I have been asked how to prove the following statement:

$f$ is integrable $\Longleftrightarrow$ for every $\epsilon > 0$, and every $\mu > 0$ there exists $\delta >0$ such that for every partition $T$ with $\Delta(T) < \delta$, $$\Sigma_{i,\omega_i \geq \epsilon} |\Delta x_i| < \mu $$.

It seems like something is missing. for example, shouldn't the term inside the $\Sigma$ be $|\Delta x_i| \cdot f(x_i) $ and shouldn't it be given that $f(x) < \mu$ or something like that?

or, maybe I am missimg something?

Thank you, Shir

  • $\begingroup$ Check the statement to be proven. The passage underlaid in gray doesn't make sense. What is $\omega_i$? $\endgroup$ Jan 3 '14 at 16:14
  • $\begingroup$ ya, it didn't make sense to me eather $\endgroup$
    – topsi
    Jan 4 '14 at 13:04

I'd guess, if $Δx_i=x_{i+1}-x_i$, that then $ω_i=\sup_{x\in[x_{i+1},x_i]}f(x)-\inf_{x\in[x_{i+1},x_i]}f(x)$ is the variation of $f$ over this interval of the subdivision.

This plays into the concept of bounded total variation and of the set of discontinuities having measure (Jordan or Lebesgue) zero.


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.