# A question concerning 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 $$O(f, e) \mathrel{:=} \inf_{\substack{m(H) \le e,\\ f\in C^0[0, 1] \setminus H}}\left\{\ \frac{1}{m(H)} \int\limits_{x \in H} \lim_{d \to 0}\ \inf_{m(G) = 0} \sup_{\substack{y, z \in B_d (x)\setminus G}} \lvert f(y) - f(z)\rvert\mathrm{d}x\right\}.$$

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.

Question:

Call a function $$f$$ tame if $$O(f, e) = 0$$ for all $$e$$. Is it true that a function is tame iff it agrees a.e. with a function that is continuous a.e.?

• Possibly Jack Brown's 1995 survey paper Restriction theorems in real analysis (preprint version here) could be of use, at least in pointing you to possibly relevant literature. Mar 20 '19 at 8:57
• I am not sure that I understood the definition, what is $O(f,e)$ for $f=\chi_{(0,1/2]}$? Mar 20 '19 at 12:01
• This would be zero for all e, since for any such H with m(H) = e, we have that the oscillation on H is 0 a.e. (everywhere except 0 and 1/2). So the integral is 0. Mar 20 '19 at 12:47
• I hate to ask a question so similar to one that was just answered, but I feel that I'm still missing something here (probably just one of those days :) ). If we took $f=\chi_{\mathbb{Q}\cap [0,1]}$, what would we get for $O(f,e)$? Mar 22 '19 at 1:53
• This function is discontinuous everywhere with oscillation 1, so the integral would be 1 as well. Mar 22 '19 at 6:51

A counterexample to this problem can be constructed as follows. Take a sequence $$(K_n)_{n\in\omega}$$ of pairwise disjoint nowhere dense compact sets $$K_n\subset[0,1]$$ of positive Lebesgue measure $$\lambda(K_n)>0$$ such that $$\sum_{n=0}^\infty\lambda(K_n)=1$$. Consider the function $$f:[0,1]\to [0,1]$$ defined by $$f(x)=\begin{cases}\frac1{2^n}&\mbox{if x\in K_n for some n\in\omega;}\\ 0&\mbox{otherwise}. \end{cases}$$
It is easy to see that the function $$f$$ is not continuous a.e.
On the other hand, for every $$\varepsilon >0$$, we can choose $$n\in\mathbb N$$ so large that $$\frac1{2^n}<\varepsilon$$ and $$\sum_{i>n}\lambda(K_i)<\varepsilon$$. Then the set $$H=[0,1]\setminus \bigcup_{i\le n}K_i$$ has measure $$\lambda(H)<\varepsilon$$ and $$f$$ has oscillation $$\le \frac1{2^n}<\varepsilon$$ at points of the open set $$H$$ (because $$f(H)\subset [0,\frac1{2^n}]$$).
• @JamesBaxter The set $H$ is open, so the oscillation of $f$ and $f{\restriction}H$ are the same at points of $H$. Mar 24 '19 at 8:53