Lemma 2

— Let $\mu$ be a positive measure on X, and let K be a compact subset of X.

(i) For every compact (resp. open) subset H of K, $\mu_K(H) = \mu(H)$.

(ii) For a subset N of K to be $\mu_K$-negligible, it is necessary and sufficient that it be $\mu$-negligible.

(iii) If S is the support of $\mu_K$, then $\mathrm{Supp}(\mu_S) = S$.

(i) We can restrict ourselves to the case that H is compact. Denote by $f$ the characteristic function of H in the space K; $f$ is upper semi-continuous, hence is the lower envelope of a decreasing directed family $(g_\alpha)$ of functions in $\mathscr{K}_+(K)$; we have $\mu_K(H) = \inf_\alpha \int g_\alpha \, d\mu_K$ (§ 4, No. 4, Cor. 2 of Prop. 5). If $g'_\alpha$ is the function equal to $g_\alpha$ on K and to 0 on $X - K$, then $g'_\alpha$ is upper semi-continuous, and the lower envelope of the decreasing directed family $(g'_\alpha)$ is the characteristic function $\varphi_H$ of H in the space X; therefore

by (1).

(ii) If N is $\mu$-negligible then, for every $\varepsilon > 0$, there exists a relatively compact open neighborhood U of N in X such that $\mu(U) \leq \varepsilon$; since $K - (U \cap K)$ is compact, it follows from (i) that $\mu_K(U \cap K) \leq \mu(U) \leq \varepsilon$, thus N is $\mu_K$-negligible. Conversely, if N is $\mu_K$-negligible then there exists an open neighborhood V of N in X such that $\mu_K(V \cap K) \leq \varepsilon$; by (i), we have $\mu(V \cap K) = \mu_K(V \cap K)$, therefore $\mu(N) = 0$ since $\varepsilon$ is arbitrary.

(iii) For every open set U in K that intersects S, by hypothesis $\mu_K(U \cap S) \neq 0$, therefore $\mu(U \cap S) \neq 0$ by (i), and since $U \cap S$ is open in S, $\mu_S(U \cap S) \neq 0$ by (i); since every nonempty open set in S is of the form $U \cap S$, where U is open in K, this proves that $\operatorname{Supp}(\mu_S) = S$.