How to evaluate the wiener measure of sets?

Question

I would like to understand how the Wiener measure of some simple sets can be evaluated. I will sketch the construction of Wiener measure I have in mind:

We denote the one point compactification of $\mathbb{R}^n$ by $\hat{\mathbb{R}}^n$. Now consider the product space $$\Omega:=\prod\limits_{0\leq t<\infty}\hat{\mathbb{R}}^n.$$ By Tychonoff's theorem this space is a compact and hausdorff space. Now the idea is to construct the wiener measure with the aid of riesz representation theorem. Consider first functions like this $\varphi:\Omega\rightarrow \mathbb{R}, \varphi(w)=F(w(t_1),...,w(t_n))$ with $n\in\mathbb{N}$, $t_1<t_2<...<t_n$ and some continuous F. Now define the following linear functional on all those functions given by:

$$\Lambda(\varphi)=\int....\int p(t_1,x,x_1)p(t_2-t_1,x_1,x_2)...\times p(t_n-t_{n-1},x_{n-1},x_n) F(x_1,...,x_n)dx_1....dx_n,$$

where $p(t,x,y)$ denotes the heat kernel on $\mathbb{R}^n$.

The space of all such $\varphi$ is dense in $C(\Omega)$ and therefore we can extend $\Lambda$ continuously on the whole space $C(\Omega)$. By Riesz representation theorem there exists a regular Borel measure $\mu$, s.t. for all continuous functions $f$ we have:

$$\Lambda(f)=\int f d\mu$$.

This measure is called Wiener measure.

Now I want to understand how to calculate measures like $\mu(G)$, whereas $G:= \lbrace{ w:[0,\infty)\rightarrow \mathbb{R}^n \vert \omega(t)\in U\rbrace}$ with fixed $U\subset \mathbb{R}^n$ open. The problem is that the characteristic function $1_{\lbrace{ w(t)\in G , 0\leq t<\infty\rbrace}}$ is not continuous, therefore I can't calculate it like $\Lambda(1_{{\lbrace{ w(t)\in G , 0\leq t<\infty\rbrace}}})$

I would really appreciate any help!

Answer 1

First let $t$ be fixed and look at the set $\{ w \mid w(t) \in U\}$. Take a sequence of continuous functions $f_k$ that are uniformly bounded and converge to the indicator function of a set $U$ in the Borel-Algebra of $\mathbb{R}^n$. Let $\pi_t$ be the projection on the $t$-th component of Omega. Then $$\int_\Omega (\pi_t)^*f_k d \mu = \int_{\mathbb{R}^n} p(t, x, y) f_k(y) d y$$ by construction. Because the $f_k$ are bounded and the measure is finite, they are dominated by an integrable function and by Lebesgue's theorem on bounded convergence, $$\mu(\{w(t) \in U\}) = \lim_{k \rightarrow \infty}\int_\Omega (\pi_t)^*f_k d \mu = \int_{\mathbb{R}^n} p(t, x, y) \lim_{k \rightarrow \infty} f_k(y) d y = \int_{U} p(t, x, y) d y $$ Hence non-continuity is no real problem.

To now tackle your problem, notice that the indicator functions of the cylinder sets $$ G_{N} := \{ w \mid w(k/N) \in U \forall k = 1, \dots, N^2 \}$$ converges a.s. to the indicator function of $G := \{ w \mid w(t) \in U \forall t \geq 0 \}$ as $N \longrightarrow \infty$: Clearly $G \subset G_N$. On the other hand, if $w \notin A$, then $w(q) \notin U$ for some rational number $q$ and therefore $w \notin G_N$ for $N$ big enough.

The dominated convergence theorem gives then $$ \mu(G) = \lim_{N \rightarrow \infty} \int_U \dots \int_U \left(\prod_{k=1}^{N^2} p(1/N, x_{k-1}, x_k) \right) d x_1 \dots d x_{N^2}$$ If this is helpful at all, I cannot say.

Answer 2

The probability $h(x,t)$ that the Brownian motion path started from $x$ stays in a given set $U$ for $0<s<t$ satisfies the heat equation: $\Delta_x h-\partial_t h=0$ with initial data $1$ and boundary data zero. The large $t$ asymptotics is governed by the leading eigenvalue of the Laplacian and by projection of the constant function $1$ to the line spanned by the leading eigenfunction.