# About modifications of right-continuous stochastic processes

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

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

• Thank you Did. Though I thought I knew why but i was looking for a more formal proof.I could not write it down . 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}