This problem is from Ziemer's Modern Real Analysis and there is a suggestion on the page which tells us to approximate $f$ in the variable $x$ by piecewise-linear continuous functions $f_n$ such that $f_n \rightarrow f$ pointwise. Then, through a corollary, since each $f_n$ is Lebesgue measurable we should conclude that $f$ is too.
I believe that approximation through "line segments" should give the result we want. To do this, we fix $y = y_o$ and given $a\in \mathbb{Z}$, define the piecewise functions: $$F_1(x)= \begin{cases} (1-x+a)f(a,y_o)+(x-a)f(a+1,y_o) & x\in [a,a+1] \\ \end{cases} \\ \vdots \\ F_n=\begin{cases} n(\frac{1}{n}-x+a)f(a)+n(x-a)f(a+\frac{1}{n}) & x\in [a,a+\frac{1}{n}] \\ n(\frac{2}{n}-x+a)f(a+\frac{1}{n})+n(x-a-\frac{1}{n})f(a+\frac{2}{n}) & x\in [a+\frac{1}{n},a+\frac{2}{n}] \\ \vdots \\ n(\frac{n-1}{n}-x+a)f(a+\frac{n-1}{n})+n(x-a-\frac{n-1}{n})f(a+1) & x\in [a+\frac{n-1}{n},a+1] \\ \end{cases} $$ So, briefly explained, we're taking the intervals between the whole numbers and at each step dividing said interval in $n$ subintervals. We then evaluate at the extremes and join them up by a straight line.
I'm pretty sure $F_n \rightarrow f$ pointwise but I'm having trouble in proving this. Any help is appreciated and also if my sequence of functions is not optimal let me know.