Proof of Lebesgue Measure on Euclidean Space is Sigma-Finite

Each result cited is universally quantified over the data in its own statement. Fix $m\in\mathbb{N}$ and write $B_{m}=B(o,\iota(m))$.

Step 1 (Borel). By claim 1 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure, applied with $n=q$, centre $o$ and radius $\iota(m)$, the set $B_{m}$ belongs to $\mathcal{B}(\mathbb{R}^{q})$; the measure $\lambda_{n}$ and the balls of that lemma are the Lebesgue measure $\lambda_{q}$ of Euclidean Space and Lebesgue Measure: Standing Notation §measure and the balls of Euclidean Space and Lebesgue Measure: Standing Notation §space, both being taken from Lebesgue Measure on $\mathbb{R}^n$ and Open Ball in a Metric Space for the Euclidean distance $d_{E}$.

Step 2 (finite measure). For every $y\in B_{m}$ we have $d_{E}(o,y)<\iota(m)$ by Open Ball in a Metric Space, hence $d_{E}(o,y)\le\iota(m)$. With the point $o$ and the positive real $R=\iota(m)$ this is the condition of Bounded Subset of a Metric Space, so $B_{m}$ is bounded, and $\lambda_{q}(B_{m})<\infty$ by claim 2 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure.

Step 3 (covering). Each $B_{m}$ is a subset of $\mathbb{R}^{q}$, so $\bigcup_{m\in\mathbb{N}}B_{m}\subseteq\mathbb{R}^{q}$. Conversely let $x\in\mathbb{R}^{q}$. By claim 1 of The Archimedean Property of the Real Numbers, applied to the real number $\lVert x\rVert$, there is $k\in\mathbb{N}$ with $\lVert x\rVert<\iota(k)$. Since $o$ is the zero vector, $x-o=x$, and $d_{E}$ is symmetric, being a metric; so $d_{E}(o,x)=d_{E}(x,o)=\lVert x-o\rVert=\lVert x\rVert<\iota(k)$ by Euclidean Space and Lebesgue Measure: Standing Notation §space. Thus $x\in B_{k}$, and $\mathbb{R}^{q}=\bigcup_{m\in\mathbb{N}}B_{m}$.

Step 4 ($\sigma$-finiteness). By Steps 1 to 3, $(B_{m})_{m\in\mathbb{N}}$ is a sequence in $\mathcal{B}(\mathbb{R}^{q})$ whose union is $\mathbb{R}^{q}$ and each of whose members has finite $\lambda_{q}$-measure. This is the definition of $\sigma$-finiteness in Measure, Measure Space, and Probability Measure. $\blacksquare$