Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am working on the following problem:

Let $(f_n)$ be a sequence of measurable real-valued functions on $\mathbb{R}$. Prove that there exist constants $c_n > 0$ such that the series $\sum c_n f_n$ converges for almost every $x$ in $\mathbb{R}$. (Hint: Use Borel-Cantelli Lemma).

I am thinking to use an approach similar to the one in Proposition 2 here http://www.austinmohr.com/Work_files/prob2/hw1.pdf , where we pick $c_n$ such that $\mu({x: c_n f_n(x) > 1}) < 1/2^n$.

But I don't understand if and why there exists such a $c_n$, i.e. consider an unbounded function.

Hence, I think a different approach is needed.

Thank you.

share|improve this question
1  
Why should unbounded functions be a problem? Here, choose $c_n$ such that $\mu(c_n|f_n|\gt1/n^2)\lt1/2^n$. –  Did Oct 9 '12 at 5:32
    
@did: If $f_n(x)=x$, then wouldn't this set have measure infinity for all positive $c_n$? –  user44081 Oct 9 '12 at 6:03
    
Are you sure $\mu$ can be unbounded (for example, Lebesgue measure)? –  Did Oct 9 '12 at 6:32
    
If $f_n(x) = x$, then choosing $c_n = 2^{-n}$ makes $\sum c_nf_n(x)$ converge (namely to $2x$), and this holds for all $x$ (not just almost every $x$). –  Amit Kumar Gupta Oct 9 '12 at 7:18
    
@did: I am dealing exclusively with Lebesgue measure in this problem. I guess I should of said that before. –  user44081 Oct 9 '12 at 12:45

1 Answer 1

up vote 2 down vote accepted

Assume that $f_n$ is Lebesgue almost everywhere finite, for every $n$ (otherwise the result fails).

For every $n$, the interval $[-n,n]$ has finite Lebesgue measure hence there exists $c_n\gt0$ such that the Lebesgue measure of the Borel set $$ A_n=\{x\in[-n,n]\,\mid\,c_n\cdot|f_n(x)|\gt1/n^2\} $$ is at most $1/n^2$. Then, Borel-Cantelli lemma shows that $\limsup A_n$ has Lebesgue measure zero, hence, for Lebesgue almost every $x$ in $\mathbb R$, $x$ is not in $A_n$ for every $n$ large enough. Since $|x|\leqslant n$ for every $n$ large enough, this means that $c_n\cdot|f_n(x)|\leqslant1/n^2$ for every $n$ large enough.

In particular, the series $\sum\limits_nc_nf_n(x)$ converges (absolutely). QED.

Edit: Assume that there exists $k$ such that $f_k$ is not Lebesgue almost everywhere finite. This means that $A=\{x\in\mathbb R\,\mid\,|f_k(x)|=+\infty\}$ has positive Lebesgue measure. Now, for every positive valued sequence $(c_n)$, the series $\sum\limits_nc_nf_n$ diverges on $A$. Hence there exists no positive valued sequence $(c_n)$ such that the series $\sum\limits_nc_nf_n$ converges Lebesgue almost everywhere.

share|improve this answer
    
Thanks. Do you know of a counterexample if $f_n$ is not Lebesgue almost surely finite for every $n$ ? –  user44081 Oct 9 '12 at 19:55
    
Yes. See Edit. $ $ –  Did Oct 11 '12 at 20:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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