CheckMyProof $\mathbb{P}(X < Y) = \int_{\mathbb{R}} \mathbb{P}(X < y) f_{Y}(y) \ \text{d}y$, where $f_Y$ is PDF of $Y$ and $X,Y$ independent 
Let $X$ and $Y$ be independent random variables on a probability space $(\Omega, \mathcal{A}, \mathbb{P})$ with the probability density functions $f_{X}$ and $f_{Y}$, respectively.
  Show that 
  $$
\mathbb{P}(X < Y)
= \int_{\mathbb{R}} \mathbb{P}(X < y) f_{Y}(y) \ \text{d}y.
$$

Here's my attempt:
We have
$$
\mathbb{P}(X < Y)
= \mathbb{P}(X - Y < 0)
= \mathbb{P}(X - Y \le 0) - \mathbb{P}(X = Y)
= \mathbb{P}(X - Y < 0),
$$
sine $X$ and $Y$ are independent.
Now, 
\begin{align}
\mathbb{P}(X - Y \le 0)
& = \int_{- \infty}^{0} f_{X + (-Y)}(x) \ \text{d}x
= \int_{- \infty}^{0} (f_{X} \ast f_{-Y})(x) \ \text{d}x = \int_{- \infty}^{0} \int_{\mathbb{R}} f_{X}(x - \tau) f_{-Y}(\tau) \ \text{d}\tau \ \text{d}x \\
& = \int_{\mathbb{R}} \int_{- \infty}^{0} f_{X}(x - \tau) \text{d}x \ f_{-Y}(\tau) \ \text{d}\tau
 = \int_{\mathbb{R}} \int_{- \infty}^{-\tau} f_{X}(x) \ \text{d}x \ f_{-Y}(\tau) \ \text{d}\tau \\
& = \int_{\mathbb{R}} \mathbb{P}(X \le - \tau) f_{-Y}(\tau) \ \text{d}\tau  = \int_{\mathbb{R}} \mathbb{P}(X \le - \tau) f_{Y}(-\tau) \ \text{d}\tau \\
& = \int_{\mathbb{R}} \mathbb{P}(X \le \tau) f_{Y}(\tau) \ \text{d}\tau 
= \int_{\mathbb{R}} \mathbb{P}(X < \tau) f_{Y}(\tau) \ \text{d}\tau 
\end{align}
Is this correct?
That $f_{-Y}(x) = f_{Y}(-x)$ follows from
\begin{equation*}
\int_{-\infty}^{-x} f_{Y}(y) \ \text{d}y
= \int_{-\infty}^{x} f_{Y}(-y) \ \text{d}y
\end{equation*}
since $f(y) \mapsto f(-y)$ is a reflection about the $y$-axis.
 A: (Migrated from the comment) Here is a simple approach:
\begin{align*}
\mathbb{P}(X<Y)
&=\int_{\mathbb{R}^2}\mathbf{1}_{\{x<y\}}f_{X,Y}(x,y)\,\mathrm{d}x\mathrm{d}y\\
&=\int_{\mathbb{R}}\int_{(-\infty,y)}f_X(x)f_Y(y)\,\mathrm{d}x\mathrm{d}y\\
&=\int_{\mathbb{R}}\mathbb{P}(X<y)f_Y(y)\,\mathrm{d}y.
\end{align*}
In the second step, we utilized Fubini-Tonelli theorem to transform the double integral into the iterated integrals. Also, $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ follows from the independence of $X$ and $Y$.
A: Here is a more complicated alternative to Sangchul's excellent answer.
If we let $V=\{ (x,y) \mid x<y \}$ and $V_n =\{ (x,y) \mid x<{1 \over n} \lfloor n y \rfloor \}$, it is not hard to show $P[(X,Y) \in V] = \lim_{n \to \infty} P [(X,Y) \in V_n]$.
Note that
$$V_n = \bigcup_k \left\{(x,y) :x < {k \over n}, {k \over n} \le y < {k+1 \over n} \right\},$$
and so from independence we have 
$$
P [(X,Y) \in V_n]
= \sum_k P\left[X < {k \over n}\right] P\left[ {k \over n} \le Y < {k+1 \over n}\right].
$$
Letting $P_Y(A) = P[Y \in A]$, we have
\begin{eqnarray}
P [(X,Y) \in V_n]
&=& \sum_k \int_{k \over n}^{k+1 \over n} P\left[X < {k \over n}\right] dP_Y(t) \\
&=& \sum_k \int_{k \over n}^{k+1 \over n} P\left[X < {1 \over n} \lfloor n t \rfloor \right] dP_Y(t) \\
&=& \int_\mathbb{R} P[X < {1 \over n} \lfloor n t \rfloor ] dP_Y(t)
\end{eqnarray}
The result follows from 
$$\lim_n \int_\mathbb{R} P\left[X < {1 \over n} \lfloor n t \rfloor \right] dP_Y(t) = \int_\mathbb{R} P[X < t ] dP_Y(t).$$
