I have a question regarding an argumentation, which is not clear to me:
Let $B_t$ be the standard Brownian motion and $\tau$ a respective stopping time with finite mean.
Fix $\epsilon>0.$ We can choose $s>0$ so small that $$(*) \quad \quad\mathbb{P}[\sup_{s < t < \epsilon} B_t > s \text{ and } \inf_{s < t < \epsilon} B_t < - s] > 1 - \frac{\epsilon}{3},$$ since $B_t$ takes positive and negative values on $(0,\epsilon)$ almost surely.
Furthermore choose $T>0$ and $M>0$ such that
$\mathbb{P}[\tau \geq T] < \frac{\epsilon}{3}\quad\quad$ and $\quad\quad\mathbb{P}[\sup_{0\leq t\leq T} |B_t| \geq M]<\frac{\epsilon}{3}.$
Now define the set $A:=A_1\cap A_2 \cap A_3,$ where $A_1:=\{\omega | \sup_{\tau+s<t<\tau+\epsilon}[B_t - B_\tau] > s\text{ and } \inf_{\tau+s<t<\tau+\epsilon}[B_t - B_\tau] < - s\}$
$A_2:=\{\omega | \tau(\omega) < T\}$
$A_3:=\{\omega | |B_{\tau(\omega)}|<M\} $
Assertion: Due to the choice of $s, T, M$ and the strong Markov property it follows that $\mathbb{P}[A]>1-\epsilon$.
Question Why is that so? I do not see how to put the 3 parts and estimates together in order to get the $\frac{\epsilon}{3}$'s summed up to $\epsilon$.
My ideas were:
With $\Delta:=t-\tau$, we can apply the strong markov property to conclude that $B_{t}-B_\tau=B_{\tau+\Delta}-B_\tau\sim B_\Delta.$ Thus, by using (*), we have $P[A_1]>1-\frac{\epsilon}{3}.$ By the assumptions on $A_2^c$ and $A_3^c$ we have also the same estimate on $A_2$ and $A_3$.
I think my problem is to properly discern the dependencies of the sets. $A_1$ should be independent of $A_3$, due to the markov property, I think. Thus, in particular, independent of $A_2\cap A_3\subset A_3$. The independence then implies $P[A]=P[A_1]*P[A_2\cap A_3]=P[A_1]*P[A_3|A_2]*P[A_2]>(1-\frac{\epsilon}{3})^3$, because of the preassumptions. But, is that correct? I'm not so satisfied. Especially, why using the fractions for the $\epsilon$'s if they do not add up in the end?
I would be glad for your support, best regards
P.s. There is some more context given, for instance, $\tau$ is explicitly defined as the first hitting time of $(t,B_t)$ into a closed set. But I hope that I gave enough information, such that the question above can be solved. If not, I can attach additional information.
Denote by $A^c := \Omega \backslash A$ the complement of $A$. Obviously, $$\mathbb{P}(A) = 1- \mathbb{P}(A^c)$$ where $$\mathbb{P}(A^c) = \mathbb{P}\{\omega: \exists i \in \{1,2,3\}: \omega \notin A_i\} \leq \sum_{i=1}^3 \mathbb{P}(A_i^c) < \varepsilon$$ since $\mathbb{P}(A_i^c) < \frac{\varepsilon}{3}$ by assumption. Hence, $$\mathbb{P}(A) > 1-\varepsilon$$