Proposition 5.2.6DAE61F3D77A3

Proof. Since the connected components of $X$ are open, it suffices to construct partitions of unity on each component; the union of the families so obtained is locally finite on $X$. So let $X$ be connected. By the preceding discussion, $X$ is covered by a sequence of coordinate balls $B_1,B_2,\ldots $ with compact closures. Put $K_1=\overline{B_1}$, and, once the compact set $K_ j$ is defined, cover it by finitely many of the balls $B_ i$ and let $K_{j+1}$ be the union of their closures and of $\overline{B_1},\ldots ,\overline{B_{j+1}}$. We obtain compact sets $K_1\subset \operatorname{Int}K_2\subset K_2\subset \operatorname{Int}K_3\subset \cdots $ with union $X$; put $K_ j=\varnothing $ for $j\leq 0$.

Now let an open cover of $X$ be given. For $j\geq 1$, the compact set $K_ j\setminus \operatorname{Int}K_{j-1}$ lies in the open set $\operatorname{Int}K_{j+1}\setminus K_{j-2}$. Cover it by finitely many coordinate balls $B'$ whose concentric enlargements $B$ have compact closures contained in a member of the given cover and in $\operatorname{Int}K_{j+1}\setminus K_{j-2}$. For each such pair choose, in the chart, a smooth function with values in $[0,1]$ and compact support in $B$ that equals one on $B'$ (lemma 2.5.1), and extend it by zero to $X$. The resulting family is locally finite, since a point of $\operatorname{Int}K_ i$ has the neighborhood $\operatorname{Int}K_ i$, which meets the supports chosen at stage $j$ only if $j\leq i+1$. Its sum is therefore smooth, and it is positive, because the balls $B'$ cover $X$. Dividing each function by this sum gives the required partition of unity.

For the cutoff function, cover $K$ by finitely many relatively compact coordinate balls with closures in $U$, and let $V$ be their union; thus $K\subset V$ and $\overline V$ is a compact subset of $U$. Let $(\chi _ a)$ be a partition of unity with compact supports subordinate to the cover $\{ V,X\setminus K\} $, and let $\chi $ be the sum of those $\chi _ a$ whose support is not contained in $X\setminus K$. Each of these has its support in $V$; by local finiteness, only finitely many supports meet the compact set $\overline V$, so $\chi $ is a finite sum, smooth, with values in $[0,1]$ and compact support in $\overline V\subset U$. The supports of the remaining $\chi _ a$ lie in $X\setminus K$, and their union $F$ is closed, as a locally finite union of closed sets. On the neighborhood $X\setminus F$ of $K$ we have $\chi =\sum _ a\chi _ a=1$.

For the last assertion, consider first a compactly supported $(2n-1)$-form $\gamma $ in a single chart. Then $\mathrm{d}\gamma $ is a sum of terms $\pm \partial _{t_ l}g\, \mathrm dt_1\wedge \cdots \wedge \mathrm dt_{2n}$, where $t_1,\ldots ,t_{2n}$ are the real coordinates $x_1,y_1,\ldots ,x_ n,y_ n$ and $g$ is smooth with compact support; the integral of each term vanishes by Fubini's theorem and the fundamental theorem of calculus on lines parallel to the $t_ l$-axis. Now let $(\chi _ a)$ be a partition of unity subordinate to a cover by charts. Only finitely many $\chi _ a$ are nonzero on the compact set $\operatorname{Supp}\alpha $, so $\alpha =\sum _ a\chi _ a\alpha $ is a finite sum of compactly supported forms in single charts, and $\int _ X\mathrm d\alpha =\sum _ a\int _ X\mathrm d(\chi _ a\alpha )=0$. $\square$