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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.

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.

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 zero)?

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

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.

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:

{there exist positive numbers $t$ and $u$ such that $B(t)<1$, $B(u)>B(t)+2$, $B(s)< B(t)$ for all $s \in [0,t)$, and $B(s) > B(t)$ for all $s \in (t,u]$}

How does the probability of this set being equal to zero imply that, the probability that Brownian motion has a point of increase is zero?

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 zero)?

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