Let $\{X_t, t\in[0,T]\}$ on {$R, \mathfrak B(R) $} be random, almost surely continuous, function. Show that $X^+=\sup_{t \in[0,T]} X_t$ is random variable.
My proof:
Let $P$ be set points of no continuity of $X_t$.
And let $\widetilde X_t =\left\{
\begin{array}{c}
X_t(\omega), \omega\in \mathbb R\setminus P \\
0, \omega\in P
\end{array}
\right. $
sup$\widetilde X_t$ will be finite and: $ \{\omega: \sup_{t \in[0,1]}\widetilde X_t>x\}=\bigcup_{t\in [0,T]\bigcap Q}\{\omega:\widetilde X_t >x\}$ Since $\{\omega:\widetilde X_t >x\}$ is in $\mathfrak B(R)$ then $\{\omega : \sup_{t \in[0,1]}\widetilde X_t>x\}$ in $\mathfrak B(R)$.
And since the intervals $(x, +\infty)$ form $\mathfrak B(R)$,$\quad$$\widetilde X_t$ is random variable.
Because sup$_{t \in[0,T]}\widetilde X_t$=sup $_{t \in[0,T]}X_t$ $\quad$ $\forall \omega\in \mathbb R\backslash P$,$\quad$ sup $_{t \in[0,T]}X_t$ is random variable too.