$(E, \mathcal{S}, \mu)$ is a $\sigma \text{-finite}$ measure space, and $f : E \rightarrow [0, \infty]$ is a $\mathcal{S} \text{-measurable}$ function. I have an argument to show that $G = \{(x, y) \in E\otimes [0, \infty]\: | \: f(x) \ge y\} \in \mathcal{S} \otimes \mathcal{B}([0, \infty])$ based on a limiting procedure. I have since realized there is a much simpler method, but I am interested in verifying the correctness of this approach:
First, let $y_n \rightarrow y$ be a decreasing sequence, then
\begin{align} &\text{liminf}\: f^{-1}([y_n, \infty]) \times [0, y_n] \\ &= \cup_{m = 1}^{\infty}\cap_{n = m}^{\infty}f^{-1}([y_n, \infty]) \times [0, y_n] \\ &= \cup_{m = 1}^{\infty}(\cap_{n = m}^{\infty}f^{-1}([y_n, \infty]) \times \cap_{n = m}[0, y_n] \\ &= \cup_{m = 1}^{\infty}(f^{-1}([y_m, \infty]) \times [0, y])\\ &= f^{-1}([y, \infty]) \times [0, y] \end{align}
The first equivalence is the definition of the lim inf, the second is the simple fact that intersection distributes over cartesian product, the third and fourth follow because $y_n$ is decreasing and hence $[0, y_n]$ is decreasing and $f^{-1}([y_n, \infty])$ is increasing.
Note that $G = \cup_{y \ge 0} f^{-1}([y, \infty]) \times [0, y]$ which is an uncountable union. I wish to "thin this out" and then apply the countable union axiom for sigma algebras.
Now, let $q_n$ be an enumeration of $\mathbb{Q}$. Since for any $y \in \mathbb{R}$ there is some decreasing sequence in $\mathbb{Q}$ that approaches $y$ we get $G = \cup_{n = 1}^{\infty}f^{-1}([q_n, \infty]) \times [0, q_n]$ (because $\text{liminf} A_n \subseteq \cup_{n = 1}^{\infty} A_n$, and any $f^{-1}([y, \infty]) \times [0, y]$ can be obtained from the liminf of a set sequence $f^{-1}([q'_n, \infty])\times [0, q'_n]$)
Now that $G$ can be written as a countable union of measurable sets, $G$ itself must be measurable.
Are there any holes in this argument?