# Integral average near a point of dispersion

Let $$\Omega\subset\subset\mathbb R^{n}$$ be a bounded domain and let $$E\subset \Omega$$ be a Lebesgue measurable set. Let $$f\in L^{1}(\Omega)$$ and let $$x\in \Omega$$ be a point of dispersion of $$E$$, that is

$$\lim_{r\to0^+}\frac{\lambda^{n}(E\cap B_{r}(x))}{\lambda^{n}(B_r(x))}=0,$$ where $$\lambda ^{n}$$ is the Lebesgue measure and $$B_{r}(x)$$ is the Euclidean ball of center $$x$$ and radius $$r$$.

Is it true that

$$\limsup_{r\to0^+}\frac{\int_{E\cap B_{r}(x)}|f|d\lambda^n}{\lambda^{n}(B_r(x))}<\infty?$$

If not, then what reasonable assumptions on $$f$$ (Higher integrability? Sobolev regularity?) would guarantee this? (The boundedness clearly implies that the limit actually exists and is $$0$$.)

$$\newcommand{\tb}{\tilde B}$$ Let $$d:=n$$. The dispersion condition $$\begin{equation*} \lim_{r\downarrow0}\frac{|E\cap B_r(x)|}{|B_r(x)|}=0 \end{equation*}$$ is of no help, where $$|\cdot|$$ denotes the Lebesgue measure on $$\mathbb R^d$$. More specifically, the following is true:
Theorem Suppose that $$f$$ is a nonnegative function in $$L^1(B_1)$$ such that $$\begin{equation*} \limsup_{r\downarrow0}\frac{\int_{B_r}f}{|B_r|}=\infty, \tag{0} \end{equation*}$$ where $$B_r:=B_r(0)$$, the open ball of radius $$r$$ centered at $$0$$. Then one can construct a measurable set $$E\subset B_1$$ such that $$\begin{equation*} \lim_{r\downarrow0}\frac{|E\cap B_r|}{|B_r|}=0 \tag{1} \end{equation*}$$ but $$\begin{equation*} \limsup_{r\downarrow0}\frac{\int_{E\cap B_r}f}{|B_r|}=\infty. \tag{2} \end{equation*}$$
So, without any conditions on $$E$$ in addition to the dispersion condition (1), the best sufficient condition for $$\begin{equation*} \limsup_{r\downarrow0}\frac{\int_{E\cap B_r}f}{|B_r|}<\infty \tag{not-2} \end{equation*}$$ is the trivial sufficient condition $$\begin{equation*} \limsup_{r\downarrow0}\frac{\int_{B_r}f}{|B_r|}<\infty. \tag{not-0} \end{equation*}$$
To simplify the presentation of the proof of this theorem a bit, assume that $$d=2$$. By (0), there is a sequence $$(r_n)$$ decreasing to $$0$$ such that $$\begin{equation*} \int_{B_{r_n}}f\ge 2^n|B_{r_n}| \end{equation*}$$ for all natural $$n$$. So, passing successively to subsequences, we can construct an increasing sequence $$(n_k)$$ of natural numbers and a sequence $$(S_k)$$ of sets such that $$\begin{equation*} n_k\ge2k, \end{equation*}$$ $$\begin{equation*} \int_{S_k}f\ge 2^{-k}2^{n_k}|\tb_k|\ge2\int_{\tb_{k+1}}f \end{equation*}$$ with $$\begin{equation*} \tb_k:=B_{r_{n_k}}, \end{equation*}$$ and, for each natural $$k$$, $$S_k$$ is a sector of the disk $$\tb_k$$ with the central angle $$2\pi/2^k$$ such that $$S_k\supset S_{k+1}$$. Let now $$\begin{equation*} E:=\bigcup_k(S_k\cap(\tb_k\setminus\tb_{k+1})) =\bigcup_k(S_k\setminus\tb_{k+1}). \end{equation*}$$ Then for any natural $$k$$ the condition $$r_{n_{k+1}}\le r\le r_{n_k}$$ implies $$E\cap B_r\subseteq S_k\cap B_r$$, so that $$|E\cap B_r|\le|S_k\cap B_r|=2^{-k}|B_r|$$, which shows that (1) holds.
On the other hand, $$E\cap\tb_k\supseteq E\cap(\tb_k\setminus\tb_{k+1})=S_k\cap(\tb_k\setminus\tb_{k+1})=S_k\setminus\tb_{k+1},$$ whence $$\begin{multline*} \int_{E\cap\tb_k}f \ge\int_{S_k\setminus\tb_{k+1}}f \ge\int_{S_k}f-\int_{\tb_{k+1}}f \\ \ge\frac12\int_{S_k}f \ge2^{-k-1}2^{n_k}|\tb_k|\ge2^{k-1}|\tb_k|. \end{multline*}$$ So, (2) also holds. $$\Box$$