3
$\begingroup$

Let $(f_i)_{i\geq 1}$ be a sequence of real-valued measurable functions on $\mathbb{R}$. I want to show that there exists a sequence of real numbers $(c_j)_{j\geq 1}$, $c_j>0$ for all $j$, such that $\sum_{n,m} \lambda(E_{n,m}^{(c_j)_{j\geq 1}})$ converges, where $E_{n,m} = \{x : |\sum_{i=1}^m c_if_i (x)| > 1/n\}$ and $\lambda$ is the Lebesgue measure.

Is it possible? In general, what are obvious bounds (or bounds that are often used) on the Lebesgue measure of sets of real numbers?

Thank you!

  • 0
    If $c_i=0$ it works, so I guess you want an additional hypothesis.2012-10-07
  • 0
    Yes, thanks! I want all $c_j>0$.2012-10-07
  • 0
    We have $E_{n,m}\subset E_{n+1,m}$ so we should have $\lambda(E_{n,m})=0$ for each $m$ and each $n$. But maybe it's more interesting taking $E_{n,m}:=\{x,|\sum_{j=1}^mc_jf_j(x)|\in \left[\frac 1{n+1},\frac 1n\right)\}$.2012-10-07
  • 0
    Davide, then the sum over $n$ just becomes the measure of the set where the absolute value of that linear combination is positive and less than 1.2012-10-07
  • 0
    Davide, why would this imply that $\lambda(E_{n,m})=0$ for each $m$ and each $n$?2012-10-08
  • 0
    Otherwise, the sum $\sum\limits_{k\geqslant n}\lambda(E_{k,m})$ diverges, hence $\sum\limits_{k}\lambda(E_{k,m})$ diverges, hence $\sum\limits_{k,\ell}\lambda(E_{k,\ell})$ diverges.2012-10-19

0 Answers 0