TITLE: Does Brownian motion immediately visit both sides of a Jordan curve?
QUESTION [11 upvotes]: Let $C$ be a Jordan curve in $\mathbb{R}^2$.  By the Jordan curve theorem, $\mathbb{R}^2 \smallsetminus C$ is uniquely partitioned into two connected regions $A$ and $B$ (the interior and exterior).
Question 1: Let $x_0$ be an arbitrary point in $A$.  Let $X$ be a planar Brownian motion starting at $x_0$.  Let 
$$\tau_C = \inf \{t: X(t) \in C \},\qquad\tau_B = \inf \{t: X(t) \in B \}.$$
Does $\tau_C = \tau_B$ a.s.?
Question 2: Let $y_0$ be an arbitrary point on $C$.  Let $Y$ be a planar Brownian motion starting at $y_0$.  Let 
$$\tau_A = \inf \{t: Y(t) \in A \},\qquad\tau_B = \inf \{t: Y(t) \in B \}.$$
Does $\tau_A = \tau_B = 0$ a.s.?
Observations:

A positive answer to question 2 gives a positive answer to question 1.  (Let $y_0 = X(\tau_C)$.)
The answers to both questions 1 and 2 are clearly true for "nice curves", for example smooth curves or curves with some kind of cone condition on each side.  However, possibly some messy curves (e.g. ones with positive 2-dimensional Lebesgue measure) would present a problem.
If the answer to either question is yes, it would be interesting to know the answer in higher dimensions as well (replacing Jordan curves with their higher dimensional analogues).

REPLY [11 votes]: As to question 2: Planar Brownian motion started at $y_0$ will almost surely loop around $y_0$, i.e., disconnect $y_0$ from $\infty$ immediately, so it has to hit $A$ and $B$ immediately, too, and $\tau_A = \tau_B = 0$ a.s., no matter what kind of Jordan curve $C$ is. This should also imply that the answer to the first question is yes, by the strong Markov property.
In higher dimensions there are easy counterexamples to question 2. E.g., if $C$ is a topological sphere, smooth except for one point where it is locally the "Lebesgue thorn", i.e., with one irregular boundary point $y_0$, then one of the stopping times is almost surely positive. I am not sure about question 1 in higher dimension.