In Chapter 6 in the book Functional analysis by Stein and Shakarchi the following theorem (Dirichlet problem) is proved:
$\mathcal{R}$ denotes a bounded open set and a point $y$ is called regular if
$$\mathbb{P}(\tau_\ast^y=0) = 1$$
where $\tau^x = \inf\{t\geq 0: x+B_t\notin \mathcal{R}\}$ and $\tau_\ast^x = \inf\{t>0: x+B_t\notin \mathcal{R}\}$ and $B_t$ is a Brownian motion started at $0$ and $B_t^x = x+B_t$.
Now my question concerns a particular step in the proof of (b).
I find the fact that $x\mapsto 1_{\{\omega':x+B_t(\omega')\in \mathcal{R}, \, \forall \epsilon\leq t \leq \delta\}}(\omega)$ should be continuous, hard to believe. Assume that $\mathcal{R} = \mathbb{D}\setminus \{0\}$, the unit disk with the $0$ removed. Then if $B_{t_0}^y(\omega) = 0$ for precisely one $t_0\in [\epsilon,\delta]$ it is not necessarily so that $B_t^x(\omega) = 0$ for some $t\in [\epsilon,\delta]$ regardless of how close $x$ is to $y$. Could someone clarify this for me? Is this an error in the book?
DISCLAIMER: THIS ANSWER IS INCOMPLETE
I think the key point is that $\mathcal{R}$ is an open set. Fix an $\omega$ such that $B_t$ is continuous. First note that $g(y,t)= B_t^y =y+B_t$ is jointly continuous.
Let us first show that $A:=\{y:y+B_t \in \mathcal{R}, \text{ all }\varepsilon\leq t \leq \delta\}$ is open. Suppsoe that $A$ is not open. Then there is a point $y \in A$ such that for every $n\geq 1$, you will get a $y_n \in (y-\frac{1}{n}, y+\frac{1}{n})$ such that $y_n \not\in A$. This means $y_n+B_{t_n} \not\in \mathcal{R}$ for some $t_n \in[\varepsilon,\delta]$. Now take a subsequence $y_{a_n}$ of $y_n$ such that for $y_{a_n}+B_{t_{a_n}}\not\in \mathcal{R}$, $t_{a_n} \in [\varepsilon,\delta]$, and $t_{a_n}\to t'$ for some $t'\in [\varepsilon,\delta]$ (you can always choose such a subsequence since $[\varepsilon,\delta]$ is compact). Now note that by construction $(y_{a_n},t_{a_n})\to(y,t')$ and therefore $y_{a_n}+B_{t_{a_n}} \to y+B_{t'} \in \mathcal{R}$ (using joint continuity of $B_t^y$). On the other hand $y_{a_n}+B_{t_{a_n}} \in \mathcal{R}^c$ which is closed, therefore the limit must be in $\mathcal{R}^c$ which is a contradiction. Therefore $A$ is open.
Coming to your problem, let $$f(y)=1\{B_t^y \in \mathcal{R}, \text{ all }\varepsilon\leq t \leq \delta\}= 1\{y \in A\}$$, and we showed $A$ is open. Now you can conclude $f(y)$ is continuous on $A$ from here.
It still does not show that $f$ is continuous everywhere.