Conditional Expectation I am learning conditional expectation for fun. I cannot solve the following problem, I think it is just my vocabulary is limited. This problem came from a book on Stochastic Processes:
Let $\Omega = [0,1]\times [0,1]$ with the usual Lebesgue measure and $\sigma$-algebra of Borel sets. Let $X$ and $Y$ be random variables on $\Omega$ with joint density
$$ f_{X,Y}(x,y) = \left\{ \begin{array}{ccc} x+y & \text{ if }& x,y\in [0,1] \\ 0 & \text{ if }& \text{ otherwise} \end{array} \right. $$
Show that $\displaystyle E(X|Y) = \frac{2 + 3Y}{3 + 6Y}$. 
The part that I do not understand is in the solutions, the author begins by writing, 
$$ Y^{-1} (B) = [0,1] \times B $$
where $B$ is a Borel set in $[0,1]$. 
How can be possibly say that? There is no reason at all why $X$ and $Y$ are defined by different coordinates. Unless I completely misunderstand what it means to be of "joint density". 

we can write 
$ E(X|Y)=\int_X xf(x|y)dx$
but $f(x|y)=\frac{f(x,y)}{f(y)}$
and
$ f(y)=\int_X f(x,y)dx=\int_{0}^{1} (x+y)dx=(x^2/2+xy |_0^1)=1/2+y$
Also
$$ E(X|Y)=\int_X xf(x|y)dx=\int_{0}^{1}x \frac{x+y}{1/2+y} dx$$
$$ =\int_{0}^{1} \frac{x^2+xy}{1/2+y} dx=\frac{1}{1/2+y} (x^3/3+x^2y/2|_0^1)$$
$$ = \frac{(1/3-y/2)}{(1/2+y)}=\frac{2+3y}{3+6y}$$
