[proofplan] We first check that the exit value $f(W_{\tau_U})$ is a well-defined bounded random variable. The existing Brownian Dirichlet theorem then gives harmonicity of $u$ in the interior. The regular-boundary hypothesis is exactly the statement that harmonic measure tends to a point mass at each boundary point, and this turns the interior harmonic function into a continuous extension with boundary values $f$. Uniqueness follows from the maximum-principle uniqueness theorem for the Dirichlet problem. [/proofplan] [step:Verify that Brownian motion exits through the boundary in finite time] Fix $x\in U$. Since $U$ is bounded, choose $R>0$ such that $U\subset B(0,R)$. If $\tau_U=\infty$, then $W_t\in U\subset B(0,R)$ for every $t\ge0$. Therefore, for every $t>0$, \begin{align*} \mathbb P^x(\tau_U=\infty)\le \mathbb P^x(W_t\in B(0,R)). \end{align*} The random vector $W_t$ has Gaussian density \begin{align*} y\mapsto \frac{1}{(2\pi t)^{n/2}}\exp\left(-\frac{|y-x|^2}{2t}\right) \end{align*} with respect to $\mathcal L^n$, so \begin{align*} \mathbb P^x(W_t\in B(0,R)) =\int_{B(0,R)}\frac{1}{(2\pi t)^{n/2}}\exp\left(-\frac{|y-x|^2}{2t}\right)\,d\mathcal L^n(y) \le \frac{\mathcal L^n(B(0,R))}{(2\pi t)^{n/2}}. \end{align*} Letting $t\to\infty$ gives $\mathbb P^x(\tau_U=\infty)=0$. Because Brownian paths are continuous and $U$ is open, the first point outside $U$ lies in $\partial U$ almost surely: \begin{align*} W_{\tau_U}\in\partial U\qquad\mathbb P^x\text{-almost surely}. \end{align*} Since $\partial U$ is compact and $f:\partial U\to\mathbb R$ is continuous, there is a constant $M_f\ge0$ such that \begin{align*} |f(y)|\le M_f\qquad\text{for every }y\in\partial U. \end{align*} Thus $f(W_{\tau_U})$ is bounded and measurable, and $u(x)=\mathbb E^x[f(W_{\tau_U})]$ is well-defined for every $x\in U$. [/step] [step:Apply Brownian Dirichlet harmonicity in the interior] The theorem [Harmonicity of the Brownian Dirichlet Solution](/theorems/1186) applies with $D=U$ and $\varphi=f$. Its hypotheses are satisfied because $U$ is a bounded domain and $f$ is continuous on the compact set $\partial U$, hence bounded and Borel measurable. Therefore the function \begin{align*} u:U&\to\mathbb R\\ x&\mapsto \mathbb E^x[f(W_{\tau_U})] \end{align*} is harmonic on $U$: \begin{align*} \Delta u=0\qquad\text{in }U. \end{align*} In particular, by the interior regularity of harmonic functions, equivalently by [Harmonic Functions are Smooth](/theorems/36) applied to the mean-value property supplied in the proof of the Brownian Dirichlet theorem, we have \begin{align*} u\in C^2(U). \end{align*} [/step] [step:Use regular boundary to prove continuity up to $\partial U$] Let $z\in\partial U$. By the regular-boundary hypothesis, the harmonic measures $\omega_U^x$ converge weakly to $\delta_z$ on $\partial U$ as $x\to z$ with $x\in U$. Since $f:\partial U\to\mathbb R$ is continuous and bounded, [weak convergence](/page/Weak%20Convergence) gives \begin{align*} \lim_{\substack{x\to z\\ x\in U}}\int_{\partial U}f(y)\,d\omega_U^x(y) =\int_{\partial U}f(y)\,d\delta_z(y) =f(z). \end{align*} By definition of $\omega_U^x$ as the law of $W_{\tau_U}$ under $\mathbb P^x$, \begin{align*} \int_{\partial U}f(y)\,d\omega_U^x(y)=\mathbb E^x[f(W_{\tau_U})]=u(x). \end{align*} Hence \begin{align*} \lim_{\substack{x\to z\\ x\in U}}u(x)=f(z). \end{align*} Since $z\in\partial U$ was arbitrary and we set $u=f$ on $\partial U$, the extended function belongs to $C(\overline U)$. [/step] [step:Conclude uniqueness from the Dirichlet maximum principle] Let $v\in C(\overline U)\cap C^2(U)$ satisfy \begin{align*} \Delta v=0\qquad\text{in }U, \qquad v=f\qquad\text{on }\partial U. \end{align*} The function $u$ is continuous on $\overline U$, harmonic on $U$, and agrees with $f$ on $\partial U$. The [Uniqueness for the Dirichlet Problem via Maximum Principle](/theorems/1187) applies because $U$ is bounded and both $u$ and $v$ are continuous on $\overline U$, harmonic in $U$, and equal on $\partial U$. Therefore \begin{align*} v=u\qquad\text{on }\overline U. \end{align*} This proves that the Brownian representation is the unique member of $C(\overline U)\cap C^2(U)$ solving the stated Dirichlet problem. [/step]