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).



Does Brownian motion immediately visit both sides of a Jordan curve?
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.


