Lemma 7.11.27.A2369B2DB317 Let $K_0\subset B$ be compact, and let $\nu _ j,\nu $ be finite positive Borel measures on $B$, supported in $K_0$, with $\nu _ j\to \nu $ weakly. Assume that there is a nondecreasing function $\omega \colon [0,+\infty ]\to [0,+\infty ]$ with $\omega (t)\to 0$ as $t\downarrow 0$ such that
\[ \nu _ j(E)\leq \omega (\operatorname {Cap}_ B^*(E)) \]
for all $j$ and every Borel set $E\subset B$. Let $f_ j,f$ be real Borel functions on $B$ bounded by a common constant, with $f$ quasi-continuous on a neighborhood of $K_0$, and suppose that $\operatorname {Cap}_ B^*(K_0\cap \{ |f_ j-f|>\delta \} )\to 0$ for every $\delta >0$. Then
\[ \int _ Bf_ j\, \mathrm d\nu _ j\longrightarrow \int _ Bf\, \mathrm d\nu . \]
Moreover, for every $c\in \mathbb R$ with $\nu (\{ f=c\} )=0$,
5F846CC6D336
7.11.28
\begin{equation} \nu _ j(\{ f_ j<c\} )\longrightarrow \nu (\{ f<c\} ). \end{equation}
Proof.
Let $C$ be a common bound for $|f_ j|$ and $|f|$. Since $\omega $ is nonnegative and nondecreasing with $\omega (t)\to 0$ as $t\downarrow 0$, we have $\omega (0)=0$ and $\omega (t_ j)\to 0$ whenever $t_ j\geq 0$ and $t_ j\to 0$. Choose $\chi \in C_ c(B)$ with $0\leq \chi \leq 1$ and $\chi =1$ on $K_0$. Since the measures are carried by $K_0$, we have $\nu _ j(B)=\int \chi \, \mathrm d\nu _ j\to \int \chi \, \mathrm d\nu =\nu (B)$, so the masses of the $\nu _ j$ and of $\nu $ have a common bound $M$. The capacity bound passes to $\nu $ on open sets: if $G\subset B$ is open, then $\operatorname {Cap}^*_ B(G)=\operatorname {Cap}_ B(G)$, and proposition 2.6.2 gives $\nu (G)\leq \varliminf _ j\nu _ j(G)\leq \omega (\operatorname {Cap}_ B(G))$.
We first consider the fixed weight $f$. Let $L\subset B$ be a compact neighborhood of $K_0$ on which $f$ is quasi-continuous, and let $\kappa >0$. Choose an open set $G\subset B$ with $\operatorname {Cap}_ B(G)<\kappa $ such that $f$ is continuous on the compact set $L\setminus G$. By lemma 7.5.3, this restriction extends to a continuous function on $\mathbb C^ n$ bounded by $C$. Multiplying the extension by a cutoff in $C_ c(\operatorname{Int}L)$ with values in $[0,1]$ and equal to one on $K_0$, we obtain $g\in C_ c(B)$ with $|g|\leq C$ and $g=f$ on $K_0\setminus G$. Since $\nu _ j$ and $\nu $ are carried by $K_0$,
\[ \left|\int _ Bf\, \mathrm d\nu _ j-\int _ Bg\, \mathrm d\nu _ j\right|\leq 2C\nu _ j(G)\leq 2C\omega (\kappa ), \]
and similarly for $\nu $, since $\nu (G)\leq \omega (\operatorname {Cap}_ B(G))\leq \omega (\kappa )$. Since $\int g\, \mathrm d\nu _ j\to \int g\, \mathrm d\nu $, we get $\varlimsup _ j|\int f\, \mathrm d\nu _ j-\int f\, \mathrm d\nu |\leq 4C\omega (\kappa )$. Letting $\kappa \downarrow 0$ proves the assertion for the fixed weight.
For the varying weights, let $\delta >0$. Since $|f_ j-f|\leq 2C$ and $\nu _ j$ is carried by $K_0$,
\[ \int _ B|f_ j-f|\, \mathrm d\nu _ j \leq \delta M+2C\omega \left( \operatorname {Cap}_ B^*(K_0\cap \{ |f_ j-f|>\delta \} )\right). \]
As $j\to \infty $, the capacity on the right tends to zero, and hence so does the second term. Letting $\delta \downarrow 0$ gives $\int |f_ j-f|\, \mathrm d\nu _ j\to 0$, which together with the fixed-weight case proves the first assertion.
For the level sets, fix $s>0$ and let $\sigma ^-_ s,\sigma ^+_ s\colon \mathbb R\to [0,1]$ be the continuous piecewise linear functions such that $\sigma ^-_ s$ equals one on $(-\infty ,c-s]$, zero on $[c,\infty )$ and is linear in between, and $\sigma ^+_ s$ equals one on $(-\infty ,c]$, zero on $[c+s,\infty )$ and is linear in between. Then
\[ \mathbf1_{(-\infty ,c-s)}\leq \sigma ^-_ s \leq \mathbf1_{(-\infty ,c)}\leq \sigma ^+_ s \leq \mathbf1_{(-\infty ,c+s]}. \]
Since $\sigma ^\pm _ s$ are Lipschitz, $\sigma ^\pm _ s\circ f_ j$ converge to $\sigma ^\pm _ s\circ f$ in the same sense as $f_ j\to f$, and $\sigma ^\pm _ s\circ f$ is quasi-continuous on a neighborhood of $K_0$ (remark following definition 7.5.1). The first assertion therefore gives
\[ \nu (\{ f<c-s\} )\leq \varliminf _{j\to \infty }\nu _ j(\{ f_ j<c\} ) \leq \varlimsup _{j\to \infty }\nu _ j(\{ f_ j<c\} )\leq \nu (\{ f\leq c+s\} ). \]
As $s\downarrow 0$, the two outer terms tend to $\nu (\{ f<c\} )$ and $\nu (\{ f\leq c\} )$, which are equal since $\nu (\{ f=c\} )=0$. This proves (7.11.28).
$\square$
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