Given a locally integrable function $f: \mathbb R \to \mathbb R$, define $ K: \mathbb R \times \mathbb R+ \to \mathbb R$ by

$$ K(x, r) := \begin{cases} 1, & \text{if }f(x) > \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ -1, & \text{if }f(x) < \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x\\ 0,& \text{if }f(x) = \dfrac{1}{2r}\displaystyle\int\limits_{B_{r}(x)}f\,\mathrm{d}x \end{cases} $$

Define $U: \mathbb R \to [-1, 1]$ and $L: \mathbb R \to [-1, 1]$ by

$$ \begin{split} U(x) &:= \limsup_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t\\ L(x) &:= \liminf_{r \to 0} \frac{1}{2r}\int\limits_{B_r(0)} K(x, t)\,\mathrm{d}t \end{split} $$

Intuitively, $U$ and $L$ represent the weighted proportion of time a function spends above or below its average value in an infinitesimal neighbourhood of a point.

i) Is it true that for any locally integrable function, $U = L$ a.e?

ii) Is it true that $U = 1, 0, or -1$ a.e.?