Suppose $z(t,k,\omega)$ is a sequence of index $k\in\mathbb N$ of stochastic process continuous in $t\in[0,T]$ for every $\omega\in\Omega$ where $\Omega$ is the sample space. Is following the correct negation of $\sup_{t\in[0,T]}|z(t,k,\omega)|\rightarrow 0$ as $k\rightarrow\infty$ almost surely?
$\sup_{t\in[0,T]}|z(t,k,\omega)|\not\rightarrow 0$ as $k\rightarrow\infty,\,\forall\omega\in A$ for some $A\subset\Omega$ with $P(A)>0$.
The main question is: is $A$ the non-converging set necessarily measurable?
If $z(t,k)$ is continuous in $t$, $\sup_{t \in [0,T]} |z(t,k)| = \sup_{t \in [0,T] \cap \mathbb Q} |z(t,k)|$ where $\mathbb Q$ is the rationals. This fails to have limit $0$ as $k \to \infty$ if and only if there exists $m \in \mathbb N$ such that for all $N \in \mathbb N$ there exist $k \in \mathbb N$ and $t \in [0,T] \cap \mathbb Q$ such that $k > N$ and $|z(t,k)| > 1/m$. Thus if each $z(t,k)$ is a measurable function, the set $A$ is a countable union of countable intersections of countable unions of measurable sets, and therefore is measurable.