# Brownian motion, harmonic functions and the Dirichlet problem

I am having trouble understanding one detail of the standard use of Brownian motion to solve the Dirichlet problem, I will write the statement and proof and then point to the detail I don't understand.

Let $U$ be a bounded domain in $\mathbb{R}^n$, and let $B_t$ be a Brownian motion in $\mathbb{R}^n$, $\tau_{2}$ be the hitting time of $\partial U$, and let $\phi(x)$ be a continuous function defined on $\partial U$. The following function is a harmonic function on $U$, $u(x) = \mathbb{E}_x \phi(B_{\tau_{2}})$. Here is the proof

Let $x\in U$, and let $r$ be small enough that $\partial D(x,r)\subseteq U$. Let $D(x,r)$ represent the ball of radius $r$ around $x$. Let $\tau_1$ be the hitting time of $\partial D(x,r)$, by the strong Markov property the process $\tilde{B}_t = B_{t+\tau_1}-B_{\tau_1}$ is a Brownian motion independent of $\mathcal{F}_{\tau_1}$. Now let $\tau_2 = \inf\{t>0:B_t\in \partial U\}$. We show that $u(x)$ satisfies the mean value property,

\begin{align} u(x) &= \mathbb{E}_x\phi(B_{\tau_2})\\ &= \mathbb{E}_x(\mathbb{E}[\phi(B_{\tau_2})|\mathcal{F}_{\tau_1}])\\ &= \mathbb{E}_x(\mathbb{E}[\phi(B_{\tau_2}-B_{\tau_1}+B_{\tau_1}|\mathcal{F}_{\tau_1}])\\ &= \mathbb{E}_x( \psi(B_{\tau_1})) \end{align}

Where $\psi(y) = \mathbb{E}_{y} \phi(\tilde{B}_{\tau_2-\tau_1})$, because Brownian motion is rotationally invariant, the distribution of $B_{\tau_1}$ is uniform over the sphere, thus if $\lambda$ is the uniform distribution over the sphere

\begin{align} u(x) &= \mathbb{E}_x( \psi(B_{\tau_1}))\\ &= \int\limits_{\partial D(x,r)} \mathbb{E}_{y} \phi(\tilde{B}_{\tau_2-\tau_1}) \lambda(dy) \end{align}

If $\tilde{B}_{\tau_2-\tau_1}$ represents the hitting time of $\partial U$ for each $y$, then this is equal to $\int\limits_{\partial D(x,r)} u(y)\lambda(dy)$, then we are done, however I don't see why $\tilde{B}_{\tau_2-\tau_1}$ necessarily represents the hitting time of $\partial U$, or maybe I have the proof wrong somewhere else

Given a ball $$\bar B_r(x)\subset D$$, consider the stopping time: $$\tau = \min\{t\geq 0;~ x+B_t\not\in B_r(x)\}$$ and also for each $$z\in D$$ let $$\tau_z = \min\{t\geq 0;~ z+B_t\not\in D\}$$. We introduce two random variables $$X_1$$ and $$X_2$$ on the probability space $$(\Omega,\mathcal{F}, \mathbb{P})$$ on which the standard Brownian motion $$\{B_t\}_{t\geq 0}$$ is defined. The first random variable is simply: $$X_1=x+B_{\tau}$$, taking values in the measurable space $$(D, \mathcal{F}_1)$$ with the usual $$\sigma$$-algebra $$\mathcal{F}_1$$ of Borel subsets of $$D$$.
The second random variable $$X_2$$ is given by the paths of $$B_{\tau+t} - B_\tau$$, almost surely belonging to $$E_0=C[0,\infty), \mathbb{R}^n)$$. We make $$E_0$$ a measurable space by equipping it with the $$\sigma$$-algebra $$\mathcal{B}(E_0)$$ of its Borel subsets with respect to the topology of uniform convergence on compact intervals. In fact, $$X_2$$ takes values in a measurable subset $$E$$ of $$E_0$$, consisting of these $$f\in E$$ which satisfy: $$f(0)=0$$ and $$|f(T)|>\mbox{diam} \, D$$ for some $$T>0$$. We name $$\mathcal{F}_2$$ the $$\sigma$$-algebra $$\mathcal{B}(E_0)$$ restricted to $$E$$.
Checking the measurability property of $$E$$ in $$E_0$$, as well as measurability of $$X_2$$ with respect to $$(E,\mathcal{F}_2)$$ relies on observing that $$\mathcal{B}(E_0)$$ is generated by the countable family of sets of the type $$A_{g,T,r} = \{f\in E_0;~ \|f-g\|_{L^\infty[0,T]}\leq r\}$$, where $$g$$ are polynomials with rational coefficients, and $$T,r>0$$ are rationals.
Now, the crucial observation is that $$X_1$$ and $$X_2$$ are independent. Indeed, $$X_1$$ is clearly $$\mathcal{F}_\tau$$-measurable, whereas $$X_2$$ is $$\mathcal{F}_\tau$$-independent. This last statement is a direct consequence of the strong Markov property for the shifted Brownian motion $$\{B_{\tau+t} - B_\tau\}_{t\geq 0}$$ and can be checked directly on the preimages of sets $$A_{g,T,r}$$.
We call $$\mu_1$$ the push-forward measure of $$\mathbb{P}$$ via $$X_1$$ and $$\mu_2$$ the push-forward of $$\mathbb{P}$$ via $$X_2$$. Thus $$(D,\mathcal{F}_1,\mu_1)$$ and $$(E,\mathcal{F}_2,\mu_2)$$ are two probability spaces, and the independence of $$X_1$$ and $$X_2$$ is equivalent to the product $$\mu_1\times \mu_2$$ equaling the push-forward of $$\mathbb{P}$$ on $$D\times E$$ via $$(X_1, X_2)$$.
Finally, consider the following random variable on $$D\times E$$, valued in $$\mathbb{R}$$: $$F(z,f) = \varphi\big(z+f(\min\{t\geq 0;~ z+f(t)\not\in D\})\big).$$ We write: $$\begin{equation*} \begin{split} u(x) & = \mathbb{E}[\varphi(x+B_{\tau_x}] = \mathbb{E}[F\circ (X_1, X_2)] = \int_{D\times E} F \;d(\mu_1\times\mu_2) \\ & = \int_D\int_E F(z,f)\;d\mu_2(f) \;d\mu_1(z)=\int_D \mathbb{E}[F(z,X_2)]\;d\mu_1(z), \end{split} \end{equation*}$$ where we simply used Fubini's theorem. In conclusion: $$\begin{equation*} \begin{split} u(x) & = \int_D \varphi(z+B_{\tau_z}) \;d\mu_1(z) = \int_D u(z) \;d\mu_1(z) \\ & = \mathbb{E}[u\circ X_1] = \mathbb{E}[u(x+B_{\tau})], \end{split} \end{equation*}$$ as claimed.