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?


2 Answers 2


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 . $\endgroup$ Commented Oct 14, 2014 at 20:32

Suppose that $X,Y$ are right-continuous, then
\begin{align} \left\{ X_t = Y_t,\ \forall t \geq 0 \right\} = \bigcap_{r \in \mathbb Q \cap [0,\infty)} \left\{ X_r = Y_r \right\} \end{align} but $P(X_r = Y_r) = 1\ (\forall r \geq 0)$, then \begin{align} P(X_t = Y_t,\ \forall t \geq 0) = P \biggl( \bigcap_{r \in \mathbb Q \cap [0,\infty)} \left\{ X_r = Y_r \right\} \biggr) = 1. \end{align}


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