While reading Lee's Introduction to Smooth Manifolds' proof of proposition 16.8, I encountered the following general statement about manifolds
(Suppose $(U,\phi)$ is a chart from a $n-$manifold with, or without boundary)
In fact, by restricting to sufficiently nice charts, we may assume that $U$ is precompact, $Y=\phi(U)$ is a domain of integration in $\mathbb{R}^n$ or $\mathbb{H}^n$, and $\phi$ extends to a diffeomorfism from $\bar U$ to $\bar Y$.
Where a domain of integration is a bounded set whose topological boundary in $\mathbb{R}^n$ has measure zero.
I certainly understand the first assumption, since smooth manifolds are localy compact, however, I don't understand why the image of each of those domains should be a domain of integration, nor why the coordinate charts can be extended to the closure the way it is asserted.
Any help will be appreciated, thanks in advance.
Fix a point $p$ in the interior of your manifold and choose a chart $(U,\phi)$ containing $p$. Now choose an open ball $B$ containing $\phi(p)$ whose closure is contained in $\phi(U)$, and let $V=\phi^{-1}(B)$. Then $(V,\phi|_V)$ is a chart containing $p$ whose image $B$ is a domain of integration, and this chart extends to a diffeomorphism $\phi^{-1}(\overline{B})\to\overline{B}$ between the closures which are compact.
For boundary points you can do the same thing, but with a half-ball instead of a ball.