Let $f: \mathbb R^n \to \mathbb R$ be a locally integrable function. We say $x \in \mathbb R^n$ is a strong Lebesgue point of $f$ if
$$\lim_{r \to 0} \frac{\int_{B_r (x)} |f(y) – f(x)| \, dy}{r^{n+1+\varepsilon}} = 0$$
for some $\varepsilon > 0$, potentially depending on $x$.
Question: Suppose every point in $\mathbb R^n$ is a strong Lebesgue point of $f$. Does it follow that $f$ is a constant function?
Best Answer
I realise I'm bumping into you again and already gave you an answer elsewhere after you posted this, but I thought I'd post my answer here for others to see. The answer is yes, $f$ has to be constant even if $\varepsilon = 0$. Here's the proof (where $\varepsilon = 0$):
Fix $x \in \mathbb{R}^n$. Then by Markov's inequality, we have:
$$\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq s \} \cap B_r(x)) \leq \frac{1}{s} \int_{B_{r}(x)} |f(y) - f(x)| dy$$
Then by setting $s = cd \space r$ (where $c$ and $d$ are arbitrary positive constants) and dividing both sides by $r^n$, we have:
$$\frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq cd \space r \} \cap B_r(x))}{r^n} \leq \frac{1}{cd \space r^{n+1}} \int_{B_{r}(x)} |f(y) - f(x)| dy$$
The right-hand side tend to $0$ as $r \rightarrow 0^+$, therefore so does the left-hand side. In other words:
$$\lim_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq cd \space r \} \cap B_r(x))}{r^n} = 0$$
Now (whenever $d < 1$) we can lower bound the above expression by:
$$\frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap [B_r(x) \setminus B_{d \space r}(x)])}{r^n}$$
and note that $\frac{\mu^n(B_{d \space r}(x))}{r^n} = V \space d^n$, where $V := \mu^n(B_1(0))$. Therefore:
$$\limsup_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap B_r(x))}{r^n} \leq V \space d^n$$
However this limit holds for all $d \in \mathbb{R}^{>0}$, and so we can actually conclude that:
$$\lim_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap B_r(x))}{r^n} = 0$$
$$\lim_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : \frac{|f(y) - f(x)|}{|y - x|} \geq c \} \cap B_r(x))}{r^n} = 0$$
Therefore $f$ is approximately differentiable at $x$ with $Df_{ap}(x) = 0$ (the definition of approximately differentiable is given here https://encyclopediaofmath.org/wiki/Approximate_differentiability).
Now I want to quote the fact that $f$ having an everywhere zero approximate derivative implies that $f$ is constant. Unfortunately I couldn't find a good source for this, expect in the case of $n=1$ where it follows from theorem 14.18 in The Integrals of Lebesgue, Denjoy, Perron, and Henstock by Russell A. Gordon.
So for arbitrary dimensions, here is my own personal proof for completeness, modified from the $n=1$ case. Feel free to let me know if there are any mistakes.