Proof of The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$

We use the notation of the statement; throughout $x\in\mathbb{R}^{n}$ and $r\in\mathbb{R}$ with $0<r$.

Claim 1. By claim 3 of Elementary Properties of the Closed Ball in a Metric Space the set $\bar{B}(x,r)$ is a closed subset of the topological space determined by $d_{E}$, and by claim 2 of that lemma it is bounded in $(\mathbb{R}^{n},d_{E})$. The two properties together are claim 2 of Heine-Borel Theorem in $\mathbb{R}^n$, which is equivalent to claim 1 of that theorem, so $\bar{B}(x,r)$ is compact. By claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure every compact subset of $\mathbb{R}^{n}$ is a Borel set of finite measure, so $\bar{B}(x,r)\in\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(\bar{B}(x,r))<\infty$.

Claim 2. Applying claim 1 with the centre $0$ and radius $1$ shows that $\bar{B}(0,1)\in\mathcal{B}(\mathbb{R}^{n})$ and that $\kappa_{n}=\lambda_{n}(\bar{B}(0,1))$ is finite; since $\lambda_{n}$ takes values in $[0,\infty]$, $\kappa_{n}$ is a nonnegative real number.

If $z\in B(0,1)$ then $d_{E}(z,0)<1$ by Open Ball in a Metric Space, hence $d_{E}(z,0)\le1$ and $z\in\bar{B}(0,1)$ by Closed Ball in a Metric Space; thus $B(0,1)\subseteq\bar{B}(0,1)$. By claim 1 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure the open ball $B(0,1)$ is a Borel set with $0<\lambda_{n}(B(0,1))$, so claim 2 of Basic Properties of a Measure gives $0<\lambda_{n}(B(0,1))\le\kappa_{n}$. Hence $0<\kappa_{n}<\infty$.

Claim 3. We first show that $\bar{B}(x,r)=\bigl(r\,\bar{B}(0,1)\bigr)+x$.

Let $y\in\mathbb{R}^{n}$ and put $z=r^{-1}(y-x)$, so that $y=rz+x$; the maps $y\mapsto r^{-1}(y-x)$ and $z\mapsto rz+x$ are mutually inverse bijections of $\mathbb{R}^{n}$. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ we have $d_{E}(y,x)=\lVert y-x\rVert$ and $d_{E}(z,0)=\lVert z\rVert$, and by claim 5 of that lemma

since $0<r$ implies $0<r^{-1}$. Multiplying by the positive number $r$ shows that $\lVert y-x\rVert\le r$ holds if and only if $\lVert z\rVert\le 1$ holds; that is, $y\in\bar{B}(x,r)$ if and only if $z\in\bar{B}(0,1)$. Since $y=rz+x$, this says exactly that $\bar{B}(x,r)$ is the set of points $rz+x$ with $z\in\bar{B}(0,1)$, which is $\bigl(r\,\bar{B}(0,1)\bigr)+x$.

By claim 1 of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$, applied with $c=r$ to the Borel set $\bar{B}(0,1)$, the set $r\,\bar{B}(0,1)$ is Borel and

using $|r|=r$. By claim 1 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$, applied with the translation vector $x$ to that Borel set,

The right-hand side does not involve $x$, and it is positive because $0<\kappa_{n}$ by claim 2 and $0<r^{n}$.