It's a homework from a past semester:
Let $f:X\rightarrow \Bbb{R}$ be a measurable function defined on the measure space $\bigl(X,\Sigma, \mu \bigl)$.
Show that the next set is measurable, meaning is in the $\sigma$-algebra that is generated by the cartesian product: $\Sigma \times \mathcal{B([0,\infty))}$
In short, show that: $\{(x,t)\in X\times[0,\infty) \text{ | }f(x)>t\} \in \sigma\bigl(\Sigma \times \mathcal{B([0,\infty))}$
Note: I know that for every $t\in \Bbb{R}:$ $\{x\in X \text{ | } f(x)>t\}=f^{-1}\bigl((t,\infty)\bigl)\in\Sigma$ becuase $f$ is measurable.
And, for every $x\in X: \{t\in [0, \infty) \text{ | } f(x)>t \}=\bigl[0,f(x)\bigl)\in \mathcal{B\bigl([0,\infty)\bigl)}$.
But, I got a feedback from the checker states that:
$\{(x,t)\in X\times[0,\infty) \text{ | }f(x)>t\} \neq \{x\in X \text{ | } f(x)>t \} \text{ } \times \{t\in [0,\infty) \text{ | } f(x)>t \}$.
So, I guess my lead wasn't in a good direction.
The checker also said to try and approximate $f$ by a sequence of simple functions, but couldn't find a solution.