Why can we consider the Brownian motion as being a mapping into the space of continuous functions, even though its paths are only a.s. continuous? Let $B=(B_t)_{t\ge 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\operatorname{P})$. By definition of $B$, for $\operatorname{P}$-almost every $\omega\in\Omega$ $$[0,\infty)\to\mathbb{R}\;,\;\;\;t\mapsto X_t(\omega)\tag{1}$$ is continuous. Generally, a stochastic process $X=(X_t)_{t\in I}$ on $(\Omega,\mathcal{A})$ with $I\subseteq\mathbb{R}$ can be viewed as a mapping $$X:\Omega\mapsto\mathbb{R}^I\;,\;\;\;\omega\mapsto \left(t\mapsto X_t(\omega)\right)\tag{2}$$ I've frequently read that $B$ is considered to be a mapping $\Omega\to C\left([0,\infty)\right)$, where $C(I)$ is the space of continuous functions $I\to\mathbb{R}$.

Why can we do that? Clearly, there exists a $\operatorname{P}$-null set $N\subseteq\mathcal{A}$ such that $(1)$ is continuous for all $\omega\in\Omega\setminus N$. Moreover, I know that we can alter measurable functions on null sets without changing their measure related properties. However, is it guaranteed that we can alter $B$ on all null sets on which $(1)$ is not continuous such that $(1)$ is continuous for all $\omega\in\Omega$?

Remark: Maybe we can use the Kolmogorov-Chentsov theorem to prove that $(1)$ can indeed be assumed as continuous for all $\omega\in\Omega$. The theorem can be formulated as follows:
Let $X=(X_t,t\ge 0)$ be a real-valued stochastic process such that for all $T>0$, there exists $\alpha,\beta,C>0$ with $$\operatorname{E}\left[\left|X_t-X_s\right|^\alpha\right]\le C|t-s|^{1+\beta}\;\;\;\text{for all }s,t\in [0,T]]$$ Then, there exists a modification of $X$ which is locally Hölder-continuous of order $\gamma\in \left(0,\frac \beta\alpha\right)$.
Stochastic processes $X,Y$ are called modifications of each other, if $X_t=Y_t$ almost surely.
 A: Strictly speaking, in probability theory you are free to excise null sets from the probability space. So you could define Brownian motion with a.s. continuous paths and then excise the null set where the paths are not continuous. Now the paths are always continuous. The process we've made is actually indistinguishable from the a.s. continuous process: their entire trajectories are equal with probability $1$. This is a stronger notion than just being modifications of one another. 
It is important in this procedure that we are not excising a null set for each $t$, we are excising a single null set which encapsulates all the paths that have any discontinuities at any point. Since this point is so important, I will rephrase it again: Brownian motion is not merely continuous at each fixed point of time a.s., it is continuous over all time a.s.
A: Besides excising your null set on which the paths are discontinuous, you can also redefine your process to take the constant value 0 (or any other fixed continuous function) on the null set. 
