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In Chapter 16 of John Lee's book Introduction to Smooth Manifolds, he defines integrals over subspaces of $\mathbb R^n$ as follows:

If $D\subseteq\mathbb R^n$ is a bounded subset whose boundary has measure zero, and if $\omega$ is a continuous $n$-form on $\overline D$, then write $\omega=fdx^1\wedge\dots\wedge dx^n$ for some continuous function $f:\overline D\to\mathbb R$. Then the integral of $\omega$ over $D$ is $$\int_D\omega=\int_DfdV.$$

My (possibly dumb) question is: Why does $\omega$ have to be defined on $\overline D$? Shouldn't it be enough for $\omega$ to be a continuous $n$-form defined on $D$?

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    $\begingroup$ What's $\int_0^1\frac{dx}x$? $\endgroup$ Commented Oct 6, 2020 at 2:35

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As mentioned by Angina Seng, when you are defining integration, you typically want to deal with functions defined on compact sets. The behavior of "escaping to infinity" is not possible if we consider compact sets. Indeed, in any dimension we encounter an issue by using a function like $\frac{1}{\lvert x\rvert}$ defined on $$\Omega=(B_1(0)\setminus \{0\})\subseteq \Bbb{R}^n.$$ Then the integral $$ \int_\Omega\frac{1}{\lvert x\rvert}dx^1\cdots dx^n$$ will not produce a reasonable result. If you really want to work with open sets, you can escape from this issue by introducing forms with compact support, but this will restrict your class of admissible forms.

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