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Lemma : Let $X$ and $X'$ be two right continuous(or left continuous) processes defined on the same probability space $(\Omega,F,P)$ be a modification then the two processes are indistinguishable.

Proof: Consider $$A=\{\omega|X_t(\omega)=X'_{t}(\omega) ,\forall t \in \mathbb{}Q^+\}.$$ Since $X$ and $X'$ are modifications, we have that $\forall t$ $X_t(\omega)=X'_{t}(\omega) $ P a.s As $A^C$ is a countable union of sets of probability measure zero ( follows from the assumption that the two process are a modification), we have that $P(A^C)=0$, hence $P(A)=1$. $\Box$

I do not understand why right continuity implies that if we sum all $\omega$ over the reals , the new set A would have a probability 1 too. How do I rigorously prove this?

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You are asking why two right-continuous functions $f$ and $g$ such that $f(q)=g(q)$ for every $q$ in $\mathbb Q$, coincide. Well, note that, for every $x$ not in $\mathbb Q$, $f(x)-g(x)$ is the limit of $f(q)-g(q)$ when $q\to x$, $q\gt x$, $q$ in $\mathbb Q$, hence $f(x)-g(x)=0$.

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  • $\begingroup$ Thank you Did. Though I thought I knew why but i was looking for a more formal proof.I could not write it down . $\endgroup$ – user3503589 Oct 14 '14 at 20:32

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