"Lebesgue" measurabillity on Riemannian manifolds

Let $X$ be a smooth oriented manifold of positive dimension; Let $g_1,g_2$ be two Riemannian metrics on $X$. Define $\Lambda_1,\Lambda_2:C_c(X) \to \mathbb{R}$ by

$$ \Lambda_i(f)=\int_X f \, Vol_{g_i},$$ where $Vol_{g_i}$ is the Riemannian volume form of $g_i$.

The discussion here implies that for each $i$, there is a unique $\sigma$-algebra $\mathfrak B_i$, and a unique measure $\mu_i$ such that $I(f)=\int_X f d \mu$ for all $f \in C_c(X)$, and the conditions (a)-(f)** are satisfied. (This is essentially Riesz Representation theorem, with the additional observation that $X$ is $\sigma$-compact $\Rightarrow$ uniqueness of the $\mathfrak B_i$).

Questions:

(1) Is it true that $\mathfrak B_1= \mathfrak B_2$? (independence of the metric chosen)

Assuming this is true, is there a way to define $\mathfrak B_i$ without passing through a Riemannian metric? (This is a natural expectation now since $\mathfrak B_i$ is an invariant of $X$ as a differentiable manifold, thus unrelated to the added Riemannian structure)

My guess is that $\mathfrak B_i$ will be the completion of the Borel $\sigma$-algebra w.r.t a suitable measure (which should be any one of the "Lebesgue" measures $\mu_i$ mentioned above. On a first glance, this does seem to go through a Riemannian metric, since the $\mu_i$ was induced by it.

However, the completion w.r.t a measure $\mu$ is dependent only on the subsets that have $\mu$-measure zero, and this is independent of the Riemannian metric chosen, and can be defined invariantly (see Lee's book on smooth manifolds, chapter 6: A subset of a manifold has measure zero if its image under every coordinate chart has measure zero in $\mathbb{R}^n$)

(2) A function $f:X \to \mathbb{R}$ is measurable w.tr.t $\mathfrak B_i$ if and only if it is measurable after (pre)composing with a coordinate chart diffeomorphism?

(3) Is this the standard way to define $L^p(X)$ spaces (from the perspective of measurability, are the elements of $L^p(X)$ exactly the measurable functions in the sense above, with the additional requirement of finiteness of the integral?)

** (a) $\mathfrak{B_i}$ contains all Borel sets,

(b) $\mu(V)=\sup \{I(f): f \in C_c(X), 0\leq f \leq 1, \operatorname{supp} f \subset V\}$ for each open $V$,

(c) $\mu(K) < \infty$ for compact $K$,

(d) $\mu(E)=\inf \{\mu(V): E \subset V, \ V \mbox{ open}\}$ for each $E \in \mathfrak{B}$,

(e) $\mu(E)=\sup \{\mu(K): K \subset E, \ K \mbox{ compact} \}$ for each open $E$ and for each $E\in \mathfrak{B}$ such that $\mu(E)< \infty$,

(f) $\mu$ is a complete measure on $\mathfrak{B}$.

Solution 1:

To put it brief, as hkr noted in the comment, "you buy the charts and topology, you buy the measurable sets".

Here is a more detailed answer:

(1) It is possible to define the Lebesgue sigma-algebra on a general manifold (with or `without' boundary) without alluding to Riemannian metrics or specific choices of densities.

Details can be found in Chapter XII of Amann, H. & Escher, J. "Analysis III". (Birkhäuser, 2009).

(2) This is implied by (1), as otherwise the construction would not work. A function is integrable with respect to a given measure---an invariant notion. (1) only gives you the sigma-algebra on the manifold, it does not give you the measure, however. Whether a function is integrable may depend on the choice of volume form/density, but I do not know of any more detailed discussion of this topic.

(3) Again, if you take a measure-theoretic perspective, the notions of Lp spaces carry over. In the mentioned reference, they even treat the case that the functions take values in some Banach space. A more interesting question would be, if one can make sense of "Lp-spaces of forms".