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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

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  • $\begingroup$ 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
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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.

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