# Probability of a sequence of events converging to 0, has an event whose probability is actually 0 in the context of Stochastic processes (BM)

Why is this true? It appears in the proof of a Holder condition for Brownian motion in Kenneth Falconer Fractal Geometry 2014 edition at page 283 (proposition 16.1)

For $$X$$ beaing a random process (Brownian motion for instance), if for each positive integer k, $$\mathbb{P}\left\{|X\left((m-1)2^{-j}\right)-X(m2^{-j})> 2^{-j\lambda} \text{ for some j\geq k and 1\leq m\leq2^j }\right\}\leq c2^{-k+1}$$ then, with probability $$1$$, there is an integer $$K$$ such that $$\left| X((m-1)2^{-j})- X(m2^-j) \right| \leq 2^{-j\lambda}$$ for all $$j>K$$ and $$1\leq m \leq 2^j$$.

This looks to me as being equivalent to: If $$\mathbb{P}(A_k) \to 0$$ then there is some $$K$$ such that $$\mathbb{P}(A_K)=0$$ which is wrong in general.

I interpret the event $$\left\{|X\left((m-1)2^{-j}\right)-X(m2^{-j})> 2^{-j\lambda} \text{ for some j\geq k and 1\leq m\leq2^j }\right\}$$ as $$\bigcup_{j=k}^\infty\ \bigcup_{m=1}^{2^j} \{|X\left((m-1)2^{-j}\right)-X(m2^{-j})> 2^{-j\lambda}\}$$ which helps me to understand what it means.

• Isn't this a Borel-Cantelli result? If so, you are in luck: all the applicaitons look the same, so all you have to do is become comfortable with one of them. – kimchi lover Apr 9 at 23:29