# Measure theory question about sum of sequence of functions

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

• Try choosing $c_n$ so that $m(\{c_n |f_n| \ge 2^{-n}\}) \le 2^{-n}$. – Nate Eldredge Nov 2 '12 at 3:15
• @NateEldredge: a small correction: choose $c_n$ so that $m\{|x|\le n:c_n|f_n(x)|\ge 2^{-n}\}\le 2^{-n}$. – 23rd Nov 2 '12 at 6:40
• @richard why don't you post this as the answer? – Norbert Nov 2 '12 at 6:47
• @Norbert: because Nate Eldredge had almost shown the answer as a comment. – 23rd Nov 2 '12 at 6:52
• @richard: Thanks! Could you go ahead and post as an answer? – Nate Eldredge Nov 2 '12 at 12:33

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.