[Notations] Sets of Continuous functions.
$C_K\,$: the set of continuous functions $f$ each vanishing outside a compact set $K(f)$.
$C_0\;\,$: the set of continuous functions $f$ such that $\lim_{|x|\rightarrow\infty}f(x)=0$.
$C_B\,$: the set of bounded continuous functions.
$C\;\;\,$: the set of continuous functions.
It is clearly that $f\in C_K\implies f\in C_0\implies f\in C_B\implies f\in C$.
[Theorem] Let $\{\mu_n\}_{n\geq1}$ and $\mu$ be a sequence of s.p.m.'s, then $$\mu_n\overset{v}{\longrightarrow}\mu\iff\forall\,f\in C_K\,(\mbox{or }C_0),\;\int f\,d\mu_n\rightarrow\int f\,d\mu.$$
$\bullet$ Proof.
$(\Longrightarrow)$ First we introduce the Approximation Lemma.
[Lemma] Approximation Lemma. Suppose that $f\in C_K$ has support in the compact interval $[a,b]$. Given any dense subset $A\subset\mathbb{R}$ and $\varepsilon>0$, there exists an $A$-valued step function $f_\varepsilon$ on $(a,b)$ such that $$\underset{x\in\mathbb{R}}{\sup}|f(x)-f_\varepsilon(x)|\leq\varepsilon.$$Note that $f_\varepsilon$ can be written as $f_\varepsilon(x)=\sum_{j}c_j\cdot I\{x\in(a_j,\,a_{j+1}]\}$ where $c_j$'s, $a_j$'s are real numbers in $A$.
Back to the proof. Suppose $\mu_n\overset{v}{\rightarrow}\mu\overset{def.}{\iff}\mu_n(a,b]\rightarrow\mu(a,b]$ for $a$, $b$ in a dense set $D\subset\mathbb{R}$. For those $a,\,b\in D$, define $f_I(x)=I\{x\in (a,b]\}$. Immediately we have for every such $D$-valued step function $f_I$, $$\int f_I(x)\,d\mu_n=\mu_n(a,b]\rightarrow\mu(a,b]=\int f_I(x)\,d\mu.$$Now, let $f\in C_K$. By Approximate Lemma, there exists a $D$-valued step function $f_\varepsilon$ such that $$\underset{x\in\mathbb{R}}{\sup}|f(x)-f_\varepsilon(x)|\leq\varepsilon.$$Then, since $f_\varepsilon$ is a $D$-valued step function, by the vague convergence, we have $$\exists\,n_0(\varepsilon)\mbox{ such that for }n\geq n_0(\varepsilon),\;\left|\int f_\varepsilon(x)\,d\mu_n-\int f_\varepsilon(x)\,d\mu\right|\leq\varepsilon.$$Thus, for $f\in C_K$, $$\begin{array}{rl}& \left|\int f(x)\,d\mu_n-\int f(x)\,d\mu\right| \\
= & \left|\int f(x)\,d\mu_n-\int f_\varepsilon(x)\,d\mu_n-\int f_\varepsilon(x)\,d\mu+\int f_\varepsilon(x)\,d\mu_n+\int f_\varepsilon(x)\,d\mu-\int f(x)\,d\mu\right| \\
\leq & \int\left| f(x)-\int f_\varepsilon(x)\right|\,d\mu_n+\left| \int f_\varepsilon(x)\,d\mu_n-\int f_\varepsilon(x)\,d\mu\right|+\int\left| f_\varepsilon(x)-f(x)\right|\,d\mu \\
\leq & \varepsilon+\varepsilon+\varepsilon=3\varepsilon, \end{array}$$where the first and third parts are due to the Approximate Lemma and the second part is due to the vague convergence of $D$-valued step functions. That is, we have $$\mu_n\overset{v}{\rightarrow}\mu\implies\forall\,f\in C_K,\;\int f(x)\,d\mu_n\rightarrow\int f(x)\,d\mu.$$
$(\Longleftarrow)$ Define $g=I(a,b]$. Given $\varepsilon>0$ and find $\delta$ such that $$a+\delta<b-\delta\mbox{ and }\mu[a-\delta,\,a+\delta]+\mu[b-\delta,\,b+\delta]<\varepsilon.$$Then we define two functions:$$\begin{array}{rl}f_1(x)&=\begin{cases}1,& x\in(a+\delta,\,b-\delta);\\ 0,&x\not\in(a,\,b);\\ \mbox{linear increase or decrease},&\mbox{o.w.},\end{cases} \\ & \\ f_2(x)&=\begin{cases}1,& x\in[a,\,b];\\ 0,&x<a-\delta\mbox{ or }x>b+\delta;\\ \mbox{linear increase or decrease},&\mbox{o.w.}\end{cases}\end{array}$$as shown in the figure.
Then we have $f_1$ and $f_2$ are both belongs to $C_K$, and $$f_1\leq g\leq f_2\leq f_1+I[a-\delta,\,a+\delta]+I[b-\delta,\,b+\delta].$$The first three parts of the inequality can be summarized as $$\begin{array}{ccccc}
\int f_1\,d\mu_n&\leq&\int g\,d\mu_n&\leq&\int f_2\,d\mu_n \\
\because\,f_1\in C_K & & & & \because\,f_2\in C_K\\
\downarrow& & & & \downarrow\\
\int f_1\,d\mu&\leq&\int g\,d\mu&\leq&\int f_2\,d\mu \end{array}$$The last part of the inequality implies $$\begin{array}{rl}&\int f_2\,d\mu\leq\int f_1+I[a-\delta,\,a+\delta]+I[b-\delta,\,b+\delta] \,d\mu \\ \implies & \int f_2\,d\mu-\int f_1\,d\mu\leq\mu[a-\delta,\,a+\delta]+\mu[b-\delta,\,b+\delta] <\varepsilon,\,\forall\,\varepsilon>0. \end{array}$$Thus, we have $$\forall\,f\in C_K\,\;\int f\,d\mu_n\rightarrow\int f\,d\mu\implies\int g\,d\mu_n\rightarrow\int g\,d\mu\iff\mu_n\overset{v}{\longrightarrow}\mu.$$
$\Box$
沒有留言:
張貼留言