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. $\Box$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.