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?


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 $f(x)-g(x)=0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.