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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 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 at 16:16

1 Answer 1

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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 $|f_n(x)| \geq \epsilon$

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Can you please explain, in the first line how did you interchange the measure and the summation ? what is the meaning of measure of the summation? –  GA316 Aug 25 at 16:34
    
@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 at 16:36
    
Sorry. I am confused. What is the meaning of measure of a function. In your last command, you have mentioned the characteristic function and you have find the measure of those characteristic function in your answer. please tell what is the meaning of it. –  GA316 Aug 25 at 16:39
    
@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 at 16:42
    
I am sorry. This is the first time I come across this notation $\mu (f)$. Thanks. –  GA316 Aug 26 at 4:17

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