Suppose that a stochastic process $X(t,\omega)$, $t\in T$, $\omega\in\Omega$, is almost surely continuous: $X(\cdot,\omega):T\to\mathbb R$ is continuous, for $\omega\in\tilde\Omega\subseteq\Omega$ with $P(\tilde\Omega)=1$. My question is whether $X$ has a version $\tilde{X}$ on $T\times\Omega$ such that $\tilde{X}(\cdot,\omega):T\rightarrow\mathbb{R}$ is continuous, for all $\omega\in\Omega$.
My idea is to redefine $X(\cdot,\omega)$ as any continuous function on $T$, for $\omega\notin\tilde{\Omega}$. Is this correct?
$\def\Ω{{\mit Ω}}$Suppose that $\widetilde{X}: T × \Ω → \mathbb{R}$ is a process defined on the same probability space as $X$ and $\widetilde{X}|_{T × \widetilde{\Ω}} = X|_{T × \widetilde{\Ω}}$ with $\widetilde{X}({\,·\,}, ω)$ being continuous for any $ω \in \Ω$.
Note that $P(A \cap \widetilde{\Ω}) = P(A)$ for any event $A$. For any $T_0 \subseteq T$ with $|T_0| < ∞$, suppose $T_0 = \{t_1, \cdots, t_n\}$. For any $B_1, \cdots, B_n \in \mathscr{B}(\mathbb{R})$,$$ P(\widetilde{X}_{t_1} \in B_1, \cdots, \widetilde{X}_{t_n} \in B_n) = P\left( \bigcap_{k = 1}^n \{\widetilde{X}_{t_k} \in B_k\} \right) = P\left( \bigcap_{k = 1}^n (\{\widetilde{X}_{t_k} \in B_k\} \cap \widetilde{\Ω}) \right)\\ = P\left( \bigcap_{k = 1}^n (\{X_{t_k} \in B_k\} \cap \widetilde{\Ω}) \right) = P\left( \bigcap_{k = 1}^n \{X_{t_k} \in B_k\} \right) = P(X_{t_1} \in B_1, \cdots, X_{t_n} \in B_n). $$ Since $\mathscr{B}(\mathbb{R}^n) = σ((\mathscr{B}(\mathbb{R}))^n)$, then for any $B \in \mathscr{B}(\mathbb{R}^n)$,$$ P((\widetilde{X}_{t_1}, \cdots, \widetilde{X}_{t_n})^{\mathrm{T}} \in B) = P((X_{t_1}, \cdots, X_{t_n})^{\mathrm{T}} \in B), $$ which implies $(\widetilde{X}_{t_1}, \cdots, \widetilde{X}_{t_n})^{\mathrm{T}} \stackrel{\mathrm{d}}{=} (X_{t_1}, \cdots, X_{t_n})^{\mathrm{T}}$. Therefore, $\widetilde{X}$ is a version of $X$.
For any $t \in T$, since $\widetilde{\Ω} \subseteq \{\widetilde{X}_t = X_t\}$ and $P(\widetilde{\Ω}) = 1$, then $P(\widetilde{X}_t = X_t) = 1$. Thus $\widetilde{X}$ is also a modification of $X$.