Weak convergence of probability mesures on metric spaces §

We discuss the weak convergence of probability mesures on metric spaces $(S,\mathcal{S})$ equipped with the borel $\sigma -$algbra from the metric induced topology. Since the objective is to prove the exintence of the Weiner mesure, you should keep in mind $S=C[0,1]$ with the supremum norm, or $S={\mathbb{R}}^d$.

Def (weak convergence) §

For $({\mu }_n{)}_{n\ge 1}$ and $\mu$ probability mesures on the mesurable space $(S,\mathcal{S})$, we say ${\mu }_n$ converges weakly to $\mu$ , ${\mu }_n\Rightarrow \mu$ if

$${\mu }_n(f)\to \mu (f)n\to \infty \forall f\in C_b(S,\mathbb{R})$$

ovvero l’integrale su $S$ rispetto alla successione delle misure converge nei reali all’integrale rispetto alla misura $\mu$, per ogni funzione continua e limitata.

Def (convergence in distribution) §

Let $(X^n{)}_{n\ge 1}$, and $X$ random variables taking values in $S$, with distribution $X^n\sim {\mu }_n$ and $X\sim \mu$ respectivley. We say that $X^n$ converges in distribution to $X$, $X^n\Rightarrow X$, if

$$\mathbb{E}[f(X^n)]\to \mathbb{E}[f(X)]n\to \infty \forall f\in C_b(S,\mathbb{R})$$

$X^n\Rightarrow X$ se e solo se ${\mu }_n\Rightarrow \mu$.

Remark §

It would be tempting to conclude that id ${\mu }_n\Rightarrow \mu$, then ${\mu }_n(A)\to \mu (A)$ for all $A$ Borel. This is in general false:

$${\mu }_n:={\delta }_{\frac{1}{n}}\mu ={\delta }_0$$

it is clear that ${\mu }_n\Rightarrow \mu$. But any set $A$ such that the origin on the boundary like $A=(0,1]$ has always mesure ${\mu }_n(A)=1\forall n$, but $\mu (A)=0$.

The following theorem characterizes weak convergence.

Theorem (Portmanteau) §

Let $({\mu }_n{)}_{n\ge 1}$, $\mu$ be probability mesures on $(S,\mathcal{S})$. The following are equivalent:

1. ${\mu }_n\Rightarrow \mu$ as $n\to \infty$,
2. ${\mu }_n(f)\to \mu (f)$ for all $f:S\mapsto \mathbb{R}$ bounded uniformly continuous.
3. ${\mathrm{limsup}}_n{\mu }_n(C)\le \mu (C)$ for all $C\in \mathcal{S}$ chiusi.
4. ${\mathrm{liminf}}_n{\mu }_n(A)\ge \mu (A)$ for all $A\in \mathcal{S}$ aperti.
5. ${\lim }_n{\mu }_n(A)=\mu (A)$ for all $A\in \mathcal{S}$ such that $\mu (\partial A)=0$.

Proof $(1⟹2)$ obvious since uniform continuity implies continuity. $(2⟹3)$ We can’t use the characteristic function of $C$, since it’s discountinuos. We can approximate with a uniform continuous function with a parameter $ϵ$. Then using 2:

$$1_C\le f_{ϵ}\le 1_{C_{ϵ}}$$

which shows that as $ϵ\to 0$ $\mu (f_{ϵ})\to \mu (C)$.

$$\underset{n}{\mathrm{limsup}}{\mu }_n(C)\le \underset{n}{\mathrm{limsup}}{\mu }_n(f_{ϵ})=\mu (f_{ϵ})\le \mu (C_{ϵ})$$

$(3⟺4)$ It follows from taking complements. $(4⟺5)$ We start from the mesure of the border:

$$\mu (\partial A)=\mu (\overline{A})-\mu (\mathring{A})\ge \underset{n}{\mathrm{limsup}}{\mu }_n(A)-\underset{n}{\mathrm{liminf}}{\mu }_n(A)$$

if $\mu (\partial A)=0$, $\mathrm{liminf}$ and $\mathrm{limsup}$ are equal, i.e. the definition of ${\lim }_n{\mu }_n(A)=\mu (A)$.

Perchè c’è la doppia implicazione? quello che mi sembra sia dimostrato è $5⟹4$!!!!

$(5⟹1)$ W.l.o.g assume $f\ge 0$, since $f$ is bounded.

$$\mu (f)={\int }_{S}fd\mu ={\int }_S{\int }_0^{M}1(f(x)\ge t)dtd\mu (x)={\int }_0^M\mu (\{f(x)\ge t\})dt$$

where $M:={\sup }_S|f|$. The same holds for ${\mu }_n(f)$. To prove 1, we need to prove only that ${\mu }_n(\{f\ge t\})\to \mu (\{f\ge t\})$ for almost all $t\in [0,\infty )$.

We use 5: for the sets $A_t:=\{f\ge t\}$ we know that ${\mu }_n(A_t)\to \mu (A_t)$ provided that $\mu (\partial A_t)=0$, where $\partial A_t=\{f=t\}$.

We need only to prove that $\mu (\partial A_t)>0$ only for a mesure zero set of $t$:

$$\{t\ge 0:\mu (\partial A_t)>0\}=\bigcup _{n\ge 1}\left\{t\ge 0:\mu (\partial A_t)\ge \frac{1}{n}\right\}$$

it is clear that $A_t\cap A_s=\varnothing$ when $t\ne s$. Then the mesure of the union is simply the sum: since $\mu (S)=1$, each term has at most $n$ elements, proving that the set is denumerable, hence has mesure zero. We are almost done:

\vert \mu_n(f)-\mu(f)\vert \leq \int \vert \mu_n(A_t)-\mu(A_t)\vert dt \to 0$$ the exchange of limits is legit since we can apply [[Teorema convegenza dominata]] (probability mesures are bounded!). $\square$ **Proposition** If $(\mu_n)_{n\geq 1}$ are probability mesures on $(\mathbb{R}, \mathcal{B})$, to check weak convergence, we need only to check if:

\mu_n((-\infty, x]) \to \mu((-\infty, x]) \qquad \forall x \in \mathbb{R} ;:; \mu({x})=0

$$i.e.tocheckconvergenceina[[Pi-system]].Inthelanguageofprobability,thismeansthat:$$

X^n \Rightarrow X \iff F_{X^n}(x)\to F_X(x)

for all $x$ where $F_X$ is continuous.