Given a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X$ defined on it, s.t. $X\geq 0$, I want to show that the mapping $$(\omega, t)\mapsto \mathbf{1}_{\{X(\omega)\geq t\}}$$ is measurable on the measure space $(\Omega\times[0,\infty),\mathcal{F}\times\mathcal{B}([0,\infty)),P\otimes\lambda)$, for $\lambda$ denoting the Lebesgue measure.
My thoughts were to simply use the definition and look at the preimages of this mapping. One would then have, e.g. $$\mathbf{1}_{\{X(\omega)\geq t\}}^{-1}(\{0\})=\{(\omega, t): X(\omega)<t\}=\{(\omega, t): \omega\in X^{-1}([0,t))\}.$$
Since $X$ is a random variable, it is measurable, so $X^{-1}([0,t))\in\mathcal{F}$. However, I cannot see at the moment why $\{(\omega, t): \omega\in X^{-1}([0,t))\}$ belongs to the product $\sigma$-algebra, since one coordinate depends on the other. I'm grateful for any hints.
If $a < b$ then there exists $q \in \mathbb{Q}$ such that $a < r < b$. Thus
$$\{(\omega, t): X(\omega)<t\} = \bigcup_{r \in \mathbb{Q}} \{(\omega, t): X(\omega)< r < t\}$$ where $$\{(\omega, t): X(\omega)< r < t\} = \{(\omega, t): X(\omega)< r\} \cap \{(\omega, t): r < t\}$$ Futher $$ \{(\omega, t): X(\omega)< r\} = \{X^{-1}((\infty, r))\}\times [0, \infty) \in \mathcal{F} \times \mathcal{B}([0,\infty))$$ $$\{(\omega, t): r < t\} = \Omega \times (r, +\infty) \in \mathcal{F} \times \mathcal{B}([0,\infty))$$ q.e.d.