Proof that conditional density is Borel measurable

Question

I want to prove that the conditional density $f_{X|Y}: \mathbb {R^2} \to \mathbb{R}$ defined by $f_{X|Y}(x|y)= \begin{cases} \frac{f_{X,Y}(x,y)}{f_{Y}(y)} & \text{ if} f_{Y}(y)>0 \\ f_{X}(x) & \text{ otherwise} \end{cases}$ is borel measurable

I know that $f_{X,Y}(x,y)$ is Borel measurable (by Radon Nikodym) and by Fubini´s theorem $f_{Y}(y)$ and $f_{X}(x)$ are also a borel measurable functions (which we can extend them in $\mathbb {R^2}$)

http://statweb.stanford.edu/~adembo/stat-310b/lnotes.pdf here is a reference about this fact in page 171 but without a proof. A friend of mine told me that this is a consequence about Fubini´s theorem, but I don´t see why this is true

I would really appreciate any hint or suggestion about this problem.

Answer 1

Firstly, $\left\{(x,y)|\frac{f_{X,Y}(x,y)}{f_Y(y)}>\alpha\right\}$ is a Borel set for every $\alpha\in\mathbb{R}$, and of course $\{(x,y)|f_X(x)>\alpha\}$ is also Borel.

Now we want to argue that $f_{X|Y}$ is Borel. Observe that \begin{align*} &\{(x,y)|f_{X|Y}(x,y)>\alpha\}\\ &=\left\{(x,y)|f_Y(y)>0, \frac{f_{X,Y}(x,y)}{f_Y(y)}>\alpha\right\} \cup \{(x,z)|f_Y(y)=0, f_X(x)>\alpha\}\\ &=\left( \{f_Y>0\}\cap\left\{\frac{f_{X,Y}}{f_Y}>\alpha\right\} \right)\cup \left( \{f_Y=0\}\cap\{f_X>\alpha\}\right). \end{align*}

Hopefully it is clear that all sets here are Borel sets.

Answer 2

Define $F:\mathbb{R}^2\to \mathbb{R}_{\ge 0}^3$ and $\psi:\mathbb{R}_{\ge 0}^3\to \mathbb{R}_{\ge 0}$ by $$ F(x,y)=(f_{X,Y},f_X\circ\pi_x,f_Y\circ \pi_y)(x,y), $$ where $\pi_x$ and $\pi_y$ are the coordinate maps, and $$ \psi(z,v,w)=\frac{z}{w}\cdot 1\{w>0\}+v\cdot1\{w=0\}. $$ Then $f_{X\mid Y}=\psi\circ F$ is $(\mathcal{B}_{\mathbb{R}^2},\mathcal{B}_{\mathbb{R}})$- measurable because the composition of Borel measruable functions is measurable.