Yes, it is the standard "surface measure" on $S^{n-1}$.
In order to prove it, it is enough to check that it gives the expected measure on some simple "surface elements" of your choosing (as long as they generate the Borel algebra). You will find the expression of the spherical volume element (in spherical coordinates) here (we are talking about the standard Euclidean sphere here).
A perhaps more instrinsic/satisfying argument is that if you denote by $\sigma_r$ the measure on $S_r^{n-1}$ ($r$ is the radius), then
- "$\mu_ n = \int_{0}^{+\infty} \sigma_r\,dr$" in the sense that $\mu_n(A) = \int_0^{+\infty} \sigma_r(A \cap S^{n-1})\,dr$ for any $A \in \mathscr{B}_{\mathbb{R}^n}$.
- "$\sigma_r = r^{n-1} \sigma_1$" in the sense that $\sigma_r (rA) = r^{n-1} \sigma_1(A)$ for any $A \in \mathscr{B}_{S^{n-1}}$
Of course, these two claims would need to be justified, I'll let you think about it. Anyway, it follows that:
$$\mu_n((0,1]A) = \int_0^1 \sigma_r(rA)\,dr = \int_0^1 r^{n-1} \sigma_1(A)\,dr = {\sigma_1(A) \over n}~.$$
I heartily welcome feedback, but I believe the proof in my question works. The first section where I thought the proof was lacking can indeed be improved. I got the idea for the improvement from the hint to Exercise 2.3 in Lee's Riemannian Manifolds, restated here for interest:
Exercise 2.3. Suppose $M\subset \widetilde M$ is an embedded submanifold.
(a) If $f$ is any smooth function on $M$, show that $f$ can be extended to a smooth function on $\widetilde M$ whose restriction to $M$ is $f$. [Hint: Extend $f$ locally in slice coordinates by letting it be independent of $(x^{n+1},\dots,x^m)$, and patch that together using a partition of unity.]
Here is the improved proof of 1:
Let $n = \dim S$. Since $S$ is embedded, each $p\in S$ is in the domain of
a slice chart $(U_p,\varphi_p)$ in $M$ such that $f|_{S\cap U_p} = f(x^1,\dots,x^n)$ and
the $x^i$ are slice coordinates. Extend $f|_{S\cap U_p}$ to $U_p$ by setting
$f_p(x^1,\dots,x^n,x^{n+1},\dots,x^m) = f|_{S\cap U_p}(x^1,\dots,x^n)$. The map
$f_p\colon U_p\to\Bbb R$ is smooth because it is independent of its last $m-n$ coordinates
and it is smooth with respect to its first $n$ coordinates.
The collection of open sets $U_p$ is an open cover of $S$, so let $(\psi_p)_{p\in S}$ be
a partition of unity subordinate to this cover, and define $\widetilde f(x) = \sum_{p\in S}
\psi_p(x)f_p(x)$. If $x\in S$, then
$$
\widetilde f(x) = \sum_{p\in S}\psi_p(x) f(x) = f(x).
$$
Hence $\widetilde f$ is an extension of $f$.
Since the collection of supports of the $\psi_p$ is locally finite, each point in
$U$ has a neighborhood in which the sum is finite, so this defines
a smooth map since every point has a neighborhood such that $\widetilde f$ restricted
to that neighborhood is smooth. Hence $\widetilde f\in C^\infty(U)$ is the desired
extension. $\square$
The proof of 2 is modified by replacing $\varphi_p^{-1}(W_p)$ with $U_p$ defined in the modified proof of 1 above. $\square$
The reason that the original proof still works is that the extensions I describe in the first proof of 1 can be taken to be the $f_p$ explicitly described in the proof in this answer. The approach taken here simplifies some of the notation and construction of the extension $f_p$ by blurring the distinction between points $p$ in the manifold and their local coordinate representations $(x^1,x^2,x^3,\dots)$. Smoothness of the coordinate representation of $f$ is equivalent to smoothness of the actual map defined on the manifold.
Best Answer
Part 1 is immediate once you understand the definitions involved (and there is no need to restrict to dimension 3). I will just leave you two hints:
(a) First prove that every closed subset of $\Bbb R^n$ belongs to the Borel sigma-algebra, hence, to the Lebesgue sigma-algebra. (You really just need to know what a sigma-algebra is and that products of open intervals belong to the Lebesgue sigma-algebra.)
(b) Prove that every (nonempty) $k$-dimensional topological submanifold $M$ of $\Bbb R^n$ can be expressed as a countable union of subsets homeomorphic to the closed unit disk $D^k$. If you have hard time with this, first verify that every subset of $\Bbb R^n$ satisfies the 2nd countability axiom (something that you typically learn in a general topology class).
Part 2 is trickier. One way to find examples of topological surfaces in $\Bbb R^3$ of positive 3-d Lebesgue measure is to start with an Osgood curve $C\subset \Bbb R^2$, which is a Jordan curve of positive 2-d Lebesgue measure. Now, take the direct product of $J$ with an open interval and obtain a topological surface in $\Bbb R^3$ that has positive 3-d Lebesgue measure. The same construction will work, of course, in higher dimensions.