# Quantitative Lusin’s theorem

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 $$e > 0$$?

• 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. – LSpice Jan 22 at 2:43
• Thanks so much! It seems accurate. I’ll learn a lot on how to Tex from this as well. – James Baxter Jan 22 at 2:44