The unit sphere is Lebesgue null

The section integral. Fix $r>0$. For every $\omega \in S^{n-1}$ one has $\mathrm{\mid }r\omega \mathrm{\mid }=r$, so $r\omega \in S^{n-1}$ if and only if $r=1$. Hence ${\mathbf{1}}_{S^{n-1}}(r\omega )={\mathbf{1}}_{\{1\}}(r)$ for every $\omega$, and by [F3] and [F1], ${\int }_{S^{n-1}}{\mathbf{1}}_{S^{n-1}}(r\omega )d\sigma (\omega )={\int }_{S^{n-1}}{\mathbf{1}}_{\{1\}}(r)d\sigma (\omega )={\mathbf{1}}_{\{1\}}(r)\sigma (S^{n-1})\mathrm{.}$ The factor $\sigma (S^{n-1})$ is finite by [F1].

The polar integral. The function $f={\mathbf{1}}_{S^{n-1}}$ is Borel and nonnegative by [F2], so the polar-coordinate formula [F1] applies and, with the section computation of step 1.1, ${\lambda }_n(S^{n-1})={\int }_{{\mathbb{R}}^n}{\mathbf{1}}_{S^{n-1}}d{\lambda }_n={\int }_0^{\mathrm{\infty }}[{\int }_{S^{n-1}}{\mathbf{1}}_{S^{n-1}}(r\omega )d\sigma (\omega )]r^{n-1}dr=\sigma (S^{n-1}){\int }_0^{\mathrm{\infty }}{\mathbf{1}}_{\{1\}}(r)r^{n-1}dr\mathrm{.}$

The radial integral vanishes. The function $r\mapsto {\mathbf{1}}_{\{1\}}(r)r^{n-1}$ is nonnegative, measurable and vanishes for every $r\mathrm{\ne }1$; the singleton $\{1\}$ is Lebesgue null in $(0,\mathrm{\infty })$ by [F4], so the function vanishes almost everywhere. By [F5] its integral over $(0,\mathrm{\infty })$ is $0$, and since $\sigma (S^{n-1})<\mathrm{\infty }$ the right-hand side of step 2.1 is $\sigma (S^{n-1})\cdot 0=0$. Therefore ${\lambda }_n(S^{n-1})=0$.

Conclusion. Steps 1.1–3.1 evaluate the polar-coordinate formula at ${\mathbf{1}}_{S^{n-1}}$ and prove ${\lambda }_n(S^{n-1})=0$ for every $n\ge 2$; Countable Choice is inherited exactly from the polar-coordinate, sigma-finite Tonelli, null-set and integral-interface suppliers.