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!



How to evaluate the wiener measure of sets?
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.


