This is a follow up to this question, which was answered in the affirmative: Are continuous functions almost completely determined by their modulus of continuity?

Note: We do not identify functions that agree a.e.

Given $f: R \to R$ in $L^{1}_{loc}$, define the approximate left modulus of continuity, $L_f (x, e): \mathbb R \times \mathbb R^+ \to [0, \infty]$ by

$$L_{f} (x, e)=\sup \{ d \geq 0\: \big|\, A(f, r) \leq e, \forall r \text{ s.t. } 0 \leq r \leq d \}.$$

where $A(f, r) := \frac{1}{r} \int\limits_{[x-r, x]} |f(t) - f(x)| {d}t.$

Similarly define the approximate right modulus of continuity by

$$R_{f} (x, e)=\sup \{ d \geq 0\: \big| I(f, r) \leq e, \forall r \text{ s.t } 0 \leq r \leq d \}.$$

where $I(f, r) := \frac{1}{r} \int\limits_{[x, x + r]} |f(t) - f(x)| {d}t.$

Note that if $f = g$ a.e., then $L_f = L_g$ a.e. and $R_f = R_g$ a.e., by which we mean for almost all $x$, for all $e$, $L_f (x, e) = R_f (x, e).$

Do $L_f$ and $R_f$ almost determine f uniquely? In the following sense:

Suppose $f$ and $g$ in $L^{1}_{loc}$ are such that $L_f = L_g$ a.e. and $R_f = R_g$ a.e. Does it follow that $f = g + c$ a.e. or $f = -g + c$ a.e. for some a.e. constant function $c$?


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