If $t\mapsto X_t$ is continuous a.s. does $\sup_{t\in [0,T]}|X_t|<\infty $ a.s.? Let $(X_t)$ a a.s. continuous process. Does $\sup_{t\in [0,T]}|X_t|<\infty $ a.s. ? 
I would say obviously yes because out of a null set, $|X_t|(\omega )\leq C$ for all $t\in [0,T]$, and thus $\sup_{t\in [0,T]}|X_t|(\omega )$ for all $\omega \notin N$, but I'm not really sure if my justification is correct. Can someone confirm ?
 A: The answer is yes, and your argument captures the reason why. With probability 1, $X_t$ is a continuous function, and all these paths are bounded. 
Note however that there you cannot expect that there is a common bound for all these paths (see e.g. Brownian motion).
A: The answer is Yes. However, there is a need to clarify notations.
Let $(\Omega,\mathcal{F},P)$ be a probability space. By an a.e. continuous
process $X$, we mean a functon $X:[0,\infty)\times\Omega\rightarrow\mathbb{R}$
such that:
(i) For each $t\in[0,\infty)$, $X(t,\cdot)$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable.
(ii) There exists $\Omega_{*}\in\mathcal{F}$ with $P(\Omega_{*})=1$
such that for each $\omega\in\Omega_{*}$, the function $t\mapsto X(t,\omega)$
is continuous.
In your argument, it is false that ``out of a null set, $|X_{t}(\omega)|\leq C$
for all $t\in[0,T]$'' because the dependence of $C$ on $\omega$
is not taken into account.
The correct argument is: Let $T\in[0,\infty)$ be fixed. Let $\Omega_{*}\in\mathcal{F}$
with $P(\Omega_{*})=1$ such that for each $\omega\in\Omega_{*}$,
the function $t\mapsto X(t,\omega)$ is continuous. For each $\omega\in\Omega_{*}$,
define $C_{\omega}=\sup_{t\in[0,T]}|X(t,\omega)|<\infty$. Now, let
$\omega\in\Omega_{*}$ be arbitrary, then $\sup_{t\in[0,T]}|X_{t}|(\omega):=\sup_{t\in[0,T]}|X(t,\omega)\leq C_{\omega}<\infty$.
