I'm not at all familiar with how such computations work.
Let $m > 0$, and $\mathbf{1}_{A}: \mathbb{R} \to \{0, 1\}$ be defined by $$\mathbf{1}_A(x) = \begin{cases} 1, & x \in A \\ 0, & x \notin A\text{.} \end{cases}$$
Let $(A_n)_{n=1}^{\infty}$ be the sequence of sets $$A_n = \begin{cases} \left[\dfrac{n-1}{m}, \dfrac{n}{m} \right], & n \leq m \\ \varnothing, & n > m\text{.} \end{cases}$$
Consider the sequence of functions $(f_n)_{n=1}^{\infty}$ where $f_n: \mathbb{R} \to \{0, 1\}$ is defined by $$f_n = \mathbf{1}_{\mathbf{A}_n}\text{.}$$
How do I calculate the value of $$\limsup_{n \to \infty} f_n(x)$$ almost everywhere (based on the Lebesgue measure) for $x \in [0, 1]$?
I have a ton of books on analysis, but none of them define the $\limsup$ of a sequence of functions (at least, as far as I can tell).
Digging around MSE, I seem to find two definitions of the $\limsup$ of sequence of functions. The first one is $$\limsup_{n \to \infty}f_n(x) = \inf_{n \geq 1}\sup_{k \geq n}f_k(x) = \inf_{n \geq 1}\sup_{k \geq n}\mathbf{1}_{A_k}\text{.}$$ So if I consider $\sup_{k \geq n}\mathbf{1}_{A_k}$, I have $$\sup_{k \geq n}\mathbf{1}_{A_k} = \sup\left\{\mathbf{1}_{\left[\frac{n-1}{m}, \frac{n}{m} \right]}, \mathbf{1}_{\left[\frac{n}{m}, \frac{n+1}{m} \right]}, \cdots \right\}$$ and I'm not sure how to handle this.
Another definition I've seen on MSE is $$\limsup_{n \to \infty}f_n(x) = \lim_{n \to \infty}\sup_{k \geq n}\mathbf{1}_{A_k}$$ and I run into the same problem here.
I haven't been provided a definition for $\limsup$ and was provided this problem. I suspect I don't know the definition because I didn't learn analysis via Baby Rudin (and I can't find it in there).
Please do not use the "infinitely often" equivalence in $\limsup$; I am not allowed to use that for this purpose.