PomeTheorem 2.6.5.F9F4241C5BD2 Let $(\mu _ j)$ be a sequence of Radon measures on $\Omega $ such that $\sup _ j\mu _ j(K)<\infty $ for every compact set $K\subset \Omega $. Then $(\mu _ j)$ has a vaguely convergent subsequence. If, moreover, the $\mu _ j$ are finite and the sequence is eventually uniformly tight, then it has a narrowly convergent subsequence with a finite limit.
Theorem 2.6.5F9F4241C5BD2
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
\[ |f_ k(x)-f(x)|\leq \sum _ n|q_{k,n}-f(x)|\varphi _{k,n}(x)\leq 2^{-k}+\sup \{ |f(y)-f(x)|:|y-x|_\infty <2^{-k}\} , \]
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 $C_ L=\sup _ j\mu _ j(L)$,
\[ \left|\int f\, \mathrm d\mu _{j_ k}-\int f\, \mathrm d\mu _{j_{k'}}\right|\leq 2C_ L\| f-f_ m\| _\infty +\left|\int f_ m\, \mathrm d\mu _{j_ k}-\int f_ m\, \mathrm d\mu _{j_{k'}}\right|, \]
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. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.