I apologize that this is short of research level but I have realized that I am not happy with my understanding of the volume form on an oriented Riemannian manifold and I was hoping to find some guidance.

For concreteness, I'll take a compact oriented manifold $X^n \subset \mathbb{R}^N$ (to be even more concrete, I am happy to take $n = 2$ and $N = 3$). I am giving $X$ the induced Riemmanian metric and I understand that together with the orientation of $X$ this gives me a volume form $dV \in \Omega^n(X)$. We then define the volume of $X$ to be the integral $\int_X dV$. I would like to understand why this deserves to be called the volume of $X$.

I'd like to see that this integral definition agrees with the more intuitive idea of taking the limit over finer and finer triangulations of $X$ of the volume of the euclidean simplices determined by these triangulations. Let me try to make that precise. Let $\Delta_1, \Delta_2,...$ be a sequence of triagulations of $X$ with the property that the maximum length of an edge in $\Delta_n$ is going to $0$ as $n \to \infty$ (here using the induced metric say on $X$). Now for each $\Delta_i$, we get a positive real number $\delta_i$ by summing over all the top dimensional cells of the triangulation and taking for each such cell the (normal euclidean) volume of the simplex in $\mathbb{R}^N$ spanned by the vertices of each cell. The volume of $X$ is then the limit of these numbers $\delta_n$. This definition I guess has the usual downside of not being clearly well-defined and not sounding easy to compute, but at least it seems like the intuitive way to define volume.

Does this intuitive definition agree with the definition using the volume form? If so, how would I go about proving this?

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    $\begingroup$ I think that Spivak has a paper about the problems of triangulation as a definition of volume. $\endgroup$
    – Ben McKay
    Commented May 25, 2022 at 9:31
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    $\begingroup$ I don't recall a Spivak paper as mentioned by @BenMcKay, but I do recall seeing an argument somewhere that triangulations inscribed in the unit 2-dimensional sphere can have arbitrarily large area. $\endgroup$ Commented May 25, 2022 at 12:29
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    $\begingroup$ See p. 129 of Spivak, Calculus on Manifolds, for the picture, and reference to the paper. $\endgroup$
    – Ben McKay
    Commented May 25, 2022 at 13:26
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    $\begingroup$ en.wikipedia.org/wiki/Schwarz_lantern $\endgroup$ Commented May 25, 2022 at 18:59
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    $\begingroup$ Thank you all for the references - I was not at all aware of the Schwarz lantern. I found this paper by Frieda Zames (linked in the above wiki) to be a nice explanation. She explains how, in the end, Lebesgue rescues the geometric limit definition by taking the limit correctly. $\endgroup$
    – Sprotte
    Commented May 25, 2022 at 19:59


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