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 Lusinâ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 âhorridâ vary with $e$?

One way to quantify âhorridâ 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 $$ O(f, e) \mathrel{:=} \inf{\text{$m(H) \le e$, $f$ continuous on $[0, 1] \setminus H$}} \int{x \in H} \lim{d \to 0} \inf{m(G) = 0} \sup{y, z \in H,\, y, z \in Bd (x)\setminus G} \lvert f(y) - f(z)\rvert/m(H). $$

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.

Question:

Call a function $f$ tame if $O(f, e) = 0$ for all $e$. Are the continuous a.e. functions the only tame functions?