For a measurable stochastic process $X_t(\omega): \Omega \times [0,T] \to \mathbb{R}$, $\tilde{X}_{t}(\omega)$ is a modification of $X_t(\omega)$ if $\mathbb{P}(X_t = \tilde{X}_t) = 1$ for all $t \in [0,T]$.
Does this imply that $$\int_0^t X_s(\omega) ds = \int_0^t \tilde{X}_s(\omega) ds ~~\text{ almost all }\omega, \forall ~t\in[0,T]? $$
I tried to prove that the set $A = \{ \omega: X_t(\omega) \neq \tilde{X}_t(\omega) \text{ for a subset of } [0,T] \text{ of positive measure} \}$ has measure 0, but I didn't know how to.
I also tried to construct a counterexample, where a modification has a different integral, but I couldn't think of a way to make sure that for a subset of $\Omega$ of positive measure the two processes are different for a subset of $[0,T]$ of positive measure.

For the case that the processes $X$ and $\tilde X$ have both right continuous paths a.s., it is true. Because two right continuous modifications are indistinguishable (see Protter page 4). And if $X$ and $\tilde X$ are indistinguishable is clear that $\int_0^tX_sds = \int_0^t \tilde X_sds$ a.s.