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Let $M$ a manifold with Atlas $\{(U,\varphi ),(V,\psi)\}$ where $U\cap V\neq \emptyset$.

If $A\subset U$, then for $f:M\to \mathbb R$, $$\int_A f:=\int_{\varphi (A)}f(\varphi ^{-1}(x))\,\mathrm d x.$$

But what happen if we integrate over $M$ ? Indeed, since $U$ and $V$ are not disjoint, we won't have that $$\int_M f=\int_Uf+\int_Vf,$$ and use the previous formula. So, how could we do ?

An idea would have been : $$\int_Mf=\int_{U\setminus V}f+\int_{V\setminus U}f+\int_{U\cap V}f$$ $$=\int_{\varphi (U\setminus V)}f(\varphi ^{-1}(x))\,\mathrm d x+\int_{\psi(V\setminus U)}f(\psi^{-1}(x))\,\mathrm d x+\int_{U\cap V}f,$$ but which chart can I use on $U\cap V$ ? (because $\varphi $ and $\psi$ are both defined on it).

joshua
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Writing $\int_Mf$ for a smooth function $f:M\to\Bbb{R}$ makes no sense until you give it a definition (and it turns out you can't give a meaningful one). We either integrate differential forms on oriented manifolds, or scalar densities in general (on not-necessarily oriented manifolds); we don't integrate functions on manifolds unless you specify somehow a measure on $M$ (eg using a Riemannian metric). The reason why we don't integrate functions is precisely because of the question you raised. If you calculate $\int_{\phi(U\cap V)}(f\circ\phi^{-1})(x)\,dx$ and compare with $\int_{\psi(U\cap V)}(f\circ \psi^{-1})(x)\,dx$, you won't get the same result.

peek-a-boo
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  • $\int_M$ is a short notation for $\int_M f(m),\mathrm d m$... which is a $1-$form (I guess, no ?). – joshua Jun 19 '22 at 09:00
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    @joshua and what does your $dm$ mean? You're supposed to integrate $n$-forms on $n$-dimensional oriented manifolds. Look up any text for the definition of integration on manifolds. – peek-a-boo Jun 19 '22 at 09:18