Let $\{B_t:t\geq0\}$ be a $d$-dimensional Brownian motion. $D\in\mathbb{R}^d$ is a bouded open set, $\sigma:=\inf\{t\geq0:B_t\in\partial D\}$.
I want to prove that $\mathrm{P}_x(\sigma<\infty)=1$. My proof is correct?
To prove this, it is enough to show $\mathrm{P}_x(\tau<\infty)=1$. $\tau:=\inf\{t\geq0:B_t\notin\mathcal{B}(x,R)\}$, $\mathcal{B}(x,R)$ is a open ball which contain $D$.
\begin{align}
\mathrm{P}(\tau=\infty)&=\mathrm{P}({}^\forall t\geq0, B_t\in\mathcal{B}(x,R))\\
&\leq \mathrm{P}(B_t\in\mathcal{B}(x,R))\\
&=\int_{|y|\leq R}\frac{1}{\sqrt[d]{2\pi t}}e^{-|y-x|^2/2t}dy.
\end{align}
By changing of variable $z_i=\frac{y_i-x_i}{\sqrt{t}}$,
\begin{align}
\mathrm{P}(\tau=\infty)
&\leq\int_{|\sqrt{t}z+x|\leq R}\frac{1}{\sqrt[d]{2\pi}}e^{-|z|^2/2}dz\\
&=\int_{\mathbb{R}^d}{\bf 1}_{|\sqrt{t}z+x|\leq R}\frac{1}{\sqrt[d]{2\pi}}e^{-|z|^2/2}dz.
\end{align}
Because of $\int_{\mathbb{R}^d}\frac{1}{\sqrt[d]{2\pi}}e^{-|z|^2/2}dz<\infty$ and $t$ is arbitrary, we can use dominated convergence theorem as $t\to\infty$. Therefore, $\mathrm{P}(\tau=\infty)=0$, this completes proof.
Thank you for your cooperation.
Best Answer
Here is a proof using martingale techniques: Set $$\tau := \inf\{t \geq 0; |B_t-x| \geq R\}.$$ By the optional stopping theorem, $$M_t := \exp\left(\xi B_{t \wedge \tau}-\frac{1}{2} \xi^2 t \wedge \tau \right),\qquad \xi \in \mathbb{R}^d,$$ is a martingale and so $$\mathbb{E}^x(M_t)=\mathbb{E}^x(M_0)=e^{\xi x}.$$ As $|B_{t \wedge \tau}| \leq R+|x|$ and
$$\lim_{t \to \infty} \exp\left(\xi B_{t \wedge \tau}-\frac{1}{2} \xi^2 t \wedge \tau \right) = \begin{cases} \exp(\xi B_{\tau}- \frac{1}{2} \tau \xi^2), &\tau<\infty,\\ 0, & \tau=\infty \end{cases}$$
it follows from the dominated convergence theorem that
$$e^{\xi x} = \mathbb{E}^x\exp\left(\xi B_{t \wedge \tau}-\frac{1}{2} \xi^2 t \wedge \tau \right) \xrightarrow[]{t \to \infty} \mathbb{E}^x\left[ 1_{\{\tau<\infty\}} \exp \left( \xi B_{\tau} - \frac{\tau}{2} \xi^2 \right) \right].$$
Letting $\xi \to 0$ yields
$$1 = \mathbb{P}^x(\tau<\infty).$$