Brownian motion: hitting times for closed sets are stopping times (and more). Let $(B_t)$ be a $d$-dimensional Brownian motion, and consider the filtrations $(\mathcal{F_t^B}) = \sigma(B_0,...,B_t)$ and $\mathcal{F_t} = \cap_{\epsilon > 0}{\mathcal{F_{t+\epsilon}^B}}$ (the right continuous one).
I am trying to prove the following statements:


*

*For a closed set $A \in \mathbb{R}^d$, the first hitting time $T_A = \inf \{ t \geq 0 : B_t \in A \}$ is a stopping time;

*For an open set $U \in \mathbb{R}^d$, $T_U$ is a stopping time w.r.t. $(\mathcal{F}_t)$ but not necessarily w.r.t. $(\mathcal{F}_t^B)$


I made a lot of attempts for solving them, unfortunately without succeeding. They that could be summed up in the following observations:


*

*First of all, if it is true that $\{ T_A \leq t \} = \cup_{ 0 \leq s \leq t }{\{ \omega \in \Omega : B_s \in A \}}$, this union is not necessarely countable so it could not be contained in the filtration (although all the sets whose union is done are);

*I tried modifying the indexes in order to obtain something countable, for example by choosing only the rational numbers, but I didn't manage to show the equality of the two sets;

*The only "result" that I have is that to take only rational numbers in the case of an open set, should make the equality $\cup_{ 0 \leq s \leq t }{\{ \omega \in \Omega : B_t \in A \}} = \cup_{ 0 \leq s \leq t, s \in \mathbb{Q} }{\{ \omega \in \Omega : B_t \in A \}}$ fail, but it is far from a solution;

*I "feel" that the key "must" be in the fact that in a Brownian motion all the paths (i.e. the functions $t \mapsto B_t(\omega)$, for a fixed omega) are continuous...and the fact that the sets under analysis are closed or open suggest looking to convergence properties...but I don't manage to go further.


Since I worked in this problem for a long time, I would be particularly grateful to anyone who would help me in any way.
Hints and suggestions are very welcome!
Thank you very much in advance.
Cheers :)
 A: Yes, you are right - the key idea is to rewrite the set $\{T_A \leq t\}$ in a clever way using the continuity of sample paths of Brownian motion.
Hints (for $T_A$):


*

*Show that $$\{T_A \leq t\} = \left\{ \omega \in \Omega; \inf_{r \in \mathbb{Q} \cap [0,t]} d(A, B_r(\omega)) = 0 \right\}$$ using the continuity of the sample paths. Here $d(A,x) := \inf\{|x-y|; y \in A\}$ denotes the distance of the set $A$ to $x$.

*Since $x \mapsto d(x,A)$ is continuous, the mapping is in particular (Borel-)measurable. Conclude that $\{T_A \leq t\} \in \mathcal{F}_t^B$.


Hints (for $T_U$):


*

*Prove that $$\{T_U < t\} = \bigcup_{r \in \mathbb{Q} \cap [0,t)} \{B_r \in U\}$$ using the (right)continuity of the sample paths and the fact that $U$ is open.

*Conclude that $\{T_U < t\} \in \mathcal{F}_t^B$ and $$\{T_U \leq t\} = \bigcap_{n \in \mathbb{N}} \{T_U < t+1/n\} \in \mathcal{F}_t.$$


For full proofs see e.g. René Schilling & Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 5.
