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?



If a function $f$ is $(1+\varepsilon)$-times Lebesgue differentiable everywhere, is $f$ a constant function?
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 $\varepsilon = 0$):




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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:


$$\lim\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq cd \space r \} \cap B\_r(x))}{r^n} = 0$$




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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:


$$\lim\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap B\_r(x))}{r^n} = 0$$


$$\lim\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : \frac{|f(y) - f(x)|}{|y - x|} \geq c \} \cap B\_r(x))}{r^n} = 0$$




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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.



> 
> Assume, seeking a contradiction, that there exists $a, b \in \mathbb{R}^n$ such that $f(a) > f(b)$. Define $g(x) := f(x) + x \cdot \frac{f(a) - f(b)}{2|b - a|} \frac{b-a}{|b-a|}$.
> 
> 
> Then first of all $g$ is approximately differentiable with $D\_{ap}g(x)(h) = h \cdot \frac{f(a) - f(b)}{2|b - a|} \frac{b-a}{|b-a|}$ for all $x \in \mathbb{R}^n$.
> 
> 
> Now without loss of generality, we can assume that $a = 0$ and $b = (1,0,...,0)$ just so I can talk about moving from $a$ to $b$ as moving from left to right. Note that $D\_{ap}g(x)$ is positive in every direction which points rightward (by which I mean the direction has a positive dot product with $b$).
> 
> 
> In what follows let $x <\_1 y$ be defined to mean that the $1^{\text{st}}$ coordinate of $x$ is less than the $1^{\text{st}}$ coordinate of $y$. Same for similar relations like $=\_1$.
> 
> 
> Let $W \subseteq \mathbb{R}^n$ be the interior and boundary of some right-angled cone with vertex at $b$, and $a$ at the middle of its base.
> 
> 
> Now define $A := \{x \in W : g(x) \geq g(a)\}$. Let $\mathcal{F}$ be a family of subsets of $A$, such that $S \in \mathcal{F}$ whenever all the following hold:
> 
> 
> 1. If $x,y \in S$ are distinct, then $x \not =\_1 y$
> 2. If $x,y \in S$ with $x <\_1 y$, then $\frac{\mu^n(A \cap B\_{|x-y|}(y))}{\mu^n(B\_{|x-y|}(y))} \geq \frac{1}{2}$
> 
> 
> Note that $\mathcal{F}$ is non-empty as $\{a\} \in \mathcal{F}$, and is partially ordered by $\subseteq$ where every non-empty linearly ordered subset has an upper bound (the union). Therefore, by Zorn's lemma, $\mathcal{F}$ has a maximal element which we can denote by $Z$.
> 
> 
> Assume, seeking a contradiction, that $Z$ does not contain an element with maximal $1^{\text{st}}$ coordinate. Then there exists $p$ which is a limit point of $Z$ and has supremum $1^{\text{st}}$ coordinate. Let $(z\_k)\_{k=1}^\infty$ be a sequence in $Z$ which converges to $p$.
> 
> 
> Then for each $x \in Z$, we have:
> $$\mu^n(A \cap B\_{|x-p|}(p))$$
> 
> 
> $$= \lim\_{k \rightarrow \infty} \mu^n(A \cap B\_{|x-z\_k|}(z\_k))$$
> 
> 
> $$\geq \lim\_{k \rightarrow \infty} \frac{1}{2} \mu^n(B\_{|x-z\_k|}(z\_k))$$
> 
> 
> $$= \frac{1}{2} \mu^n(B\_{|x-p|}(p))$$
> 
> 
> (The justification for the first and last equalities comes from the fact that $\lim\_{k \rightarrow \infty} \mu^n(B\_{|x-z\_k|}(z\_k) \Delta B\_{|x-p|}(p)) = 0$, where $\Delta$ denotes the symmetric difference).
> 
> 
> So $\frac{\mu^n(A \cap B\_{|x-p|}(p))}{\mu^n(B\_{|x-p|}(p))} \geq \frac{1}{2}$ for all $x \in Z$. Also note that $g$ is approximately continuous at $p$, so we must have that $p \in A$.
> 
> 
> Then $Z \cup \{p\} \in \mathcal{F}$, which contradicts the maximality of $Z$. Therefore $Z$ contains an element with maximal $1^{\text{st}}$ coordinate. Denote the element as $q$.
> 
> 
> 
> 
> ---
> 
> 
> Now we must have that $q =\_1 b$, otherwise we could use the fact that $g$ is approximately differentiable at $q$ (with positive derivative in any rightward direction) to extend $Z$, again contradicting its maximality. But due to $q \in W$, we must have that $q = b$.
> 
> 
> So $b = q \in Z \subseteq A$. Hence $g(a) \leq g(b)$, which contradicts the assumption that $f(a) > f(b)$ from the beginning.
> 
> 
> Hence $f$ is constant, as $a$ and $b$ are arbitrary.
> 
> 
> 


