Prove that: $P\{X < Y\} = \int^{+\infty}_{0}F_{X}(x)f_{Y}(x)dx$ 
Prove that: $P\{X < Y\} = \int^{+\infty}_{0}F_{X}(x)f_{Y}(x)dx$ 
When X and Y are two non-negtive independent r.v.s

Here is What I have tried:
L.H.S = $\int _{-\infty}^{+\infty}dy\int_{-\infty}^{x}f(x, y)dx$
= $\int _{-\infty}^{+\infty}dy\int_{-\infty}^{x}f_X(x)f_Y(y)dx $
= $\int _{-\infty}^{+\infty}f_Y(y)dy\int_{-\infty}^{x}f_X(x)dx $
= $\int _{-\infty}^{+\infty}f_Y(y)F_X(x)dy $
= $\int _{0}^{+\infty}f_Y(y)F_X(x)dy $
Then I got stuck.
Could You give me some suggestions or hints?
I'll really appreciate it. Thanks a lot.
 A: It should be $$\mathbb{E}[\mathbb{E}(\mathbf{1}_{\{X<Y\}}|Y=y)]=\\\int_{0}^{\infty}P(X<Y|Y=y)f_{Y}(y)dy=\\\int_{0}^{\infty}P(X<y)f_{Y}(y)dy= \\\int_{0}^{\infty} F_{X}(y)f_{Y}(y) dy$$.
( As $P(X<y)= F_{X}(y)$ )
Now $y$ is just a variable in integration. You can obviously change it to $x$ or $t$ or $\theta$ or anything you want.
So it is indeed $$ \int_{0}^{\infty} F_{X}(y)f_{Y}(y) dy=\int_{0}^{\infty} F_{X}(t)f_{Y}(t) dt=\int_{0}^{\infty} F_{X}(x)f_{Y}(x) dx$$ . Hence proof is complete.
The 3rd step was possible as $X$ and $Y$ are independent.
Here $f_{X}$ denotes the pdf and $F_{X}$ denotes the cdf.
In your method you do not need to take the integral from $-\infty$ as they are non negative rv's.
If you want to use the joint density $f(x,y)$. Then you should integrate wrt to $x$ first . i.e. $$\int_{0}^{\infty}\int_{0}^{y}f(x,y)\,dx\,dy=\\ \int_{0}^{\infty}\int_{0}^{y}f_{X}(x)f_{Y}(y)\,dx\,dy=\\\int_{0}^{\infty}P(X<y)f_{Y}(y)\,dy=\\ \int_{0}^{\infty}F_{X}(y)f_{Y}(y)\,dy$$. This would give you the same result.
If you want to integrate wrt $y$ first. $$\int_{0}^{\infty}\int_{x}^{\infty}f(x,y)\,dy\,dx$$  This will obviously coincide with what I have done above as the joint pdf is just product of marginal pdf's and you just need to change the order of integration. Then following your steps you would arrive at $$\int_{0}^{\infty}(1-F_{Y}(x))f_{X}(x)\,dx$$ which is also correct but not what's required.
