Equip $[0, \infty)$ with the Borel $\sigma$-algebra. Let $(X, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space. Let $f: X \to [0, \infty)$ be a measurable function. Define $g: X \times [0, \infty) \to [0, \infty)$ by
$g(x, t) = 1$ if $t \ge f(x)$
$g(x, t) = 0$ if $t < f(x)$
I want to show that $g$ is measurable.
It is clear that $g = \chi_{E}$ where $E = \{(x, t): t \ge f(x)\}$. So it suffices to prove that $E$ is measurable under the product $\sigma$-algebra.
Fix $t \ge 0$. Define $E_t = \{(x, t): x \in f^{-1}(-\infty, t)\}$. Then $E = \cup_{t \in [0, \infty)}E_t$.
Each $E_t$ is measurable as singletons are measurable in the Borel $\sigma$-algebra, and the preimage of $f$ is measurable. The cartesian product of measurable sets is measurable. However, where this fails is that I am taking an uncountable union, which does not necessarily preserve measurability. I am thinking it is perhaps $\sigma$-finiteness of $X$ that may help here, but I don't know how.
Any hints are appreciated.