact, we apply the Cauchy-Schwarz inequality here, for $A,B$ two random variable
$$E(AB) \le \sqrt{E(A^2)}\sqrt{E(B^2)} \tag{1}$$
Just replacing $A = X$ and $B = \mathbb{I}_{\{X \le a \}} $ in $(1)$ and noticing 2 things below:
$$B^2 = B = \mathbb{I}_{\{X \le a \}} $$
and
$$E(\mathbb{I}_{\{X \le a \}})= P(X \le a )$$
A: Thie is very very false. Take $X$ to be  an exponetial r.v. with parameter $1$. Then LHS is $\int_0^{a} xe^{-x}dx$ and RHS is $\int_0^{a} e^{-x}dx$. You can compute these integrals and see that these two are not equal for any $a \in (0,\infty)$.
$EX1_{X\leq a}=\int_0^{a}xf_X(x)dx$ is the correct formula.
A: The equation you provided is false. The correct one is
$$ \mathbb{E}[\mathbb{1}_{X\leq a}] = \mathbb{P}(X ≤ a) .$$
In fact,
$$ \mathbb{E}[\mathbb{1}_{X ≤ a}] = \int_{-\infty}^{\infty} \mathbb{1}_{X ≤ a} f_X(x) dx = \int_{-\infty}^a f_X(x) dx = \mathbb{P}(A) . $$
A: That is not quite correct.  It is that : $\mathsf E(\mathbf 1_{X\leqslant a})=\mathsf P(X\leqslant a)$

What $\mathsf E(X\,\mathbf 1_{X\leqslant a})$ equals is $\int\limits_0^a\mathsf P(y\lt X\leqslant a)\,\mathrm d y$
The key is that $X$ is a positive continuous random variable; meaning that what ever the support for $f$ may be, it does not intercept the negative reals.  So for any positive $a$...
$$\begin{align}\mathsf E(X\,\mathbf 1_{X\leqslant a})&=\int_0^a x~f(x)\,\mathrm d x\\&=\int_0^a \left(\int_0^x \mathrm d y\right)~f(x)~\mathrm d x\\&=\int_0^a\int_y^a f(x)\,\mathrm d x\,\mathrm d y\\&=\int_0^a\mathsf P(y\lt X\leqslant a)~\mathrm d y\end{align}$$
