Lemma 16.12.1EC77D6F30D3D

Proof. (1) Cover $X$ by finitely many coordinate charts $U_1,\ldots ,U_ N$, with homeomorphisms $\kappa _ i$ of $U_ i$ onto open sets $\Omega _ i\subset \mathbb R^{2n}$, and let $E\subset X$ be Borel and $\varepsilon >0$. The Borel sets $E_ i=(E\cap U_ i)\setminus (U_1\cup \cdots \cup U_{i-1})$ partition $E$. The image of the restriction of $\lambda $ to $U_ i$ under $\kappa _ i$ is a finite Borel measure on $\Omega _ i$, hence a Radon measure by proposition 2.2.2, so there is a compact set $L_ i\subset \kappa _ i(E_ i)$ whose complement in $\kappa _ i(E_ i)$ has measure less than $\varepsilon /N$. Then $K=\bigcup _ i\kappa _ i^{-1}(L_ i)$ is a compact subset of $E$ with $\lambda (E\setminus K)<\varepsilon $.

(2) Fix a metric $d$ on $X$ that induces its topology (lemma 5.2.7). Let $K\subset W$ be compact and nonempty; by compactness there is $r>0$ such that every point at distance at most $r$ from $K$ lies in $W$. For $k>1/r$ the continuous functions $\chi _ k=\max \{ 0,1-k\, d(\cdot ,K)\} $ take values in $[0,1]$, equal $1$ on $K$, have compact support in $W$, and decrease pointwise to $\mathbf1_ K$ as $k\to \infty $. By dominated convergence for the finite measures $\lambda $ and $\lambda '$,

If $E\subset W$ is Borel, then by (1), $\lambda (E)=\sup _ K\lambda (K)\leq \sup _ K\lambda '(K)\leq \lambda '(E)$, the suprema being taken over the compact sets $K\subset E$. $\square$