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Lemma 17.5.3B408410BD203

Lemma 17.5.3.B408410BD203 Let $\lambda $ and $\sigma $ be finite positive Borel measures on $X$.

  1. If $\int _ Xg\, \mathrm d\lambda \geq \int _ Xg\, \mathrm d\sigma $ for every continuous function $g\geq 0$ on $X$, then $\lambda (E)\geq \sigma (E)$ for every Borel set $E\subset X$. In particular, two finite positive Borel measures on $X$ with the same integrals of all continuous functions coincide.

  2. If finite positive Borel measures $\lambda _ j$ converge weakly to $\lambda $ and satisfy $\lambda _ j\geq \sigma $ for every $j$, then $\lambda \geq \sigma $.

Proof. (1) This is lemma 16.12.1(2) with $W=X$; the second assertion follows by applying the first in both directions.

(2) Since $X$ is compact, weak convergence of measures on $X$ is convergence of the integrals of all continuous functions (see the discussion following definition 5.3.5). Hence $\int _ Xg\, \mathrm d\lambda =\lim _ j\int _ Xg\, \mathrm d\lambda _ j\geq \int _ Xg\, \mathrm d\sigma $ for every continuous function $g\geq 0$ on $X$, and (1) gives $\lambda \geq \sigma $. $\square$

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