The standard way to do this is first to show that the sup and inf of any sequence are measurable, then apply this result twice, but I was trying to think of an alternate proof using the definition of lim sup, to solidify my understanding of the function. I have a feeling what I came up with is wrong, so I want to understand where I made my error. Define the lim sup of the sequence as function $g$.
I want to show that $\{x: g(x) \geq c \}$ is measurable. My strategy is to imagine what it would mean if the lim sup were strictly less than $c$, then negate that string of quantifiers to get the opposite claim. If the lim sup were strictly less than $c$, I believe this would mean that for some natural $m \geq 1$, there exists $N \geq 1$ such that for all $n \geq N$, $f_{n} \leq c - 1/m$: a strictly smaller "superior" or "dominating" number for the sequence.
Now negate the chain: for all natural $m \geq 1$, for all $N \geq 1$, for some $n \geq N$, $f_{n} > c - 1/m$. So we have the following:
$\{x: g(x) \geq c \} = \cap_{m \geq 1} \cap_{N \geq 1} \cup_{n \geq N} \{x: f_{n}(x) > c - 1/m \}$.
This looks different from other proofs I've seen, but then again, other proofs aren't always starting out with the same statement of measurability for the lim sup, so perhaps this is one way to do it after all?
