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Let $\left(f_n\right)_{n\ge1}$ be a sequence of measurable real valued functions. Prove that there exist a sequence of constants $c_n$ $>0$ such that $\sum_{n=1}^{\infty} c_nf_n $ converges for almost every x $\in \mathbb R$.

Any hints is appreciated. Thanks

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  • $\begingroup$ Try choosing $c_n$ so that $m(\{c_n |f_n| \ge 2^{-n}\}) \le 2^{-n}$. $\endgroup$ – Nate Eldredge Nov 2 '12 at 3:15
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    $\begingroup$ @NateEldredge: a small correction: choose $c_n$ so that $m\{|x|\le n:c_n|f_n(x)|\ge 2^{-n}\}\le 2^{-n}$. $\endgroup$ – 23rd Nov 2 '12 at 6:40
  • $\begingroup$ @richard why don't you post this as the answer? $\endgroup$ – Norbert Nov 2 '12 at 6:47
  • $\begingroup$ @Norbert: because Nate Eldredge had almost shown the answer as a comment. $\endgroup$ – 23rd Nov 2 '12 at 6:52
  • $\begingroup$ @richard: Thanks! Could you go ahead and post as an answer? $\endgroup$ – Nate Eldredge Nov 2 '12 at 12:33
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This work when the involved measure space $(X,\mathcal A,\mu)$ is $\sigma$-finite. Let $\left(A_n\right)_{n\geqslant 1}$ be a non-decreasing sequence of measurable subsets of finite measure such that $\bigcup_{n\geqslant 1}A_n=X$. Notice that for all fixed $n$, the following convergence holds: $$ \lim_{R\to +\infty}\mu\left(A_n\cap\left\{x\in X\mid \left\lvert f_n(x)\right\rvert\gt 2^{-n} R\right\}\right)=0 $$ hence using the definition of the limit with $\varepsilon=2^{-n}$, we see that we can choose $R_n\gt 0$ such that $$ \mu\left(A_n\cap \left\{x\in X\mid \left\lvert f_n(x)\right\rvert\gt 2^{-n} R_n\right\}\right)\lt 2^{-n}. $$ Choose $c_n=1/R_n$. Then for all $n$, the following inequality holds $$ \mu\left(A_n\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right)\lt 2^{-n}. $$ Fix a $N\geqslant 1$. Since $A_N\subset A_n$ for $n\geqslant N$, it follows that for such $n$'s, $$ \mu\left(A_N\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right)\lt 2^{-n}. $$ hence exploiting the convergence of the series $\sum_n \mu\left(A_N\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right) $, we derive that there exists $E_N\subset A_N$ of measure $0$ such that for all $x\in A_N\setminus E_N$, there exists $m(x)$ such that for all $n\geqslant m(x)$, $c_n\left\lvert f_n(x)\right\rvert\leqslant 2^{-n}$. This proves that for all $x\in A_N\setminus E_N$, the series $\sum_n c_n\left\lvert f_n(x)\right\rvert$ converges. Finally, let $E:=\bigcup_{N\geqslant 1}E_N$. For all $x\in X\setminus E$, there exists a $N$ such that $x\in A_N\setminus E_N$ and $E$ has measure zero.

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I think this is too general a statement. If f(n)=n^m and c(n)=1/n^(m+2) , the sum will be =1/1^2+1/2^2+1/3^2+...=pi^2/6 . If c(n)=1/n^(m+s) , then the sum will be the Zeta function Z(s) --B.Sahu

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  • $\begingroup$ The statement is true, and the proof is basically given in the comments. Did you read them? $\endgroup$ – Lukas Geyer Nov 9 '12 at 15:38

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