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 $\mathbb{R}^n$ §borel every closed ball is a Borel set of finite measure, and by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ §value 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 $\mathbb{R}^n$ §constant.

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

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

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 Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, this says exactly that

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 Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$,

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

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

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

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

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

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