TITLE: right continuous continuous function is measurable
QUESTION [5 upvotes]: Let $f: S \times [0, \infty)\rightarrow \mathbb{R}$ satisfy $f(x, t)$ is continuous in $x$ for each $t$ and right continuous in $t$ for each $x \in S$.  Here $S$ is a metric space.  Why is $f$ Borel measurable? (In the joint sense)  I am vaguely familiar with the idea of proof I did a long time ago for if $S$ is instead a Euclidean space, and we have full continuity instead of only right.  Then one interpolated.  But you can't do that here anyway, since $S$ is not Euclidean, and nevermind the fact that you only have right continuity anyway.  I also tried to write $f$ as a limit of joint measurable functions using the right continuity, but failed.

REPLY [2 votes]: Edit: The stuff in quotes is not true. For example, there are bijections of reals whose graphs are non measurable. Thanks to Jeff for pointing this out.
"Suppose $S$ is separable locally compact metric space and $Y = [0, \infty)$ under right open topology. Let us show that if every vertical and horizontal section of $B \subseteq S \times Y$ is open then $B$ is Borel."
Fix $f: [0, \infty) \times S \rightarrow \mathbb{R}$. Define $f_n(x, t) = f(m/n, t)$ where $m, n \geq 1$ are such that $(m-1)/n \leq x < m/n$. Then $f_n$'s are all continuous. Now observe that for fixed $(x, t) \in [0, \infty) \times S$, as $n \rightarrow \infty$, $f_n(x, t) \rightarrow f(x, t)$.