Let $(f_n)_{n\geq1}$ be a sequence of measurable almost everywhere finite real-valued functions on $(X,M,\mu)$, where $µ$ is a $σ$-finite measure. Prove that there exist constants $c_n > 0$ such that the series $\sum c_nf_n(x)$ converges for $\mu$-a.e. $x\in X$

  • $\begingroup$ when you say "a.e. finite", do you mean "a.e. bounded"? $\endgroup$ – Giovanni Oct 7 '15 at 23:23
  • $\begingroup$ @Giovanni: Probably not, the problem with "bounded" instead of "finite" is very easy, whereas this version is actually a nice challenging problem for an introductory measure theory class. $\endgroup$ – Lukas Geyer Oct 7 '15 at 23:57
  • $\begingroup$ How is the "bounded" case easy? Hints please? $\endgroup$ – user275834 Oct 8 '15 at 0:36
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    $\begingroup$ For the bounded case you can just take $c_n = 2^{-n}\|f_n\|_{\infty}$. For the finite case the only approach I have in mind involves the Borel Cantelli lemma. $\endgroup$ – Giovanni Oct 8 '15 at 0:41
  • $\begingroup$ I was thinking along the lines of Borel Cantelli too. If I first assume $\mu$ to be finite I can show the existence of such $c_n$ but I am stuck where I have to go from $\mu$ finite to $\sigma$-finte $\endgroup$ – user275834 Oct 8 '15 at 14:45

Hint: Because $\mu$ is $\sigma$-finite, there is a strictly positive measurable function $g$ such that $\int g\,d\mu\le 1$. Now choose constants $c_n>0$ so small that $\int \left[1-\exp(-c_n|f_n|)\right]g\,d\mu\le 2^{-n}$ for $n=1,2,\ldots$. The series $\sum_n c_n f_n(x)$ converges absolutely for $\mu$-a.e $x$.

Fix $n$. To see that $c_n$ exists, notice that $\lim_{c\to 0+}\int \left[1-\exp(-c|f_n|)\right]g\,d\mu=0$ by Monotone Convergence (the integrand $[1-\exp(-c|f_n|)g$ decreases to $0$ as $c$ decreases to $0$, and $\int \left[1-\exp(-|f_n|)\right]g\,d\mu\le\int g\,d\mu<\infty$).

Because $\int \left[1-\exp(-c_n|f_n|)\right]g\,d\mu\le 2^{-n}$ for each $n$ we have $$ \int\sum_n \left[1-\exp(-c_n|f_n|)\right]g\,d\mu=\sum_n\int \left[1-\exp(-c_n|f_n|)\right]g\,d\mu<\infty, $$ which means that $$ \sum_n \left[1-\exp(-c_n|f_n|)\right]<\infty $$ $\mu$-a.e. (because $g>0$). Let $G$ be the subset of $X$ where this series converges. Notce that if $x\in G$ then $\lim_nc_n|f_n(x)|=0$; therefore there exists $N(x)$ with $\mu(\{x\in G: N(x)=\infty\})=0$ such that $c_n|f_n(x)|\le 1$ for all $n\ge N(x)$. For $x\in G\cap\{N<\infty\}$ we have $$ \eqalign{ \infty&>\sum_n \left[1-\exp(-c_n|f_n(x)|)\right]\ge\sum_{n=N(x)}^\infty \left[1-\exp(-c_n|f_n(x)|)\right]\cr &\ge \sum_{n=N(x)}^\infty (1-e^{-1})c_n|f_n(x)|,\cr } $$ because $1-e^{-x}\ge (1-e^{-1})x$ for $0\le x\le 1$. Thus $\sum_{n=N(x)}^\infty c_n|f_n(x)|<\infty$ for all $x\in G\cap\{N<\infty\}$, which guarantees that $\sum_nc_n|f_n|<\infty$, $\mu$-a.e.

  • $\begingroup$ I'd be happy to give a +1 if you could elaborate on how these constants give convergence, and why they exist $\endgroup$ – Anthony Peter Jan 26 '16 at 0:43
  • $\begingroup$ Details now added to my answer. $\endgroup$ – John Dawkins Jan 26 '16 at 14:53
  • $\begingroup$ Much Appreciated $\endgroup$ – Anthony Peter Jan 26 '16 at 16:26

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