2.6 Weak convergence of measuresA1AF2491CD1A

Proof. For a nonnegative test function $\chi $, the numbers $\langle T_ i,\chi \rangle $ are nonnegative, hence so is their limit $\langle T,\chi \rangle $; likewise $\langle T,\chi \rangle $ is real for real $\chi $. Thus $T$ is positive, and $T$ and the $T_ i$ are given by Radon measures $\mu $ and $\mu _ i$. Let $f\in C_ c(\Omega )$. By lemma 2.5.1(5), $f$ is the uniform limit of test functions supported in a fixed compact set $L$; in particular, $f$ vanishes outside $L$. If $\eta $ is a smooth cutoff function for $L$ in $\Omega $, then $\mu _ i(L)\leq \langle T_ i,\eta \rangle \to \langle T,\eta \rangle $. A convergent net of real numbers need not be bounded, but it is bounded from some index on; hence there is $i_0$ with $\mu _ i(L)\leq C=\langle T,\eta \rangle +1$ for $i\geq i_0$, and also $\mu (L)\leq \langle T,\eta \rangle <C$. For every test function $\chi $ supported in $L$ and every $i\geq i_0$,

so $\varlimsup _ i|\int f\, \mathrm d\mu _ i-\int f\, \mathrm d\mu |\leq 2C\| f-\chi \| _\infty $, which can be made arbitrarily small. Hence $\mu _ i\to \mu $ vaguely. The converse holds because test functions belong to $C_ c(\Omega )$. $\square$

Proof. If $f\in C_ c(\Omega )$, $0\leq f\leq 1$ and $\operatorname{Supp}f\subset U$, then $\int f\, \mathrm d\mu =\lim _ i\int f\, \mathrm d\mu _ i\leq \varliminf _ i\mu _ i(U)$; taking the supremum over such $f$ and using (2.3.4) gives the first inequality. For the second, let $\varepsilon >0$, choose by outer regularity an open set $V\supset K$ with $\mu (V)<\mu (K)+\varepsilon $, and let $\eta $ be a cutoff function for $K$ in $V$. Then $\mu _ i(K)\leq \int \eta \, \mathrm d\mu _ i\to \int \eta \, \mathrm d\mu \leq \mu (V)$, so $\varlimsup _ i\mu _ i(K)\leq \mu (K)+\varepsilon $. Finally, for $E$ as in the last assertion, the closure $\overline E$ is a compact subset of $\Omega $, and $\overline E\setminus \operatorname{Int}E=\partial E$ is $\mu $-null, so $\mu (\operatorname{Int}E)=\mu (E)=\mu (\overline E)$. Since $\mu _ i(\operatorname{Int}E)\leq \mu _ i(E)\leq \mu _ i(\overline E)$, the two inequalities, applied to $\operatorname{Int}E$ and $\overline E$, give

Proof. Let $\varepsilon >0$, and choose $K$ and $i_0$ as above. Since $\Omega \setminus K$ is open, proposition 2.6.2 gives $\mu (\Omega \setminus K)\leq \varepsilon $, and since $\mu (K)<\infty $, the measure $\mu $ is finite. Let $f$ be a bounded continuous function on $\Omega $ and $\eta $ a cutoff function for $K$ in $\Omega $. Then $f\eta \in C_ c(\Omega )$, so $\int f\eta \, \mathrm d\mu _ i\to \int f\eta \, \mathrm d\mu $, while $1-\eta $ takes values in $[0,1]$ and vanishes on $K$, so that for $i\geq i_0$

Hence $\varlimsup _ i|\int f\, \mathrm d\mu _ i-\int f\, \mathrm d\mu |\leq 2\| f\| _\infty \varepsilon $. Letting $\varepsilon \downarrow 0$ proves narrow convergence, and $f=1$ gives the last assertion. $\square$

Proof. Step 1: a countable dense family. Let $\varphi (t)=\max \{ 0,1-|t|\} $, so that $\sum _{n\in \mathbb Z}\varphi (t-n)=1$ for $t\in \mathbb R$. For $k\geq 1$ and $n\in \mathbb Z^ d$ put $\varphi _{k,n}(x)=\prod _{l=1}^ d\varphi (2^ kx_ l-n_ l)$. These are nonnegative continuous functions with $\sum _ n\varphi _{k,n}=1$, and $\varphi _{k,n}(x)\neq 0$ only if $|x-2^{-k}n|_\infty <2^{-k}$, where $|\cdot |_\infty $ is the maximum norm. Let $\mathcal F$ be the countable family of all finite sums $\sum _ nq_ n\varphi _{k,n}$, $k\geq 1$, with rational coefficients $q_ n$, whose support is a compact subset of $\Omega $. Given $f\in C_ c(\Omega ,\mathbb R)$, extended by zero to $\mathbb R^ d$, choose rational numbers $q_{k,n}$ with $|q_{k,n}-f(2^{-k}n)|\leq 2^{-k}$ and $q_{k,n}=0$ if $f(2^{-k}n)=0$, and put $f_ k=\sum _ nq_{k,n}\varphi _{k,n}$. Then

which tends to zero uniformly in $x$ because $f$ is uniformly continuous. Moreover, $f_ k$ vanishes at points of $|\cdot |_\infty $-distance more than $2^{-k}$ from $\operatorname{Supp}f$, so for large $k$ all the $f_ k$ belong to $\mathcal F$ and are supported in a fixed compact set $L\subset \Omega $.

Step 2: diagonal extraction. Enumerate $\mathcal F=\{ g_1,g_2,\ldots \} $. For each $l$, the numbers $\int g_ l\, \mathrm d\mu _ j$ are bounded by $\| g_ l\| _\infty \sup _ j\mu _ j(\operatorname{Supp}g_ l)$. Choosing successively convergent subsequences for $g_1,g_2,\ldots $ and passing to the diagonal subsequence, we find $(\mu _{j_ k})$ such that $\int g_ l\, \mathrm d\mu _{j_ k}$ converges for every $l$. If $f\in C_ c(\Omega ,\mathbb R)$ and $L$, $f_ m$ are as in Step 1, then with $C_ L=\sup _ j\mu _ j(L)$,

so $(\int f\, \mathrm d\mu _{j_ k})_ k$ is a Cauchy sequence. Its limit is a positive linear functional of $f$, hence integration against a Radon measure $\mu $ by theorem 2.3.3, and $\mu _{j_ k}\to \mu $ vaguely; complex $f$ are handled by linearity. If the sequence is eventually uniformly tight, so is every subsequence, and proposition 2.6.3 gives the last assertion. $\square$