If $y \to f(x,y)$ is Leb-measurable, $x \to f(x,y)$ is continuous, show $f$ is measurable w.r.t. completion of $L \times L$ Problem statement: Let $f(x,y) : [0,1]^2 \to R$ be such that for every $x \in [0,1]$, the function $y \to f(x,y)$ is Lebesgue measurable on $[0,1]$, and for every $y \in [0,1]$ the function $x \to f(x,y)$ is continuous on $[0,1]$. Prove that $f$ is measurable with respect to the completion of the product $\sigma$-algebra $L \times L$ on $[0,1]^2$. Here, $L$ is the Lebesgue $\sigma$-algebra on $[0,1]$.
My attempt at a solution: I'm not entirely sure how to start this. We want to show that $f^{-1}((a,\infty)) = \{(x,y) : f(x,y) > a\} \in L \times L \cup N$, where $N$ is the collection of all null sets with respect to $m \times m$. Now, we have that the function $S_x f (y) = f(x,y)$ is Lebesgue measurable, and the function $T_y f(x) = f(x,y)$ is continuous, so the pre-image of any open set is open. I've been playing around with some ideas but nothing I've tried has worked.

Standard trick: Define $$f_n(x,y)=f\left(\frac jn,y\right)\quad\left(x\in\left[\frac jn,\frac{j+1}n\right)\right).$$Show that $f_n$ is measurable and note that $f_n\to f$ pointwise.
