I am reading Burdzy's paper on the points of increase of Brownian motion:
[Burdzy's Paper](https://projecteuclid.org/journals/annals-of-probability/volume-18/issue-3/On-Nonincrease-of-Brownian-Motion/10.1214/aop/1176990732.full)


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.



Proving that Brownian motion has no points of increase
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 $B\_{t+\epsilon}(w)-B\_t(\omega)\ge 2$.


