# Suppose that the infinite series $\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\}$ converges for each $\epsilon > 0.$

Let $\{f_n\}$ be a sequence of measurable functions on a measure space $(X, \mathcal{M}, \mu)$. Suppose that the infinite series $\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\}$ converges for each $\epsilon > 0.$ Prove that $f_n(x) \rightarrow 0$ a.e.

I am not really sure how to approach this problem. Some help would be awesome. Thanks. It is a past qual problem.

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Have you tried something? –  Tomás Aug 25 '14 at 16:09
I let $E_n^\epsilon = \{x \in X : |f_n(x)| \geq \epsilon\}$ I noticed that the intersection of all $E_n^\epsilon$ for fixed epsilon must be zero if the statement is true for all epsilon. Is proving this helpful? –  kingkongdonutguy Aug 25 '14 at 16:16

We have $$\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\} = \mu(\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}) < \infty$$ which means $\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}$ is finite $\mu$-a.e.
That is to say for any $\epsilon$, there are only finitely many $n$ such that $|f_n(x)| \geq \epsilon$
@GA316 Since $1_{ |f_n(x)| \geq \epsilon}$ is non-negative, we can use monotone convergence theorem. $\mu(f)$ means the integration of $f$ w.r.t $\mu$ –  Petite Etincelle Aug 25 '14 at 16:36
@GA316 $\mu(A)$ for a set $A$ means the measure of $A$. $\mu(f)$ for a function $f$ means the integration of $f$. With these two definitions we have $\mu(A) = \mu(1_A)$ –  Petite Etincelle Aug 25 '14 at 16:42
I am sorry. This is the first time I come across this notation $\mu (f)$. Thanks. –  GA316 Aug 26 '14 at 4:17