I am practising the application of
Theorem (Lebesgue Density).
Let $E \in \mathcal B^n \subset \mathcal P(\mathbb R^n)$, meaning that $E$ is Borel. Let $m$ be Lebesgue measure, with $\text{dom}(m)=\mathcal L^n \supset \mathcal B^n.$
Define, where it exists, $$D_E(x) = \lim_{r\to 0} \frac{mE\cap B_r(x)}{mB_r(x)}.$$
Then $D_E(x)=1$ for a.e. $x\in E$ and also $D_E(x)=0$ for a.e. $x\in \mathbb R^n\setminus E.\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (2)$
Firstly, I wonder if we can replace $\mathcal B^n$ with $\mathcal L^n$ in the theorem? (This is not my main question.)
Exercise (1) shown next was posted on the site years ago, and answered. Below, I try to flesh out the answer given by Anonymous.
(1). A Borel set $E\subset \mathbb{R}^2$ has the property that each point of $[0,1]^2$ is a density-one point for $E$. Show that the Lebesgue measure of $E$ is not smaller than $1$.
Suppose instead that $mE<1.$ Define $X=[0,1]^2\setminus E\in \mathcal B^n$, and note that $mX>0.$ Define, where it exists, $$D_X(x) = \lim_{r\to 0} \frac{mX\cap B_r(x)}{mB_r(x)}.$$ For a.e. $x\in X$ we have $D_X(x) =1$, which means $m\{x\in X: D_X(x)\text{ is not defined or is }\not =1\}=0.$
Therefore some $t \in X$ is a density-one point for $X$ (since $mX>0$).
Now Anonymous asserts it follows that this point $t\in [0,1]^2$ cannot be a density-one point for $E$.
If so the proof is done.
It seems to me that the Theorem does not prohibit $t$ to be such that $D_E(t)=1,$ because of the "almost everywhere." So my question is: how to justify the last assertion? By going to the definition (see below), I believe I have justified it for the case when $t$ is an interior point of the little unit square. But I wonder if there is an easier way to see it; also, I do not see how to show it in the case $t$ is a corner of the square, or on the edge.
Sketch: Assume this point $t\in [0,1]^2$ is a density-one point for $E$.
If $t$ is interior to the square, we can choose $s>0$ that simultaneously satisfies
- $mE\cap B_s(t) > \frac12 mB_s(t)$,
- $m([0,1]^2\setminus E)\cap B_s(t) > \frac12 mB_s(t)$,
- $B_s(t)\subset[0,1]^2$.
This implies $mB_s(t)>mB_s(t).$
Bibliography
(1) Two exercises on Lebesgue density points (math.stackexchange.com/questions/276255)
(2) Application of Lebesgue density theorem (math.stackexchange.com/questions/3468349)
Best Answer
Yes, the theorem also holds for Lebesgue measurable sets. In fact, Lebesgue's density theorem is the special case of Lebesgue's differentiation theorem:
$L^1_{\text{loc}}(\Bbb{R}^n)$ means $f$ is Lebesgue measurable and for every compact set $K\subset\Bbb{R}^n$, $\int_K|f(y)|\,dy<\infty$. As a consequence, we have for a.e $x\in\Bbb{R}^n$ that $\lim\limits_{r\to 0^+}\frac{1}{m(B_r(x))}\int_{B_r(y)}f(y)\,dy=f(x)$. Now, by taking $f=\mathbf{1}_E$, you obtain the Lebesgue density theorem.
For your second question, I believe you're right, but note that since the boundary of the square has measure zero, this doesn't really affect our conclusion. Consider $X'=(0,1)^2\setminus E$ instead. By the density theorem, a.e point of $X'$ is a density-1 point for $X'$. Since $m(X')>0$, there exists some point $t\in X'$ which has density $1$ for $X'$. Since this point $t$ belongs to the interior of the square, for sufficiently small $r>0$, we have \begin{align} E\cap B_r(t)&=E\cap (0,1)^2\cap B_r(t) = [(0,1)^2\setminus X']\cap B_r(t)= B_r(t)\setminus [X'\cap B_r(t)] \end{align} So, taking measures, we find that \begin{align} \lim_{r\to 0^+}\frac{m(E\cap B_r(t))}{m(B_r(t))}&=\lim_{r\to 0^+}\frac{m(B_r(t))- m(X\cap B_r(t))}{m(B_r(t))}=1-1=0. \end{align} This contradicts that each point of $[0,1]^2$ (and hence $(0,1)^2$) has density-1 for $E$.
Note that we can also supply a proof without contradiction. Consider $E'=E\cap (0,1)^2$. Then, from the assumption, $D_{E'}(x)$ is $1$ for $x$ in the interior of the square (since inside the square, $D_{E'}(x)=D_E(x)=1$ by hypothesis), while for $x$ outside the closed square, $D_{E'}(x)=0$ (the numerator becomes measure of empty set). Since the boundary of the square has measure zero, this says that the density function $D_{E'}$ equals the indicator function $\mathbf{1}_{(0,1)^2}$ a.e. On the other hand, Lebesgue's density theorem tells us that $D_{E'}=\mathbf{1}_{E'}$ a.e. Since equality a.e is a transitive relation, it follows that $\mathbf{1}_{E'}=\mathbf{1}_{(0,1)^2}$ a.e, and hence $m(E)\geq m(E')=m((0,1)^2)=1$.