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 $0$.)
 A: $\newcommand{\tb}{\tilde B}$
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. $\Box$
