Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is integrable. Is it true that
$$
\lim_{r\to 0}\frac{\displaystyle\int_{B_r(0)}f(y)~\mathrm dy}{r^{n-1}}=0 \quad ?
$$
This is obvious if $0$ is a Lebesgue point of $f$ or if $n=1$, but I would like to know if it's true in general.
How Badly Can the Lebesgue Differentiation Theorem Fail?
geometric-measure-theoryintegrationlebesgue-measure
Related Solutions
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.
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:
- If $x,y \in S$ are distinct, then $x \not =_1 y$
- 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.
The answer is no. Let me present a counterexample for $n=2$.
Fix some bump function $h(x)$ supported in $[-1,1]$ and such that $|h(x)|\leq 1$ and $h(0)=1$ (though those choices don't matter much). Define $f(x,y)=h(y/x^2)$ for $x>0$ and $f(x,y)=0$ elsewhere. This function is easily seen to be smooth at all points other than $(0,0)$, and discontinuous at the origin. I claim $f$ is $\frac{1}{2}$-times Lebesgue differentiable at the origin.
$f$ is $0$ except for the region $0<|y|<x^2$, and on that region it is bounded by $1$. Therefore the integral of $|f(x,y)-f(0,0)|=|f(x,y)|$ over a ball $B_r(0)$ is bounded by the volume of the intersection of this region with the ball, which is contained inside the rectangle with vertices $(0,\pm r^2),(r,\pm r^2)$ of volume $2r^3$. We are done as $\frac{2r^3}{r^{2+1/2}}=2r^{1/2}$ tends to $0$.
Best Answer
Metafune has given an example of the limit failing to be $0$ at a particular point - namely for $n > 1$, the function $|x|^{-\alpha}$, with $1 \leq \alpha < n$ has that limit equal to $\infty$ at $0$.
However, you can still get some kind of affirmative result.
In general the limit in question is zero $\mathcal H^{n-1}$-a.e, where $\mathcal H^{n-1}$ denotes the $n-1$ dimensional Hausdorff measure.
This is Theorem 2.10 in Measure Theory and Fine Properties of Functions by Evans and Gariepy (2015 version).