Let $f\,$ be a bounded function and $D=\bigcup D_\alpha$ be a measure zero set where

$$D_\alpha :=\{x_0\in[a,b]: \omega_f(x_0)\geq \alpha, \},\quad \alpha >0. $$ Setting $\alpha := \frac{\epsilon}{2(b-a)}$, it needs to be shown that there exists a finite collection of open intervals $\{G_1,G_2,...,G_N\}$ such that the collection covers $D_\alpha$ and has $$\sum_{n=1}^N |G_i|\leq \frac{\epsilon}{4M}$$ where $|G_i|$ is the length of the interval $G_i$ and $M = \sup\{f(x): x\in [a,b]\}$.

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    $\begingroup$ What's $f$? What's $M$? Are the $x_m$ members of a sequence or something? $\endgroup$ – Dimitrije Kostic Dec 9 '11 at 2:27
  • $\begingroup$ Sorry, I'll edit this so all of those things are clear. $\endgroup$ – Henry Shearman Dec 9 '11 at 4:53
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    $\begingroup$ The definition of $D_{\alpha}$ still doesn't make sense; in the condition $|f(x_0)-f(x)| \geq \alpha$, what point is $x$? Is it a fixed point or something? And what's the union you're taking to get $D$, and how do you know it's measure $0$? $\endgroup$ – Dimitrije Kostic Dec 9 '11 at 5:14
  • $\begingroup$ $D_\alpha$ is the set of all alpha discontinuities of the function $f$ on $[a,b]$ i.e. points where the oscillation of $f$ exceeds $\alpha$. Taking the union of these sets gives all the dicontinuities of $f$ which is known to be a meausre zero set. $\endgroup$ – Henry Shearman Dec 9 '11 at 6:28
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    $\begingroup$ I expect that $D_\alpha$ is supposed to be $\{x_0\in[a,b]:\omega_f(x_0)\ge\alpha\}$, where $$\omega_f(x)\triangleq\limsup\limits_{x\to x_0}f(x)-\liminf\limits_{x\to x_0}f(x)$$ is the oscillation of $f$ at $x_0$. Your expression for $D_\alpha$ simply makes no sense as written. $\endgroup$ – Brian M. Scott Dec 9 '11 at 6:38

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