3
$\begingroup$

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. $x\in X$

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.