For any smooth manifold $M$, does there exist some coordinate chart $(U,(x^i))$ such that $M\setminus U$ has measure zero?

Question

Assume that $M$ is connected. Ideally, I would like $U$ to be connected, though that's not strictly necessary.

If this were true, it would instantly make integration on compact oriented manifolds much easier, since we can simply pass to Euclidean space $\mathbb{R}^n$ in order to evaluate the integral. Intuitively, it seems like it must be true. Consider the following examples:

• $\mathbb{S}^n$ – we use the stereographic projection
• $\mathbb{T}^n \cong (\mathbb{R}/\mathbb{Z})^n$ – we consider $U=(0,1)^n$
• $\mathbb{RP}^n$ – we consider $U_0=\{[1:x^1:\cdots:x^n]\, \big\vert\, x^1,\ldots,x^n\in\mathbb{R}^n\}$
• If $M$ and $N$ have smooth charts $(U,(x^i)$ and $(V,(y^j))$ defined almost everywhere, respectively, then $M\times N$ has smooth chart $(U\times V, (x^1,\ldots,x^n,y^1,\ldots,y^m))$ defined almost everywhere.

However, I have no clue about how to prove a statement like this, and I suspect that any counterexample would be very difficult to find.

Answer 1

A $n$-manifold is an n-CW complex. If you remove the n-1 cells, you obtain a disjoint union of $n$ cells which are open subsets of $R^n$.