Proof of The Supremum of a Bounded Right-Continuous Process is a Random Variable

Step 1: $\overline{Z}$ is a random variable. For each $n$ let $D_n=\{kT2^{-n}:k\in\{0,1,\dots,2^n\}\}$, a finite set, so that $D=\bigcup_nD_n$ and $D_n\subseteq D_{n+1}$, the latter because $kT2^{-n}=(2k)T2^{-(n+1)}$. Enumerate $D$ as a sequence $(t_i)_{i\in\mathbb{N}}$ by listing the finitely many points of $D_0$, then those of $D_1$, and so on, each in increasing order; every point of $D$ occurs. Each $|Z_{t_i}|\,\mathbf{1}_{\Omega_0}$ is a random variable. Indeed $|Z_{t_i}|$ is one, being the composition of the sequentially continuous map $c\mapsto|c|$ with a random variable, by measurability of sequentially continuous functions of measurable Euclidean maps; and for a real number $c$ the set $\{|Z_{t_i}|\,\mathbf{1}_{\Omega_0}>c\}$ equals $\Omega$ when $c<0$, since the function is nonnegative, and equals $\Omega_0\cap\{|Z_{t_i}|>c\}$ when $c\ge0$, so it lies in $\mathcal{F}$ in either case. Moreover $|Z_{t_i}|\,\mathbf{1}_{\Omega_0}$ is bounded in absolute value by $K$ at every point of $\Omega$: on $\Omega_0$ by hypothesis 1, and off $\Omega_0$ it vanishes. Hence, by measurability of countable suprema,

is a random variable, and $0\le\overline{Z}\le K$.

Step 2: the supremum over $D$ is the supremum over $[0,T]$ on $\Omega_0$. Fix $\omega\in\Omega_0$. There $\mathbf{1}_{\Omega_0}(\omega)=1$, so $\overline{Z}(\omega)=\sup_{t\in D}|Z_t(\omega)|$, and since $D\subseteq[0,T]$ this gives $\overline{Z}(\omega)\le\sup_{t\in[0,T]}|Z_t(\omega)|$.

For the reverse inequality, let $t\in[0,T]$ and $\varepsilon>0$. If $t=T$ then $T\in D_0\subseteq D$, so $|Z_t(\omega)|\le\overline{Z}(\omega)$. If $t=0$ then $0\in D_0$ and the same holds. Suppose $0<t<T$. By right-continuity at $t$ choose $\eta>0$ such that $|Z_s(\omega)-Z_t(\omega)|\le\varepsilon$ whenever $t\le s\le\min(t+\eta,T)$. Choose a natural number $n$ with $T2^{-n}<\min(\eta,T-t)$, and let $k$ be the least element of $\{0,1,\dots,2^n\}$ with $kT2^{-n}\ge t$; such a $k$ exists because $2^nT2^{-n}=T\ge t$, and $k\ge1$ because $t>0$. Minimality gives $(k-1)T2^{-n}<t$, hence

using $T2^{-n}<\eta$ and $T2^{-n}<T-t$. Therefore $s=kT2^{-n}$ lies in $D$ and satisfies $t\le s\le\min(t+\eta,T)$, so

As $\varepsilon>0$ was arbitrary, $|Z_t(\omega)|\le\overline{Z}(\omega)$ for every $t\in[0,T]$, and taking the supremum over $t$ gives the reverse inequality. $\blacksquare$