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I was wondering about the sense of the following definition:

Let (M,g) be a Riemmanian manifold and $\varphi\in C^{\infty}(M)$ a smooth function, such that $\varphi\geq 0$ or $supp(\varphi)\subset M$ is compact. Let $A\subset M$ be mesuarable. If there is a chart $x:U\rightarrow\mathbb{R}^n$ with $\overline{A}\subset U$, then the following makes sense (since the right hand side is integrable, i.e. less than infinity):

$\int\limits_{M}\varphi dV:=\int\limits_{x(A)}(\varphi \circ x^{-1})\cdot \sqrt{det(g_ij)}dV$

I don't know why the right hand side is integrable if $\varphi\geq 0$. If the support is compact, then of course we have $\int\|\|<\infty$ But what is the explanation in the other case?

Thank you a lot.

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Your accept rate is showing, in case you didn't know. –  rschwieb Sep 25 '12 at 19:56
    
excuse me, what do you mean? I do not understand. –  Braten Sep 25 '12 at 20:18
    
We usually remind new users that their accept rates are visible because low accept rates deter people from answering your question. I see you moved it up from 30% to 71%, so that's progress! –  rschwieb Sep 25 '12 at 21:02

1 Answer 1

up vote 1 down vote accepted

Probably they allow $+\infty$ as the result for the integral.

Take for example $M:=\mathbb R$, $A:=\displaystyle\bigcup_n (n,\ n+\frac 1{n^2})$, measurable, and $\varphi(x):=e^x$, smooth.

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