Question about Karatzas & Shreve - Brownian motion and stochastic calculus : Problem 2.7 chapter 1 (page 7 and solution page 39)
Let $X$ a continuous process, $\Gamma$ a closed set and we have :
- $\Gamma_n=\{x\in\mathbb R^d; \rho(x,\Gamma) < 1/n\}$ with $\rho(x,\Gamma) = \inf\{\|x-y\|; y\in\Gamma\}$
- $(T_n)_{n\geq 1}$ the reaching time of $\Gamma_n$ (it is an optional time)
- $T:=\lim_{n \to \infty} T_n$
- $H_{\Gamma} :=\inf \{t\geq 0; X_t \in \Gamma\}$ an hitting time.
The question of the exercise is to show that $H_{\Gamma}$, is a stopping time.
All the argumention of Karatzas et al. is to show that : $H_{\Gamma} = T$ (*).
My problem:
Nevertheless, I do not understand the conclusion written after having shown (*) : $\{H\leq t\}=\cap^\infty_{n=1}\{T_n<t\}$ (copied/pasted)
If I split up it, thanks to (*) : $\{H\leq t\} = \{T\leq t\}= \{T < t\} \cup \{T = t\}= \cap^\infty_{n=1}\{T_n<t\}$ (**)
However something disturb me, I can say : $\{T < t\} = \cap^\infty_{n=1}\{T_n<t\}$
But we cannot say : $\{T = t\} \subseteq \cap^\infty_{n=1}\{T_n<t\}$ because it means that $\{T = t\} \subseteq \{T_\infty=T<t\}$ ?
Does my split (**) is correct ? $$ %"{\cap^\infty_{n=1}\{T_n<t\} = \{T_n<t\}\cap\{T_n<t\}\cap...\cap\{T_\infty=T<t\}}" $$
The key to the equality $\{H\le t\}=\cap_{n=1}^\infty\{T_n<t\}$, which is asserted by Karatzas & Shreve for $t>0$, is the fact that $\{T_n\}$ approximates $H$ strictly from below on $\{H>0\}$, as detailed in the "dichotomy" listed in their solution to the problem. (Plus the fact that $\{H=0\}=\cap_{n=1}^\infty\{T_n=0\}$.)