Proving that Brownian motion has no points of increase

Question

I am reading Burdzy's paper on the points of increase of Brownian motion: Burdzy's Paper

He is proving that, almost surely, a Brownian motion, has no points of increase. What he actually proves is that the probability of the following set is $0$, and that is supposed to be sufficient for the original problem. The set is:

{$\omega$; $\exists t,u, 0<t<u$ such that $B_t(\omega)\leq1$, $B_u(\omega)\geq B_t(\omega)+2$, $B_s(\omega)< B_t(\omega) \forall s \in [0,t)$, and $B_s(\omega) > B_t(\omega) \forall s \in (t,u]$}

So, this is the set of all $\omega$ such that there exist positive numbers $t<u$(which may depend on $\omega$) for which $0<t<u$ such that $B_t(\omega)\leq1$, $B_u(\omega)\geq B_t(\omega)+2$, $B_s(\omega)< B_t(\omega) \forall s \in [0,t)$, and $B_s(\omega) > B_t(\omega) \forall s \in (t,u]$.

The original probelm is asking to prove that a Brownian motion almost surely has no points of increase, for which we'll have to prove that the probability of the set

{$\omega$; $\exists t , \epsilon > 0$ such that $B_t(\omega)\geq B_s(\omega) \forall s \in (t-\epsilon, t]$ and $B_s(\omega) \geq B _t(\omega) \forall s \in [t, t + \epsilon)$}

How does the probability of the first set I wrote being equal to zero imply that, the probability that Brownian motion has a point of increase is zero(i.e., the probability of the second set that I wrote is also zero)?

I have linked above the original paper in which these things are introduced, and that can be seen for any clarifications.

Answer 1

If we have a point of increase at $t$, witnessed by $\epsilon$, the question is threefold: why can we assume $t-\epsilon=0$, $B_t(\omega)\le 1$, and $B_{t+\epsilon}(w)-B_t(\omega)\ge 2$. These conditions are somewhat similar, so here is an argument for the last of these conditions.

Suppose we have a point of increase at $t$, witnessed by $\epsilon$. If we look at the process $B_{q+t+\epsilon}$, $q\ge 0$ then it starts out strictly above $B_t$. Therefore, there is a positive probability that our process reaches $B_t+2$ before it reaches $B_t$, at a time $T$. In that event, we can redefine $\epsilon$ so that $t+\epsilon=T$, and we have ensured that $B_{t+\epsilon}(w)-B_t(\omega)\ge 2$.