Suppose $A \subseteq \mathbb{R} $, $m(A) > 0 $. Then for almost all $x \in A $ we have $$ \lim_{\epsilon \to 0^+ } \frac{ m(A \cap (x - \epsilon, x + \epsilon))}{2 \epsilon} = 1.$$

Can someone help me with this problem? $m$ is the Lebesgue measure


  • $\begingroup$ So you want a proof of the theorem? $\endgroup$ – user99680 Dec 7 '13 at 10:41
  • $\begingroup$ yeah, an elementary one $\endgroup$ – ILoveMath Dec 7 '13 at 10:41
  • 1
    $\begingroup$ $m(A)>0$ is not required. $\endgroup$ – Emanuele Paolini Dec 7 '13 at 11:12
  • $\begingroup$ you need a covering theorem to prove something like that. Not very elementary, I would say... $\endgroup$ – Emanuele Paolini Dec 7 '13 at 11:22

Here is a nice proof due to Sierpiński, Démonstration élémentaire du théorème sur la densité des ensembles, Fundamenta Mathematicae, 4 (1), (1923), 167-171. I learned it from Appendix D in van Rooij, and Schikhof, A second course on real functions.

The argument is indeed elementary, and applies to all sets, even non-measurable ones:

Theorem. Given $E\subseteq\mathbb R$, almost every point of $E$ is an exterior density point of $E$, that is, for almost every $a\in E$, we have $$ \lim_{r\to0^+}\frac{m^*(E\cap(a-r,a+r))}{2r}=1, $$ where $m^*$ denotes Lebesgue outer measure.

The usual version follows, since if $E$ is measurable, then so is $E\cap(a-r,a+r)$, and so its outer measure is just its measure.

To prove the result, note that we may assume that $E$ is bounded. It is enough to show that if $t\in(0,1)$, then $$ A=\left\{a\in E\mid \liminf_{r\to0^+}\frac{m^*(E\cap(a-r,a+r))}{2r}<1-t\right\} $$ is null. (Recall that if $g$ is defined on $(a,a+\eta)$, then $$\liminf_{y\to a^+}g(y)=\sup_{0<\epsilon<\eta}\inf\{g(y):a<y<a+\epsilon\}.)$$

Fix $\epsilon>0$. We show that $m^*(A)<\epsilon(1+3/t)$. Since $t$ is fixed, this gives the result. To prove this, start by fixing $U$ open covering $A$ and of measure $m(U)<m^*(A)+\epsilon$ (which exists, by definition of outer measure, and the fact that $A$ is bounded). Note first that if $X\subseteq U$ is measurable, then $$ \begin{array}{cl} m^*(A)&\le m^*(A\cap X)+ m^*(A\setminus X)\le m^*(A\cap X)+m(U\setminus X)\\ &=m^*(A\cap X)+m(U)-m(X)\le m^*(A\cap X)+m^*(A)+\epsilon-m(X),\end{array} $$ so

$$m^*(A\cap X)\ge m(X)-\epsilon. $$

For each $a\in A$, pick an open interval $I$ with rational endpoints such that $a\in I\subseteq U$ and $m^*(E\cap I)<(1-t)m(I)$. This gives us a covering of $A$ by countably many open intervals $(I_n)_{n\in\mathbb N}$, all contained in $U$, and such that

$$ m^*(E\cap I_n)<(1-t)m(I_n) $$

for all $n$. Write each $I_n$ as $(x_n-\delta_n,x_n+\delta_n)$, and let $J_n=(x_n-3\delta_n,x_n+3\delta_n)$. Now, since $m^*(A)\le m(\bigcup_n I_n)$, we can find $N$ large enough that $$m^*(A)-\epsilon<m(\bigcup_{n\le N} I_n).$$ If needed, rearrange the intervals $I_1,I_2,\dots,I_N$ so $\delta_1\ge\delta_2\ge\dots\ge\delta_N$.

By a straightforward recursion, define a subset $L$ of $\{1,\dots,N\}$ with the property that, for each $i$ with $1\le i\le N$, the set $\{I_n\mid n\in L\cap[1,i]\}$ is parwise disjoint and maximal with this property. That $\{I_n\mid n\in L\cap[1,i]\}$ is maximal disjoint for each $i\le N$ implies that, for each $i\le N$, there is an $n\in L\cap[1,i]$ such that $I_n\cap I_i\ne\emptyset$. Since $\delta_n\ge\delta_i$, it follows that $I_i\subseteq J_n$. Therefore, $$ \bigcup_{i\le N}I_i\subseteq \bigcup_{n\in L}J_n. $$

Now set $X=\bigcup_{n\in L}I_n$ and note that, since the $I_n$ for $n\in L$ are disjoint, then $m(X)=\sum_{n\in L}m(I_n)$. This gives us that

$$ m^*(A)-\epsilon<m(\bigcup_{n\le N} I_n)\le m(\bigcup_{n\in L}J_n)\le\sum_{n\in L}m(J_n)=3\sum_{n\in L}m(I_n)=3m(X). $$

Since $$ m^*(E\cap I_n)<(1-t)m(I_n), $$ then

$$ m^*(A\cap X)\le\sum_{n\in L}m^*(A\cap I_n)\le \sum_{n\in L}m^*(E\cap I_n)\le\sum_{n\in L}(1-t)m(I_n)=(1-t)m(X).$$

But $$m^*(A\cap X)\ge m(X)-\epsilon, $$ so $m(X)\le\epsilon/t$ and, since $$ m^*(A)-\epsilon<3m(X),$$ then $$ m^*(A)<\epsilon+3\epsilon/t,$$ as we wanted.

(Note we are indeed using a covering result, namely that $\bigcup_{n\in L}J_n$ covers $\bigcup_{i\le N}I_i$. This is a particular instance of Vitali's covering lemma. A nice presentation of this and related covering results can be found in Krantz-Parks, The geometry of domains in space.)


This is nothing more than Lebesgue differentiation theorem applied to the function $f=\chi_A$. Follow this link to read the proof.


What about this excellent article in The American Mathematical Monthly?

They essentially only use properties of the outer measure.


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.