Proof continuous conditional expectation Let $(X,Y)\in \mathbb{R}^2$ be a random vector with joint density $f(x,y)$, this means that for any bounded Borel function $\phi$ on $\mathbb{R}^2$,
\begin{align}
\mathbb{E}[\phi(X,Y)]=\int \int_{\mathbb{R}^2} \phi(x,y) \cdot f(x,y)\ dx\ dy.
\end{align}
The marginal density of $Y$ is defined by $f_Y(y)=\int_{\mathbb{R}} f(x,y)\  dx$. Let,
\begin{align}
f(x|y) = \begin{cases}
\frac{f(x,y)}{f_Y(y)} &\text{if $f_Y(y)>0$}\\
0 &\text{if $f_Y(y)=0$}.
\end{cases}
\end{align}
Now I want to prove that:
• $f(x|y)=\frac{f(x,y)}{f_Y(y)}$ for almost every $(x,y)$ with respect to the Lebesgue measure on $\mathbb{R}^2$. 
• $f(x|y)$ functions as conditional density of $X$, given that $Y=y$, for a bounded Borel function $h$ on $\mathbb{R}$ yields:
\begin{align}
\mathbb{E}[h(X)|Y](\omega) = \int_{\mathbb{R}} h(x)\ f(x|Y(\omega))\ dx.
\end{align}
To start with the first bullet, we can introduce the set $H:=\{(x,y):\ f(x|y)\neq \frac{f(x,y)}{f_Y(y)}\}$. With the idea to show that $m(H_y)=0$ for each $y$-section of $H$ (how?) and thereafter use Tonelli's theorem. (why?) 
 A: I will give a proof for the second bullet. Let $h: \Bbb R \to \Bbb R$ be a bounded measurable function. Fix a Borel set $B \subset \Bbb R$. Note that since $f_Y$ is the density of $Y$, we have $$\Bbb E \bigg[ \bigg( \int_{\Bbb R} h(x)f(x|Y)dx\bigg) \cdot 1_{\{Y\in B\}} \bigg] =\int_B \bigg( \int_{\Bbb R}  h(x) f(x|y) dx \bigg)f_Y(y) dy $$
where $f(x|Y)$ is a shorthand notation for the random variable $\omega \mapsto f(x|Y(\omega))$. But, $f(x|y)f_Y(y)=f(x,y)$ for all $x,y \in \Bbb R^2$ (by definition), and thus this last expression reduces to $$\int_B \int_{\Bbb R}  h(x) f(x,y) dx \;dy = \Bbb E[h(X)\cdot 1_{\{Y \in B\}}]$$ Thus we have shown that $$\Bbb E \bigg[ \bigg( \int_{\Bbb R} h(x)f(x|Y)dx\bigg) \cdot 1_{\{Y\in B\}} \bigg] = E[h(X)\cdot 1_{\{Y \in B\}}]$$ for any Borel set $B \subset \Bbb R$. Thus by the definition of conditional expectation it follows that $$\Bbb E[h(X)|Y]=\int_{\Bbb R} h(x)f(x|Y)dx$$ because the right-hand side is clearly $\sigma(Y)$-measurable. This is exactly what you wanted to prove.
