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Let $M$ be a non-oriented Riemannian manifold of dimension $m$. Nash embedding theorem implies that there exists an isometric embedding $\phi: M\longrightarrow \mathbb{R}^n$ for $n$ sufficiently large. The Lebesgue measure of $\mathbb{R}^n$ induces a measure on $\phi(M)$ (we induce the measure on $m$-submanifold $\phi(M)$ locally, similar as in Fubini Theorem). Hence we obtain a measure on $M$. If $\phi$, $\psi$ are two such isometric embeddings of $M$ into $\mathbb{R}^n$, $\mathbb{R}^N$ resp., then since they are isometries, the induced measure from $R^n$ is the same with the induced measure from $\mathbb{R}^N$. Thus we get a well-defined measure on the non-oriented manifold $M$.

Is my argument valid or not? I am very confused...

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  • $\begingroup$ I think you can define the measure in terms of metric tensor, without referring to any embeddings. In maps, the measure is absolutely continuous with respect to the Lebesgue measure and the Radon-Nikodym derivative is the norm of the metric tensor, and the result should be the same. That said, my differential geometry is not very fresh, so I might be mixing some things up. $\endgroup$ – tomasz Mar 29 '14 at 1:52
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The argument is valid, but you don't need Nash's theorem to come up with the canonical measure on a nonorientable Riemannian manifold $M$. There are other ways:

  1. $M$ is a metric space, and thus carries the $m$-dimensional Hausdorff measure. This measure (up to normalization) agrees with what you get from the volume form when $M$ is orientable. See Riemannian measure and Hausdorff measure in a general Riemannian Manifold.
  2. $M$ has an orientable double cover. You can push the measure from the cover to $M$ by the covering map, which is a local isometry. Divide by $2$ for proper normalization.
  3. Use the concept of $m$-dimensional density.
  4. Use a partition of unity $(\varphi_i)$ subordinate to coordinate patches. Define the measure on each patch using its volume form, multiply by $\varphi_i$, sum over $i$.
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