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.

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    $\begingroup$ 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. $\endgroup$ – kimchi lover Apr 9 at 23:29

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