First hitting locations of Brownian motion gets arbitrarily close Let $\Omega$ be an open topological disk. Let $\delta > 0$. I want to show that there exists $\epsilon > 0$ and $K_\epsilon \subseteq_c \Omega$, where $d(y,\partial \Omega) < \epsilon$ for all $y \in \partial K_\epsilon$, such that 
$\mathbb P(|B_T - B_{T_\epsilon}| \geq \delta) < \delta$, where $T$ and $T_\epsilon$ are the first hitting times of $\partial \Omega$ and $\partial K_\epsilon$, respectively. 
This argument looks intuitive. If $\epsilon$ decreases sufficiently small, $B_{T_\epsilon}$ gets very close to the boundary of $\Omega$, and thus $B_T$ is expected to be very close to $B_{T_\epsilon}$. But how do I put together a proof for this? I think I may be missing some property of Brownian motions so that I have a hard time doing it.
 A: Here's a rigorous proof. Assume $\Omega \subset \Bbb R^n$ is bounded and has $C^1$ boundary (or just satisfies the usual "cone condition"). Let us fix $\delta>0$.
For $x \in \partial \Omega$, let us define $U_{\delta}(x) = \{y \in \partial \Omega: |y-x|<\delta/2\}$. Next, let us define $f_{x,\delta}:\overline \Omega \to [0,1]$ to be the unique harmonic function with boundary data $f_{x,\delta}(y)=1_{U_{\delta}(x)}(y)$, for $y \in \partial \Omega$. Define the set $V_{\delta}(x):=f_{x,\delta}^{-1}\big((1-\delta,1]\big) \cap B(x,\delta/2)$.
By compactness, we may choose $x_1,...,x_n \in \partial \Omega$ such that $\partial \Omega = \bigcup_1^n U_{\delta}(x_i)$. We claim that there exists some $\epsilon>0$ (obviously depending on $\delta)$ such that for any $z \notin \bigcup_1^n V_{\delta}(x_i)$, one has $d(z,\partial \Omega)>\epsilon$. Indeed, the set $E:=\overline \Omega \backslash \bigcup_1^n V_{\delta}(x_i)$ is a closed (hence compact) subset of $\overline \Omega$, since each $V_{\delta}(x)$ is an open subset of $\overline \Omega$. Thus the function from $E \to [0,\infty)$ given by $z \mapsto d(z, \partial \Omega)$ achieves a strict minimum. If this minimum was $0$, then $E$ would intersect $\partial \Omega$ which is impossible.
Let $\Bbb P_z$ denote the law of BM started from $z\in \Omega$. If $z \in V_{\delta}(x_i)$, then by the fact that $|z-x_i|<\delta/2$ and by the potential theory of Brownian motion, we have $$\Bbb P_z(|B_T-z|>\delta) \leq \Bbb P_z(|B_T-x_i|>\delta/2) = 1-f_{x_i,\delta}(z) < 1-(1-\delta)=\delta.$$Note that this is true uniformly over all $z \in \bigcup_1^n V_{\delta}(x_i)$. In particular, if we define $K_{\epsilon}:=\{z \in \Omega: d(z,\partial \Omega)\ge\epsilon/2\}$, then $\partial K_{\epsilon} \subset \bigcup_1^n V_{\delta}(x_i)$, and thus for any point $a \in K_{\epsilon}$, the strong Markov property and the preceding calculation tells us that $$\Bbb P_a(|B_T-B_{T_{\epsilon}}|>\delta) = \Bbb E_a\big[ \Bbb P_{B_{T_{\epsilon}}}(|B_T-B_{0}|>\delta)\big] < \Bbb E_a[\delta]=\delta.$$
