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 $0$.)



Integral average near a point of dispersion
$\newcommand{\tb}{\tilde B}$
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. $\Box$


