Probability kernel Brownian motion

Question

I have a question about the construction of Brownian Motions. In our lecture notes it is stated that the map $$\left(x,A\right) \mapsto \mu_x(A)$$ is a measurable probability kernel. Here $\mu_x$ is the Wiener measure for a Brownian Motion starting at $x$, i.e. $ \mu_x=\mathbb{P}\circ \mathbb{B}_{x}^{-1}$. How can I see measurability?

Answer 1

You can start by showing that $x\mapsto \mu_x(A)$ is continuous for sufficiently many $A$s, and then apply the monotone class theorem.

In more detail, the class $\mathcal A$ of events of the form $$ A=\cap_{k=1}^n\{B_{t_k}\in C_k\}, $$ for $n$ a positive integer, $0<t_1<t_2<\cdots<t_n$, and $C_1, C_2,\ldots,C_n$ Borel subsets of the real line, forms a $\pi$-system that generates $\mathcal F=\sigma(B_t: t\ge 0)$. You can show that $x\mapsto \mu_x(A)$ is continuous (hence Borel measurable) for each $A\in\mathcal A$. At this point you can deduce from the monotone class theorem that $x\mapsto\mu_x(A)$ is Borel measurable for all $A\in\mathcal F$.

To get you started on the continuity assertion, consider $A=\{B_t\in C\}$ for a single $t>0$ and Borel set $C$. Then $$ \mu_x(A)=\int_{\Bbb R}p(t,x,y)1_C(y)\,dy, $$ where $p(t,x,y)=(2\pi t)^{-1/2}\exp(-(x-y)^2/2t)$. Because $x\mapsto p(t,x,y)$ is continuous for each fixed $y$, and $\int_{\Bbb R} p(t,x,y)\,dy=1$ for all $x$, Scheffé's lemma implies that $x\mapsto p(t,x,\cdot)$ is continuous in $L^1(\Bbb R)$; that is, $$ \int|p(t,x,y)-p(t,x_0,y)|\,dy\to 0\quad\hbox{as }x\to x_0, $$ for a fixed $x_0$. From this it follows easily that $x\mapsto\mu_x(B_t\in C)$ is continuous for each Borel set $C$.

For general $A\in\mathcal A$, first use the Markov property to condition on $B_{t_1}$.