Lemma 17.2.25F05F95AAF1F

Proof. We write the difference of the two integrals as

and show that both terms tend to zero. For the first term fix $\varepsilon >0$ and put $A_ j=\{ |f_ j-f|>\varepsilon \} $. Splitting the integral over $A_ j$ and its complement gives

Since $\operatorname {Cap}_\omega (A_ j)\to 0$, uniform absolute continuity gives $\nu _ j(A_ j)\to 0$. Thus the upper limit of the first term is at most $\varepsilon \sup _ k\nu _ k(X)$, and $\varepsilon $ is arbitrary.

For the second term let $\eta >0$, and choose an open set $G$ with $\operatorname {Cap}_\omega (G)<\eta $ such that the restriction of $f$ to the compact set $X\setminus G$ is continuous. By lemma 5.2.7, there is a metric $d$ on $X$ which induces its topology, and by lemma 7.5.3, applied in the metric space $(X,d)$, the restriction of $f$ to $X\setminus G$ extends to a continuous function $g$ on $X$ with $|g|\leq B$. Since $X$ is compact, $g$ has compact support, so weak convergence of measures (see the discussion following definition 5.3.5) gives $\int _ Xg\, \nu _ j\to \int _ Xg\, \nu $. Since $f=g$ outside $G$, we have $|\int _ Xf\, \lambda -\int _ Xg\, \lambda |\leq 2B\lambda (G)$ for each $\lambda \in \{ \nu ,\nu _1,\nu _2,\ldots \} $. The upper limit of the second term is therefore at most $4B$ times the supremum of $\lambda (G)$ over the family, which tends to zero as $\eta \downarrow 0$ by uniform absolute continuity. $\square$