# Prove that $f_n$ converges to $f$ a.e.

Suppose that $f_n$ and $f$ are measurable functions such that for each $\varepsilon>0$ we have $$\sum_{n=1}^\infty\mu(\left\lbrace x:|f_n(x)-f(x)|>\varepsilon\right\rbrace)<\infty.$$ Prove that $f_n\rightarrow f$ a.e.

I can only get it for a subsequence so far. For if $$\sum_{n=1}^\infty\mu(\left\lbrace x:|f_n(x)-f(x)|>\varepsilon\right\rbrace)<\infty,$$ then the divergence test implies that $$\mu(\left\lbrace x:|f_n(x)-f(x)|>\varepsilon\right\rbrace)\rightarrow 0.$$ Thus $f_n$ converges to $f$ in measure. Thus there exists a subsequence $f_{n_j}$ such that $f_{n_j}\rightarrow f$ a.e.

How can I improve this to get the desired result of $f_n\rightarrow f$ a.e.?

The proof is standard: Let $E=\{x|f_n(x) \;\text{do not converge to }f(x)\}$. Then we can write $E$ as $$E=\cup_{n\geq 1}\cap_{K\geq 1}\cup_{k\geq K}\{x||f_k(x)-f(x)|>1/n\}$$ So $$\mu(E)\leq \Sigma_{n\geq 1}\;\mu(\cap_{K\geq 1}\cup_{k\geq K}\{x||f_k(x)-f(x)|>1/n\})$$ By the assumption, for any $n$ and $\epsilon>0$, we can find $K_0$ sufficiently large, such that $$\Sigma_{k\geq K_0}\;\mu(\{x||f_k(x)-f(x)|>1/n\})<\epsilon$$ thus $\mu(\cap_{K\geq 1}\cup_{k\geq K}\{x||f_k(x)-f(x)|>1/n\})\leq \mu(\cup_{k\geq K_0}\{x||f_k(x)-f(x)|>1/n\})<\epsilon$. This implies that $$\mu(\cap_{K\geq 1}\cup_{k\geq K}\{x||f_k(x)-f(x)|>1/n\})=0$$ So $\mu(E)=0$.