Let $(X, S, \mu)$ be a finite measure space. Suppose $g: X \rightarrow \mathbb{R}$ is integrable and that $g(x) \geq 0 \ \forall x \in X$. Let $A = \{(x,y) \in X \times [0,\infty) : 0 \leq y < g(x)\}$. Show that A is $\mu \times \lambda $-measurable for $\mu$ given and Lebesgue measure $\lambda$ on the set $[0,\infty)$.
So here is what I think. Since $g$ is integrable, there is a sequence of step functions, $\{\phi_n \} $, such that $ 0 \leq \phi_n \uparrow f$. I was going to generate $A$ as a sequence of step functions like this:
$A = \bigcup_{n = 1}^{\infty} \{(x,y) \in X \times [0,\infty) : 0 \leq y < \phi_n(x)\ \}$. This is what I got so far. What does $\mu \times \lambda $-measurable set mean?
Also, is this the way to go or there is another way to approach this? Thank you very much for your help!
Since $g$ is measurable w.r.t. $\mu$ there exists a set $E \in S$ and a function $h$ measurable w.r.t. $S$ such that $\mu (E^{c})=0$ and $g(x)=h(x)$ for all $x \in E$. Now $\{(x,y): 0\leq y <g(x)\} \Delta \{(x,y): 0\leq y <h(x)\} \subseteq \{(x,y): g(x) \neq h(x)\} \subseteq [0,\infty) \times E^{c}$. Note that $(\mu \times \lambda ) ([0,\infty) \times E^{c})=0$. Hence it is enough to show that $\{(x,y): 0\leq y <h(x)\}$ belongs to the product of $S$ and the Borel sigma algebra of $[0, \infty)$.
$A=\cup_{r\in \mathbb Q} \{(x,y): 0\leq y <r<h(x)\}$ and $\{(x,y): 0\leq y <r<h(x)\}=\{(x,y): 0\leq y <r\} \cap \{(x,y): h(x) >r\}$. Can you finish?