I have $E_n = \{x : |f_n(x)| > \frac{1}{n}\}$ and I know that
$$m(\limsup_{n\to\infty} E_n) = 0$$
and
$$m(E_k)< \frac{1}{2^k}$$
and I want to claim that
$$\lim_{n\to \infty} f_n = 0 $$ almost everywhere, so in other words
$$m(\{x : \lim_{n\to\infty} |f_n(x)| > 0 \}) =0$$
Is it valid to say that
$$m(\limsup_{n\to\infty} E_n) \geq m(\lim_{n\to\infty} E_n) = m(\{x : \lim_{n\to\infty} |f_n(x)| > 0 \}) $$
Can this be made more rigorous?
Taking a limsup of a sequence of sets is fine, but the limit of a sequence of sets is not a common notion and I think you should avoid it unless you really find it useful. Note that the opposite of $f_n(x)\to 0$ is not $\lim_n|f_n(x)|>0,$ but $\limsup_n|f_n(x)|>0.$ So you have made a logical leap where you introduce lim's that aren't limsup's.
The original problem is just unwrapping the definitions. The set $\limsup E_n$ is the set of points where $|f_n(x)|>1/n$ infinitely often. So the complement of $\limsup E_n$ is the set of points $x$ such that $|f_n(x)|\leq 1/n$ eventually. But for each $x$ such that $|f_n(x)|\leq 1/n$ eventually, clearly $f_n(x)\to 0.$ So $f_n(x)\to 0$ on the complement of $\limsup E_n,$ which is almost everywhere.