I don't think this problem should be too difficult, but I'm not sure on some details. Give a sequence of functions $f_{n} : \mathbb{R} \rightarrow \mathbb{R}$ converging for any $x \in \mathbb{R}$ to $f(x)$, show the following two sets coincide:
$$ S = \Big \{f(x) < y \Big \} = \bigcup_{k \geq 1} limsup_{n \to \infty} \Big \{ f_{n}(x) \leq y - \frac{1}{k} \Big \} = T $$
I understand concentually what is being said here, that writing limsup as $\bigcup_{N \geq 1} \bigcap_{n \geq N}$ we can see this as the definition of convergence, and thus (I think I have to assume $f_{n}$ postive here to get rid of the absolute values?), we have
$$ k \geq 1, \exists N \geq 1 : |f(x) - f_{n}(x)| < \frac{1}{k}, \forall n \geq N $$ $$ \implies \forall x \in S, |f_{n}(x) - f(x) + f(x)| <\frac{1}{k} + |f(x)| < y + \frac{1}{k} $$
This almost shows the inclusion $S \subseteq T$, but I want a different sign on the fraction over k. I don't see how to get the right sign here.
Similarly, in the other direction (assuming everything is positive), $$ \forall x \in T, f(x) < \frac{1}{k} + f_{n}(x) \leq \frac{1}{k} + y - \frac{1}{k} = y $$
Which appears to hold, but again I had to assume everything was non-negative. I wonder then how to prove this if $f_{n}, f$ are not non-negative as well as how to get the right sign in the first inclusion?
It is handsome to know that: $$x\in\limsup A_n\iff\{n\in\mathbb N\mid x\in A_n\}\text{ is infinite}$$ We also say then that $x\in A_n$ infinitely often, or shorter $x\in A_n$ i.o.
If $z\in S$ or equivalently $f(z)<y$ then we can find a positive integer $k$ such that also $f\left(z\right)<y-\frac{1}{k}$ and consequently $f_{n}\left(z\right)\leq y-\frac{1}{k}\text{ i.o.}$
So this implies that $z\in T$.
Conversely the statement $z\in T$ is equivalent with the statement that some positive integer $k$ exists such that $f_{n}\left(z\right)\leq y-\frac{1}{k}\text{ i.o.}$
This implies that $f(z)\leq y-\frac1k<y$ or equivalently $z\in S$.