Let $D \subset \mathbb{R}^n$ be a bounded open set and $f: \partial D \rightarrow \mathbb{R}$ a continuous function. The Dirichlet problem consists of finding a continuous function $U: \overline{D} \rightarrow \mathbb{R}$ satisfying the mean value property in $D$ and being equal to $f$ on $\partial D$. In the lecture on "Brownian Motion and Stochastic Calculus" I saw the following Proposition:
Proposition: If a solution $U$ to the Dirichlet problem exists, then necessarily we have for all $x \in \overline{D}$ that $U(x)=E_x [f(B_T)]$.
Let me say first something about notation. The lower index $x$ means that $B=(B_t)_{t \geq 0}$ is a Brownian motion started from $x$ and $T$ is the first hitting time of $D^c$ (a closed set). More precisely $T=\inf\{t \geq 0 : B_t \in D^c\}$, which is almost surely finite.
To prove the Proposition we first define some stopping times: Let $T_0 = 0$ and for each $n \geq 0$ let $r_n=d(B_{T_n},\partial D)$ and $$ \begin{equation} T_{n+1}=\inf\{t > T_n : d(B_t, B_{T_n}) = r_n / 2\} \end{equation} $$
Now this is where I am stuck. I do not see why the $T_n$'s are stopping times, since they are so complicatedly nested. Can someone help me out to see why they are stopping times?
Thanks a lot in advance!
They are not that complicatedly nested. You just need the following result:
Let $T$ be a stopping time, $r$ be a $\mathcal F_T$-measurable positive random variable and $$ S=\inf\{t>T\mid d(B_t,B_T)=r\}. $$
Then $S$ is a stopping time.
Proof:
$\begin{align*} \{S\le t\}&=\{T<t\}\cap\left(\bigcup_{u\in]T,t]}\{d(B_u,B_T)=r\}\right)\\ &=\{T<t\}\cap\left(\bigcap_{n\in\mathbb N^*}\bigcup_{u\in]T,t]\cap\mathbb Q}\{d(B_u,B_T)\ge r-\frac1n\}\right)\\ &=\{T<t\}\cap\left(\bigcap_{n\in\mathbb N^*}\bigcup_{u\in[0,t]\cap\mathbb Q}\{T<u\}\cap\{d(B_u,B_T)\ge r-\frac1n\}\right) \end{align*} $
$\{T<t\}\in\mathcal F_t$.
For all $n\in\mathbb N^*$ and $u\in[0,t]\cap\mathbb Q$, $\{d(B_u,B_T)\ge r-\frac1n\}\in\mathcal F_{T\vee u}$, so $\{T<u\}\cap\{d(B_u,B_T)\ge r-\frac1n\}\in\mathcal F_{u}\subset\mathcal F_t$.