First, \forall x \in \mathbb{R} and \forall n \in \mathbb{N} define x_{r,n} = \frac{r}{2^n} where x \in
\big(\frac{r}{2^n}, \frac{r+1}{2^n} \big].
Define f_n(x,y) = f(x_{r,n},y). Are these f_n's measurable? Observe that:
\lbrace (x,y) \in \mathbb{R}^2 \big| f_n(x,y) > \alpha \rbrace
= \bigcup_{r = -\infty}^{\infty} \bigg(\frac{r}{2^n}, \frac{r+1}{2^n} \bigg] \times \lbrace y \in \mathbb{R} \big| f(x_{r,n},y) > \alpha \rbrace
Certainly, the intervals \big(\frac{r}{2^n}, \frac{r+1}{2^n} \big] are measurable, and further, the sets \lbrace y \in \mathbb{R} \big| f(x_{r,n},y) > \alpha \rbrace are measurable, given that f(x,y) is continuous in the y-direction for any fixed x. Thus, their product is measurable. Lastly, since the measurable sets form a \sigma-algebra, we have that f_n must be a measurable function \forall n \hspace{0.25cm} \in \mathbb{N}.
Next, if we can confirm that f_n \to f as n \to \infty on \mathbb{R}^2, we're done (given property 4 of measurable functions on p. 29.)
Fix an arbitrary (x,y) \in \mathbb{R}^2. Since f is continuous in both directions, we have that since x_{r,n} \to x as n \to \infty, that f_n(x,y) = f(x_{r,n},y) \to f(x,y) as n \to \infty. Thus, for any (x,y) in \mathbb{R}^2, f_n converges to f point-wise.
Thus, f is measurable on \mathbb{R}^2.
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