Is this random Lebesgue-integral well-defined? Let 
$$ X : [0,T] \times \Omega \rightarrow \mathbb{R} $$
be an almost-surely continuous stochastic process.
Then how is the random Lebesgue-integral
$$ \omega \mapsto \int_{0}^{T} X_t(\omega ) dt \quad (\dagger) $$
well defined for all (not merely almost all) $\omega \in \Omega$? What happens with $(\dagger)$ if $\omega$ is such that $X_t(\omega)$ is not continuous in $t$ ?
 A: It seems to me that the question is motivated by a fundamental misunderstanding of the specifics of probability theory, of its tools and of its aims.
We usually work with continuous functions whenever we are interested in their values at various points of the space under consideration. In order to obtain results that would be difficult to get in the continuous setting, which proves to be too constraining now, one loosens further this already loose setting and escapes in the surrounding, much larger and weaker Borel (or just measurable, in general) setting. Here, though, the tools available are too weak to give pointwise results, but are perfectly adapted to give results involving integrals, i.e. averages of functions. And since integration does not "feel" negligible sets, it is always of great practical value to modify our functions on them; it might look like a minor thing to do, but this turns out to be both natural in this context, and the key to success.
Using your notations, we know that the subset $C = \{ \omega \in \Omega \mid [0,1] \ni t \mapsto X_t (\omega) \in \mathbb R \text{ is continuous} \}$ has measure $P(C) = 1$, where $P$ is the probability on $\Omega$. We would like to define a random variable (i.e. a measurable function) $I : \Omega \to \mathbb R$ that captures the idea of "Riemann integrals of the trajectories". The problem is that not all the trajectories are Riemann-integrable, so what do we do? Well, we remember what has been said in the previous paragraph: for our purposes (that later on will involve integration) we don't really care about what happens on negligible sets, and fortunately $\Omega \setminus C$ is so! This allows us to ignore this subset and define $I (\omega) = \int _0 ^1 X_t (\omega) \ \mathrm d t$ for $\omega\in C$, and $I(\omega) = 0$ for $\omega \in \Omega \setminus C$.
Was this the only possible choice? No, for sure, there are infinitely many of them - but any two of them differ only on a negligible subset, that integration does not "feel". Why did we work with $0$ on $\Omega \setminus C$? Because that is the simplest value to use. If we had had another measurable function $f : \Omega \to \mathbb R$ available, then $f \big| _{\Omega \setminus C}$ would have been measurable, so we could have defined $I$ to coincide with $f$ on $\Omega \setminus C$ - this wouldn't have made any difference from the point of view of integration on $\Omega$. But did we have such a function available? No. Plus, why use a more complicated function that the simplest one imaginable, i.e. the constant $0$, given that it doesn't bring any benefit?
Was the decomposition of $\Omega$ into $C$ and its complementary the only possible one for the definition of $I$? Of course not, there are infinitely many other decompositions available, but any useful two of them will again differ by a negligible subset. We could have defined $R = \{ \omega \in \Omega \mid [0,1] \ni t \mapsto X_t (\omega) \in \mathbb R \text{ is integrable} \}$ (notice that $R \supseteq C$) and defined $\tilde I (\omega) = \int _0 ^1 X_t (\omega) \ \mathrm d t$ for $\omega\in R$, and $\tilde I(\omega) = 0$ for $\omega \in \Omega \setminus R$. This, though, doesn't change the story much, because $R \setminus C \subseteq \Omega \setminus C$ which is negligible, so $I$ and $\tilde I$ coincide on $C$ (which is of full measure) and differ, again, only on a negligible subset.
Two conclusions are to be retained from the above discussion:


*

*for all the practical purposes of probability theory we don't care if two random variables differ on a negligible subset, because sooner or later you will integrate them anyway (which forgets about negligible subsets), so we just treat them as being equal; in particular, if $P(\{\omega\}) = 0$ then it makes no sense to ask about the value of a random variable at $\omega$ - it could be anything, since $\{\omega\}$ is negligible;

*if you have a measurable function defined on some subset of full measure, extend it to the whole of $\Omega$ by defining it on the complementary to be equal to some arbitrary point in the state space; if your state space is a vector space, define it to be $0$.
