Proof of The Lebesgue Density Theorem in $\mathbb{R}^n$

We use the notation of the statement. By The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ §borel every closed ball is a Borel set of finite measure, so the numbers $\lambda_{n}(\bar{B}(y,r))$ below are defined and finite; by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ §value they equal $\kappa_{n}r^{n}$, where $0<\kappa_{n}$ by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ §constant.

Step 1 (Reduction to countably many bounded pieces). For a natural number $p$ with $2\le p$ put $\alpha_{p}=1-1/p$, so that $0<\alpha_{p}<1$, and let $N_{p}$ be the set of those $x\in E$ with the following property: for every $\rho\in\mathbb{R}$ with $0<\rho$ there is $r\in\mathbb{R}$ with $0<r<\rho$ and

We claim that every $x\in E$ which is not a density point of $E$ belongs to $N_{p}$ for some such $p$. Indeed, negating Density Point of an Arbitrary Subset of $\mathbb{R}^n$ §density-point gives $\alpha\in\mathbb{R}$ with $\alpha<1$ such that for every $\rho>0$ there is $r$ with $0<r<\rho$ and $\lambda_{n}^{\ast}(E\cap\bar{B}(x,r))<\alpha\,\lambda_{n}(\bar{B}(x,r))$. Since $0<1-\alpha$, The Archimedean Property of the Real Numbers provides a natural number $p$ with $1/p<1-\alpha$, and replacing $p$ by a larger natural number only decreases $1/p$, so we may assume $2\le p$. Then $\alpha<1-1/p=\alpha_{p}$, and since $0\le\lambda_{n}(\bar{B}(x,r))$ we get $\alpha\,\lambda_{n}(\bar{B}(x,r))\le\alpha_{p}\,\lambda_{n}(\bar{B}(x,r))$; hence $x\in N_{p}$.

For natural numbers $p\ge2$ and $m$ put $N_{p,m}=N_{p}\cap\bar{B}(0,m)$. Given $x\in\mathbb{R}^{n}$, The Archimedean Property of the Real Numbers provides a natural number $m$ with $\lVert x\rVert\le m$, and $d_{E}(x,0)=\lVert x\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so $x\in\bar{B}(0,m)$; hence $N_{p}=\bigcup_{m\in\mathbb{N}}N_{p,m}$. By claim 5 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$ a countable union of null sets is null, and a set is null exactly when its outer measure vanishes. Applying this twice, first over $m$ and then over the natural numbers $p\ge2$ re-indexed as $p=q+1$ with $q\in\mathbb{N}$, it suffices to prove that $\lambda_{n}^{\ast}(N_{p,m})=0$ for all $p\ge2$ and all $m$.

Step 2 (The covering estimate). Fix such $p$ and $m$ and write $P=N_{p,m}$, $\alpha=\alpha_{p}$ and $\alpha'=1-1/(2p)$, so that $0<\alpha<\alpha'<1$. Suppose, for contradiction, that $0<\lambda_{n}^{\ast}(P)$.

Since $P\subseteq\bar{B}(0,m)$, claims 2 and 1 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$ give $\lambda_{n}^{\ast}(P)\le\lambda_{n}^{\ast}(\bar{B}(0,m))=\lambda_{n}(\bar{B}(0,m))=\kappa_{n}m^{n}<\infty$, so $\lambda_{n}^{\ast}(P)$ is a positive real number.

Put $\varepsilon_{0}=(\alpha'/\alpha-1)\,\lambda_{n}^{\ast}(P)$, a positive real number because $\alpha<\alpha'$ and $0<\alpha$. By Outer and Inner Regularity of Lebesgue Measure on $\mathbb{R}^n$ §outer-arbitrary, applicable because $\lambda_{n}^{\ast}(P)<\infty$, there is an open set $U\subseteq\mathbb{R}^{n}$ with $P\subseteq U$ and

Let $\mathcal{F}$ be the set of all pairs $(y,r)$ with $y\in P$, $r\in\mathbb{R}$, $0<r$, $\bar{B}(y,r)\subseteq U$ and $\lambda_{n}^{\ast}(E\cap\bar{B}(y,r))\le\alpha\,\lambda_{n}(\bar{B}(y,r))$.

We check that $\mathcal{F}$ finely covers $P$. Let $x\in P$ and let $\eta\in\mathbb{R}$ with $0<\eta$. Since $U$ is open and $x\in U$ there is $\rho_{1}\in\mathbb{R}$ with $0<\rho_{1}$ and $B(x,\rho_{1})\subseteq U$. If $0<r<\rho_{1}$ and $z\in\bar{B}(x,r)$ then $d_{E}(z,x)\le r<\rho_{1}$, so $z\in B(x,\rho_{1})$; hence $\bar{B}(x,r)\subseteq U$ for every such $r$. Let $\rho$ be the smaller of $\eta$ and $\rho_{1}$. Since $x\in N_{p}$ there is $r$ with $0<r<\rho$ and $\lambda_{n}^{\ast}(E\cap\bar{B}(x,r))<\alpha\,\lambda_{n}(\bar{B}(x,r))$. Then $(x,r)\in\mathcal{F}$, while $x\in\bar{B}(x,r)$ and $r<\eta$, which is what fine covering requires.

Put $\varepsilon=\tfrac{1}{2}(1-\alpha')\,\lambda_{n}^{\ast}(P)$, a positive real number. Since $\lambda_{n}^{\ast}(P)<\infty$, The Vitali Covering Theorem in $\mathbb{R}^n$ §finite provides a natural number $N$, possibly $0$, and pairs $(y_{1},r_{1}),\dots,(y_{N},r_{N})\in\mathcal{F}$ whose closed balls $\bar{B}_{i}=\bar{B}(y_{i},r_{i})$ are pairwise disjoint and satisfy $\lambda_{n}^{\ast}\bigl(P\setminus\bigcup_{i=1}^{N}\bar{B}_{i}\bigr)\le\varepsilon$.

Now $P$ is the union of $P\setminus\bigcup_{i=1}^{N}\bar{B}_{i}$ and the sets $P\cap\bar{B}_{i}$ for $1\le i\le N$. Applying claim 4 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$ to the sequence consisting of these $N+1$ sets followed by empty sets, and then claim 2 of that lemma together with $P\subseteq E$, we obtain

Each pair $(y_{i},r_{i})$ lies in $\mathcal{F}$, so $\lambda_{n}^{\ast}(E\cap\bar{B}_{i})\le\alpha\,\lambda_{n}(\bar{B}_{i})$. The balls $\bar{B}_{i}$ are pairwise disjoint Borel sets contained in $U$, so claim 1 of Basic Properties of a Measure and then claim 2 of that lemma give

Combining the two displays, $\lambda_{n}^{\ast}(P)\le\varepsilon+\alpha'\lambda_{n}^{\ast}(P)$, that is

Since $0<1-\alpha'$ and $0<\lambda_{n}^{\ast}(P)$, the left-hand side is a positive real number and the right-hand side is half of it, which is impossible.

Therefore $\lambda_{n}^{\ast}(P)=0$ for all $p\ge2$ and all $m$, and by Step 1 the set of those $x\in E$ which are not density points of $E$ is null.