Proof of Near a Density Point Every Direction Meets the Set Closely

We use the notation of the statement. By The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (i) every closed ball is a Borel set of finite measure, and by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (iii) its measure is $\kappa_{n}$ times the $n$-th power of its radius, where $0<\kappa_{n}$ by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (ii).

Since $0<\delta$, the numbers $\delta^{n}$ and $(1+\delta)^{n}$ are positive; put

$$\gamma=\frac{\delta^{n}}{(1+\delta)^{n}},\qquad \alpha=1-\tfrac{1}{2}\gamma .$$

Then $0<\gamma$ and $\alpha<1$. Because $x$ is a density point of $G$, Density Point of an Arbitrary Subset of $\mathbb{R}^n$ (i) provides $\rho_{0}\in\mathbb{R}$ with $0<\rho_{0}$ such that

$$\alpha\,\lambda_{n}\bigl(\bar{B}(x,R)\bigr)\le\lambda_{n}^{\ast}\bigl(G\cap\bar{B}(x,R)\bigr)\qquad\text{for every }R\in\mathbb{R}\text{ with }0<R<\rho_{0}.$$

Put $\rho=\rho_{0}/(1+\delta)$, a positive real number.

Let $w\in\mathbb{R}^{n}$ with $0<t<\rho$, where $t=\lVert w-x\rVert$, and suppose for contradiction that no $g\in G$ satisfies $\lVert g-w\rVert\le\delta t$. Since $d_{E}(g,w)=\lVert g-w\rVert$ by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, this says exactly that

$$G\cap\bar{B}(w,\delta t)=\varnothing .$$

Put $R=(1+\delta)t$; then $0<R<(1+\delta)\rho=\rho_{0}$.

If $z\in\bar{B}(w,\delta t)$ then, by claim 2 and claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square,

$$\lVert z-x\rVert\le\lVert z-w\rVert+\lVert w-x\rVert\le\delta t+t=R,$$

so $\bar{B}(w,\delta t)\subseteq\bar{B}(x,R)$. Combining this with $G\cap\bar{B}(w,\delta t)=\varnothing$ gives

$$G\cap\bar{B}(x,R)\subseteq\bar{B}(x,R)\setminus\bar{B}(w,\delta t).$$

Both balls are Borel sets of finite measure, so claim 3 of Basic Properties of a Measure applies to the right-hand side, and claims 2 and 1 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$ give

$$\lambda_{n}^{\ast}\bigl(G\cap\bar{B}(x,R)\bigr)\le\lambda_{n}\bigl(\bar{B}(x,R)\bigr)-\lambda_{n}\bigl(\bar{B}(w,\delta t)\bigr)=\kappa_{n}R^{n}-\kappa_{n}\delta^{n}t^{n}.$$

Together with the density inequality at radius $R$ and $\lambda_{n}(\bar{B}(x,R))=\kappa_{n}R^{n}$ this yields

$$\alpha\,\kappa_{n}R^{n}\le\kappa_{n}R^{n}-\kappa_{n}\delta^{n}t^{n},\qquad\text{that is}\qquad \kappa_{n}\delta^{n}t^{n}\le(1-\alpha)\,\kappa_{n}R^{n}.$$

Now $R^{n}=(1+\delta)^{n}t^{n}$ and $1-\alpha=\tfrac{1}{2}\gamma$, so dividing by the positive number $\kappa_{n}t^{n}$ gives

$$\delta^{n}\le\tfrac{1}{2}\gamma\,(1+\delta)^{n}=\tfrac{1}{2}\,\frac{\delta^{n}}{(1+\delta)^{n}}\,(1+\delta)^{n}=\tfrac{1}{2}\delta^{n},$$

which is impossible because $0<\delta^{n}$.

Hence for every $w$ with $0<\lVert w-x\rVert<\rho$ there is $g\in G$ with $\lVert g-w\rVert\le\delta\,\lVert w-x\rVert$.