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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 $0$.)

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$\newcommand{\tb}{\tilde B}$ 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. $\Box$

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