Proposition 1

Let $(Y_\alpha)_{\alpha \in A}$ be an open covering of $X$ and suppose given, on each subspace $Y_\alpha$, a measure $\mu_\alpha$, in such a way that for every pair $(\alpha, \beta)$, the restrictions of $\mu_\alpha$ and $\mu_\beta$ to $Y_\alpha \cap Y_\beta$ are identical. Under these conditions, there exists one and only one measure $\mu$ on $X$ whose restriction to $Y_\alpha$ is equal to $\mu_\alpha$ for every index $\alpha$.

Let us first show that every function $f \in \mathscr{K}(X; \mathbf{C})$ may be written in the form of a finite sum $f = \sum_i f_i$ where, for each of the functions $f_i \in \mathscr{K}(X; \mathbf{C})$, there exists an index $\alpha_i$ such that $\operatorname{Supp}(f_i) \subset Y_{\alpha_i}$. If $K = \operatorname{Supp}(f)$, there exists a finite number of indices $\alpha_i$ ($1 \leq i \leq n$) such that the $Y_{\alpha_i}$ form a covering of $K$; let $h_i$ ($1 \leq i \leq n$) be continuous mappings of $X$ into $[0, 1]$ such that the support of $h_i$ is compact and is contained in $Y_{\alpha_i}$ for $1 \leq i \leq n$, and such that $\sum_{i=1}^n h_i(x) = 1$ on $K$ (§ 1, No. 2, Lemma 1); the functions $f_i = f h_i$ meet the requirements. This shows in the first place that if there exists a measure $\mu$ meeting the requirements then it is unique, because for every finite sum $f = \sum_{i=1}^n f_i$, where $f_i \in \mathscr{K}(Y_{\alpha_i}; \mathbf{C})$, necessarily $\mu(f) = \sum_{i=1}^n \mu_{\alpha_i}(f_i)$. Moreover, we will have shown the existence of a linear form $\mu$ on $\mathscr{K}(X; \mathbf{C})$ whose restriction to each subspace $\mathscr{K}(Y_\alpha; \mathbf{C})$ is $\mu_\alpha$, provided we demonstrate the following property: if $(g_i)_{1 \leq i \leq m}$ and $(h_j)_{1 \leq j \leq n}$ are two finite sequences of functions in $\mathscr{K}(X; \mathbf{C})$ such that $g_i \in \mathscr{K}(Y_{\alpha_i}; \mathbf{C})$ for $1 \leq i \leq m$, $h_j \in \mathscr{K}(Y_{\beta_j}; \mathbf{C})$ for $1 \leq j \leq n$ and

on $K$, then Now,

whence

Similarly, But since the support of $f g_i h_j$ is contained in $Y_{\alpha_i} \cap Y_{\beta_j}$, we have $\mu_{\alpha_i}(f g_i h_j) = \mu_{\beta_j}(f g_i h_j)$, which establishes our assertion.

It remains to see that $\mu$ is a measure on $X$; now, every point of $X$ admits a compact neighborhood contained in one of the $Y_\alpha$; the conclusion therefore follows at once from the definition of $\mu$ and from Prop. 6 of §1, No. 3.