# Do we have full control of the oscillation of a function by modifying it on a small set?

Definitions and some motivation:

Let $$\mathcal B$$ be the set of bounded measurable functions from $$[0, 1]$$ to $$\mathbb R$$. Denote by $$\mathcal N$$ the set of measurable subsets of $$[0, 1]$$ with Lebesgue measure $$0$$.

Given a function $$f \in \mathcal B$$, define the function $$\mathcal Of$$ by

$$\mathcal Of(x) := \inf_{N \in \mathcal N} \lim_{\delta \to 0} \sup_{y, z \in B_\delta (x) \setminus N}\ |f(y) - f(z)|$$.

Thanks to Lusin’s theorem, we know that we can modify $$f$$ on an arbitrarily small set and get a continuous function, and so we force the oscillation to be $$0$$ everywhere. But can we force it to be whatever we want, and in a controlled way?

Question:

Does there exist, for any given $$f, g \in \mathcal B$$ and $$\varepsilon > 0$$, a function $$f’ \in \mathcal B$$ such that the following conditions are satisfied?

i) $$f’ = f$$ everywhere except for a set of measure at most $$\varepsilon$$.

ii) $$\mathcal Of’ = \mathcal Og$$.

iii) $$\mathcal O(f’ - f) \leq |\mathcal O(g) - \mathcal O(f)|$$.

Remark: Condition (iii) intuitively says that we do not modify the values of $$f$$ any more than strictly necessary.

Note: All functions are genuine functions and not equivalence classes modulo null sets of such.

• Yes, that’s true! But I think you can modify $f$ on a small (less than measure $\varepsilon$) interval around $0$ to achieve any value of $\mathcal Of’(0)$. Of course this is only at one point though.. Commented Apr 19, 2021 at 7:31
• Yes Martin Hairer, but I don’t see immediately how this answers to the OP. Commented Apr 19, 2021 at 7:32
• I don’t think it’s meant to be an answer... @A.DellaCorte Commented Apr 19, 2021 at 7:33

Let $$f,g: [0,1] \to \mathbf{R}$$ be two given, bounded measurable functions and $$\epsilon > 0$$ be an arbitrary constant. By Lusin's theorem there exist continuous functions $$F,G: [0,1] \to \mathbf{R}$$ so that together $$$$\lvert \{ F \neq f \} \rvert + \lvert \{ G \neq g \} \rvert < \epsilon.$$$$
In other words $$F - f$$ and $$G - g$$ are zero except on a set of measure at most $$\epsilon$$. Note also that $$\mathcal{O}(G) = \mathcal{O}(F) = 0$$ by continuity, so they can be added and subtracted with impunity. Therefore we may set $$$$f' = F + g - G = f + (F - f) + (g - G).$$$$
• $$\mathcal{O}(f') = \mathcal{O}(F + g - G) = \mathcal{O}(g)$$ because $$F,G$$ are continuous,
• $$\{ f' \neq f \} \subset \{ g \neq G \} \cup \{ f \neq F \}$$ so $$\lvert \{ f' \neq f \} \rvert < \epsilon$$,
• $$\mathcal{O}(f - f') = \mathcal{O}(f - F - g + G) = \mathcal{O}(f - g)$$ and taking absolute values this is bounded by $$\lvert \mathcal{O}(f) - \mathcal{O}(g) \rvert$$ by the triangle inequality.