# 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. – Dave L Renfro Mar 20 '19 at 8:57
• I am not sure that I understood the definition, what is $O(f,e)$ for $f=\chi_{(0,1/2]}$? – Fedor Petrov 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. – James Baxter 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 $O(f,e)$? – Gary Moon Mar 22 '19 at 1:53
• This function is discontinuous everywhere with oscillation 1, so the integral would be 1 as well. – James Baxter 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 $H$. – Taras Banakh Mar 24 '19 at 8:53