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I want to show that if $\{f_n\}$ is a sequence of measurable functions, then there are positive $\{c_n\}$s such that $\sum c_nf_n$ converge almost everywhere.

I don't really see how to approach this... any tips?

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  • $\begingroup$ Let $f_n=x^n$. I am wondering how can we find $c_n$ such that $\sum c_n f_n$ is convergent a.e.? $\endgroup$ – polfosol Nov 2 '16 at 7:49
  • $\begingroup$ @polfosol: This is not too hard, even with convergence everywhere: Choose $c_n = \frac{1}{n!}$, then $\sum c_n f_n(x) = e^x$ for all $x$. $\endgroup$ – PhoemueX Nov 2 '16 at 8:27
  • $\begingroup$ @PhoemueX thanks. I needed some intuition $\endgroup$ – polfosol Nov 2 '16 at 9:56
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I will assume that the underlying measure space has finite measure. This will also prove the claim for $\sigma$-finite measure spaces, since on each such measure space, there is an equivalent measure (with the same null sets) that is finite. I don't think the claim is true for general (non $\sigma$-finite) measure spaces.

The proof is an application of the Borel-Cantelli lemma (cf. https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma#Statement_of_lemma_for_probability_spaces).

First think about this. If it does not help, read on.

For each $n$, we have$$0 = \mu\left(\bigcap_N \{x : |f_n(x)| \geq N\}\right) = \lim_N \mu(\{x : |f_n(x)| \geq N\}). $$ Hence, we can choose $N_n$ such that $M_n := \{x : |f_n(x)| \geq N_n\}$ satisfies $\mu(M_n) \leq 2^{-n}$. We have $\sum_n \mu(M_n) < \infty$, so the Borel-Cantelli lemma shows that for each $x \in X \setminus N$ (where $X$ is the underlying measure space and $N$ is a set of measure zero), we have $|f_n(x)| \leq N_n$ for all but finitely many (depending on $x$) $n$. Now let $c_n := \frac{1}{N_n 2^n}$ and note that $c_n |f_n(x)| \leq 2^{-n}$ for all but finitely many (depending on $x$) $n$. This easily implies the claim.

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