Hi everyone I'd like to know if the following is correct and also I'd appreciate any suggestion to improve the argument. Thanks in advance
For every positive integer $n$, let $f_n:{\bf{R}}\to [0,\infty)$ be a non-negative measurable function such that $$\int_{\bf{R}}f_n\le \frac{1}{4^n}$$ show that for every $\varepsilon>0$ there exists a set $E$ of lebesgue measure $m(E)\le \varepsilon$ such that $f_n$ converges pointwise to zero for all $x\in R\setminus E$.
Pf: Let $A_n=f_n^{-1}\big(\big(1/2^n\varepsilon,\infty\big)\big)$ we will show that $m(A_n)\le \varepsilon/2^n$ for all $n\in {\bf{N}}$. Suppose for sake of contradiction that $m(A_n)>\varepsilon/2^n$ for some $n$. Let $s=1/2^n\varepsilon\cdot 1_{A_n}$ so $$\int_{\bf{R}}f_n\ge \int_{\bf{R}}s=\frac{1}{2^n\varepsilon}m(A_n)>\frac{1}{4^{n}}$$ Contradiction. So for all $n\in {\bf{N}}$ we thus have $m(A_n)\le \varepsilon /2^n$. Let $E =\bigcup_{n\in \bf{N}}A_n$ using subadditivity of $m$ we have $$m(E)\le \sum_{n\in \bf{N}}m(A_n)\le \varepsilon$$
We claim that $E$ is the set we're looking for. Suppose $x\notin E$ this means that $x\notin A_n$ for all $n\in {\bf{N}}$. Thus $f_n(x)\in[0,1/2^n\varepsilon]$ for all $n$, i.e., $f_n(x)\to 0$ as $n\uparrow \infty$ and $x\in {\bf{R}}\setminus E$, i.e., $f_n$ converges pointwise to zero as long as $x\in {\bf{R}}\setminus E$ which concludes the proof.
Your argument works fine.
Another way to do this is noting that $f=\sum f_n$ defines an $L^1(\mathbb{R})$ function by the monotone convergence theorem, and is thus finite a.e. But then $f_n$ must tend to zero a.e. (at least at every point where the series converges).