I have read that an equivalent Definition of almost surely convergence may look like
$$\lim_{k\rightarrow\infty} P(\max_{k\leq n<\infty}\lVert \tilde{\theta}_n-\theta_0\rVert)>\epsilon)=0 \tag 1$$ where $\theta$ is a estimator and $\theta_0$ is a true value. I guess k comes from Lemma 1 in Chow and Teicher (1978, p. 66), which states that for k large enough, for any random variable $Y_n\rightarrow Y$ almost surely, for any $\epsilon, \delta>0$ $$P(\bigcap_{n=k}^\infty|Y_n-Y|\leq\epsilon)\geq 1-\delta,\ \forall k\geq k(\epsilon, \delta)$$
I'm trying to show this definition. What I have found is that $$Y_n\rightarrow Y \text{almost surely} \tag 2$$ iff $$P(\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty[|Y_n-Y|>\epsilon]) ... [ = 0 ] \tag 3$$ iff $$\lim_{k\rightarrow\infty}P(\sup_{n\geq k}|Y_n-Y|>\epsilon). ... [ = 0 ] \tag 4 $$
But I cant understand where this maximum come from since I only get a supremum
First, I have a minor suggestion to rewrite the $\max$ in (1) as $$\max_{n\in \mathbb N\cap [k,\infty)}$$ to emphasize that $k$ is fixed by the context and $\max$ is over all $n$ specified.
Now, when we clearly see the expression, it is maximum over an infinite which might not be defined as you correctly suggest.
Considering the context a hasty answer would be to say it is wrong since there is no guarantee the maximum exists and that ...
Most likely, the $\max$ is a typo or a result of ignorance.
Let now complete the hasty answer: there is no guarantee the maximum exists ... and an example is easy to construct -- !!?? -- Well, we fail to construct the counterexample. It turns out that, for arbitrary numbers $a_n, a$, we have $$a_n \to a$$ if an only if, $$\text{for every $k$, $m_k:=\max \{ |a_n-a|: n\ge k\}$ is well defined and $\lim m_k =0$.}$$
You can try yourself to prove that this equivalence holds true. You should also be able to apply this in the probability context. BTW, you forgot to explain what all that $Y$ and $\theta$ are and what is the meaning of $\|..\|$ and $|..|$.
Conclusion: Although (1) probably turns out to be correct, most likely the $\max$ is a typo or a result of ignorance.