1.23

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 $\mathbb{R}^2$.