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.

  • 1
    $\begingroup$ Why should unbounded functions be a problem? Here, choose $c_n$ such that $\mu(c_n|f_n|\gt1/n^2)\lt1/2^n$. $\endgroup$ – Did Oct 9 '12 at 5:32
  • $\begingroup$ @did: If $f_n(x)=x$, then wouldn't this set have measure infinity for all positive $c_n$? $\endgroup$ – user44081 Oct 9 '12 at 6:03
  • $\begingroup$ Are you sure $\mu$ can be unbounded (for example, Lebesgue measure)? $\endgroup$ – Did Oct 9 '12 at 6:32
  • $\begingroup$ 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 $x$). $\endgroup$ – Amit Kumar Gupta Oct 9 '12 at 7:18
  • $\begingroup$ @did: I am dealing exclusively with Lebesgue measure in this problem. I guess I should of said that before. $\endgroup$ – user44081 Oct 9 '12 at 12:45

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.

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

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