In exercise 6.1 in Billingsley's Probability and Measure, one is asked to prove that
$Z_n\rightarrow Z$ with probability 1 if and only if for every positive $\epsilon$, there exists and $n$ such that $P[|Z_k-Z|<\epsilon, n\leq k \leq m] > 1-\epsilon$ for all $m$ exceeding $n$.
I'm unclear what the notation $P(|Z_k - Z|<\epsilon, n\leq k\leq m)$ means. Does it mean \begin{align*} P\left(\bigcap_{n=k}^m |Z_k-Z| < \epsilon\right) \end{align*} ? Thank you in advance.
i think you're on the right track but a little bit wrong. The probability expression $ P(|Z_k - Z| < \epsilon, n \leq k \leq m) > 1-\epsilon $ is essentially stating that as you consider random variables $ Z_k $ with indices in the range from $ n $ to $ m $, the probability that the difference between $ Z_k $ and the limit $ Z $ is less than $ \epsilon $ is is siginificant so the convergence holds true
For convergence to hold true, you want this probability to be close to 1 even if $ m $ becomes arbitrarily large compared to $ n $. In other words, as you consider more and more terms in the sequence $ Z_k $, the probability that $ Z_k $ gets close to $ Z $ within a margin of $ \epsilon $ will remain significant .
as someone mentioned in comments the right way to interpret it as
$$ P\left(\bigcap_{n=k}^m \{|Z_k - Z| < \epsilon\}\right) $$
i hope it helps :)