We consider only the set $M$ of a.e. essentially locally bounded measurable functions $[0, 1] \to \mathbb R$. Here $m(S)$ denotes the Lebesgue measure of $S$.

Let $f$ be measurable. For every $e$ in $(0, 1]$, by Lusin’s theorem, we can write our measurable function as continuous on $[0, 1]-H$, and horrid on a set $H$ of measure $e$. How does “horrid” vary with $e$?

One way to quantify “horrid” is to ask how discontinuous the function is on $H$. Inspired by this, we calculate the average pointwise oscillation of the function of $H$. Formally this is the integral of the essential oscillation of $f$ on $H$ divided by $m(H)$. Since oscillation is upper semi continuous, it is integrable. Further we take the infimum over all such $H$ of measure less than or equal to $e$.

Thus $$ O(f, e) \mathrel{:=} \inf_{\text{$m(H) \le e$, $f$ continuous on $[0, 1] \setminus H$}} \int_{x \in H} \lim_{d \to 0} \inf_{m(G) = 0} \sup_{y, z \in H,\, y, z \in B_d (x)\setminus G} \lvert f(y) - f(z)\rvert/m(H). $$

The end result is that for every $e$, we get a function $O(f): (0, 1] \to [0, \infty) $ describing how horrible the discontinuity behaviour is on the best behaved $H$ we can find.

How does the operator taking $f$ in $M$ to $O(f, e)$ behave? More concretely, what is the image of $M$ under $O$? Also, for what class of functions $P$ does $O(f, e) = 0$ for every $f$ in $P$ and $e > 0$?

  • $\begingroup$ I tried to convert your plain-text definition of O(f, e) to TeX, but I had a hard time understanding it. Please re-edit as necessary. $\endgroup$ – LSpice Jan 22 at 2:43
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    $\begingroup$ Thanks so much! It seems accurate. I’ll learn a lot on how to Tex from this as well. $\endgroup$ – James Baxter Jan 22 at 2:44

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