$f_x$ is Borel measurable and $f^y$ is continuous then $f$ is Borel measurable I have to prove the following:
Let $f: \mathbb{R^2}\to \mathbb{R}$ such that $f_x:y\to f(x,y)$ is Borel measurable for all $x\in\mathbb{R}$ and that $f^y:x\to f(x,y)$ is continuous for all $y\in\mathbb{R}$. Prove that $f$ is Borel measurable.
What I have tried to do is to find a sequence of functions $f_n(x,y)$ s.t for a fixed $y$ $f_n(.,y)$ is a linear approximation of $f(.,y)$..
 A: By the continuity of $f^y$ we have
$$f(x,y) = \lim_{n \to \infty}f(\lfloor nx \rfloor / n, y).$$
By the measurability of $f^x$, we see that $f$ is the pointwise limit of a sequence of Borel measurable functions, and hence is itself Borel measurable.  
A: There is a general result that can be applied in this situation:

Definition C:
Suppose $(S,\mathscr{S})$ is a measurable space, and $(X,\tau_X)$ and $(Y,\tau_Y)$ topological spaces equipped with the Borel $\sigma$-algebras $\mathscr{B}(\tau_X)$ and $\mathscr{B}(\tau_Y)$. A function $f:S\times X\rightarrow Y$ is called a Caratheodory function if

*

*$f_s:x\mapsto f(s,x)$ is continuous for each $s\in S$,

*$f_x:s\mapsto f(s,x)$ is $\mathscr{S}-\mathscr{B}(\tau_Y)$ measurable.


Here is a well known result by Carathéodory.

Theorem: If $f$ is as in definition C, $(X,\tau_X)$ is metrizable and separable and $(Y,\tau_Y)$ is metrizable, then $f$ is $\mathscr{S}\otimes\mathscr{B}(\tau_X)-\mathscr{B}(\tau_Y)$ measurable.

A proof of this can be found in Aliprantis, C. and Border, K., Infinite dimensional analysis: A hitchhiker's guide., Third Edition, Springer. pp 153.
In the context of the OP, $S=\mathbb{R}$, $\mathscr{S}=\mathscr{B}(\mathbb{R})$, $X=Y=\mathbb{R}$ also equipped with the Borel $\sigma$-algebra.
