Let $\{f_n\}$ be a sequence of measurable functions on $X$, and suppose that, for every $x \in X$:
$$0 \le f_1(x) \le f_2(x) \le \cdots \le \infty $$ and $f_n(x) \rightarrow f(x) $ as $n \rightarrow \infty$ for every $x \in X$.
Let $s$ be any simple measurable function such that $0 \le s \le f$, let $c$ be a constant, $0 < c < 1$, and define
$$E_n = \{x : f_n(x) \ge cs(x) \} \ \ (n = 1,2,3 ...)$$
Then each $E_n$ is measurable. Why is this?
It would suffice to prove that the set $\{x : f_n(x) \ge cs(x) \}$ is open but this does not seem true to me.
Let $g_n = f_n - cs$. As $f_n$ and $s$ are measurable, so is $g_n$. Now note that
$$E_n = \{x \mid f_n(x) \geq cs(x)\} = \{x \mid f_n(x) - cs(x) \geq 0\} = \{x \mid g_n(x) \geq 0\} = g_n^{-1}([0, \infty)).$$
As $g_n$ is measurable and $[0, \infty)$ is a Borel set, $E_n = g_n^{-1}([0, \infty))$ is measurable.