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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 '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
up vote 3 down vote accepted

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 '14 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 '14 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 '14 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 '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

Denote $E =\{x : f_n(x) \to 0\}$. We prove that $$E = \bigcap_{k=1}^\infty \bigcup_{m =1}^\infty \bigcap_{l=m}^\infty \left\{x: |f_l(x)| \leq \frac 1k\right\}.$$ Denote the set in right hand side by $F$. If $x\in E$ then $f_n(x) \to 0$, hence for any $k\geq 1$, there exists $m$ dependings on $k$ such that $|f_l(x)| \leq 1/k$ for any $l\geq m$. This shows that $x\in F$. Conversely, if $x\in F$, for any $\epsilon >0$, choosing $k_0$ such that $1/k_0 < \epsilon$ since $x\in F$ then $$x\in \bigcup_{m =1}^\infty \bigcap_{l=m}^\infty \left\{x: |f_l(x)| \leq \frac 1{k_0}\right\}.$$ This means that there exists $m$ such that $|f_l(x) |\leq 1/k_0 < \epsilon$ for any $l\geq m$. This means that $f_n(x) \to 0$, or $x\in E$. Hence $E =F$.

We have $$E^c = \bigcup_{k=1}^\infty \bigcap_{m=1}^\infty \bigcup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}. $$ Hence $$\mu(E^c) \leq \sum_{k=1}^\infty \mu\left(\bigcap_{m=1}^\infty \bigcup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}\right)\leq \sum_{k=1}^\infty \lim_{m\to\infty} \mu\left(\cup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}\right).$$ From our assumptions, we have $$\lim_{m\to\infty} \mu\left(\cup_{l=m}^\infty \left\{x : |f_l(x)| >\frac1k\right\}\right) \leq \lim_{m\to \infty} \sum_{l=m}^\infty \mu\left(\left\{x : |f_l(x)| >\frac1k\right\}\right) = 0.$$ Therefor $\mu(E^c) = 0$.

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You might want to use \bigcup and \bigcap rather than \cup and \cap. Also you can use \left and \right to ensure the parantheses and braces have the correct height. – Michael Albanese Jan 9 at 14:14

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