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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.?

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  • $\begingroup$ 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. $\endgroup$ – Dave L Renfro Mar 20 '19 at 8:57
  • $\begingroup$ I am not sure that I understood the definition, what is $O(f,e)$ for $f=\chi_{(0,1/2]}$? $\endgroup$ – Fedor Petrov Mar 20 '19 at 12:01
  • $\begingroup$ 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. $\endgroup$ – James Baxter Mar 20 '19 at 12:47
  • $\begingroup$ 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 $O(f,e)$? $\endgroup$ – Gary Moon Mar 22 '19 at 1:53
  • $\begingroup$ This function is discontinuous everywhere with oscillation 1, so the integral would be 1 as well. $\endgroup$ – James Baxter Mar 22 '19 at 6:51
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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 $f(H)\subset [0,\frac1{2^n}]$).

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  • $\begingroup$ Nice! A well deserved bounty for you sir. $\endgroup$ – James Baxter Mar 24 '19 at 8:51
  • $\begingroup$ Well actually, the oscillation is counted over the whole set, not just on H. But i believe this bug might be fixable? $\endgroup$ – James Baxter Mar 24 '19 at 8:52
  • $\begingroup$ @JamesBaxter Thanks! So, quick reaction! $\endgroup$ – Taras Banakh Mar 24 '19 at 8:52
  • $\begingroup$ @JamesBaxter The set $H$ is open, so the oscillation of $f$ and $f{\restriction}H$ are the same at points of $H$. $\endgroup$ – Taras Banakh Mar 24 '19 at 8:53
  • $\begingroup$ Oh, right I missed that. Very nice.. $\endgroup$ – James Baxter Mar 24 '19 at 8:53

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