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