# Example for two stochastic processes $X$ and $Y$ such that $X_t = Y_t$ a.s. for all $t \in \mathbb{R}_+$, but not indistinguishable.

According to page 3 of J. Jacod and A. Shiryaev's Limit Theorems for Stochstic Processes (2013),

Note that if $$X$$ and $$Y$$ are indistinguishable, one has $$X_t = Y_t$$ a.s. for all $$t \in \mathbb{R}_+$$, but the converse is not true. This converse is true, however, when both $$X$$ and $$Y$$ are càd, or càg.

Also, they define indistinguishablility in 1.10 of their book:

1.10 A random set $$A$$ is called evanescent if the set $$\{ \omega: \exists{t \in \mathbb{R}_+} \text{ with }(\omega, t) \in A \}$$ is $$\mathbb{P}$$-null; two $$E$$-valued processes $$X$$ and $$Y$$ are called indistinguishable if the random set $$\{ X \neq Y\} = \{ (\omega, t): X_t(\omega) \neq Y_t (\omega) \}$$ is evanescent, i.e. if almost all paths of $$X$$ and $$Y$$ are the same.

However, I'm confused that $$X_t = Y_t$$ a.s. for all $$t \in \mathbb{R}_+$$ does not imply indistinguishability. Is there any example to show that the converse is not true?

Here, what I have done.

"$$X_t = Y_t$$ a.s. for all $$t \in \mathbb{R}_+$$" exactly means

\begin{align*} &\mathbb{P} \left\{ X_t = Y_t \text{ for all } t \in \mathbb{R}_+ \right\} = 1 \\ \iff & \mathbb{P} \left\{ X_t = Y_t \text{ for all } t \in \mathbb{R}_+ \right\}^c = 0 \\ \iff & \mathbb{P} \left\{ \exists{t \in \mathbb{R}_+}\text{ such that } X_t \neq Y_t \right\} = 0. \end{align*} Since the event set $$\left\{ \exists{t \in \mathbb{R}_+}\text{ such that } X_t \neq Y_t \right\}$$ is the same for $$\{ \omega: \exists{t \in \mathbb{R}_+} \text{ with } X_t(\omega) \neq Y_t(\omega) \}$$, I think they should be equivalent. Is there any mistake that I have? For example, I may misunderstood the statement "$$X_t = Y_t$$ a.s. for all $$t \in \mathbb{R}_+$$" which is actually meaning that for every fixed $$t \in \mathbb{R}_+$$, $$X_t = Y_t$$ a.s. for each. In this case, the indistinguishability is "$$X_t = Y_t$$ for all $$t \in \mathbb{R}_+$$ a.s.".

Thanks,

My definition of indistinguishable for $$X,Y$$ is $$P(\exists t \in T : X_t \ne Y_t) = 0$$. More precisely is $$\exists N \in \mathcal{A} : P(N) = 0$$ and $$\bigcup\limits_{t \in T} \{X_t \ne Y_t\} \subseteq N$$.
An other definition I have is : $$Y$$ is a modification of $$X$$ if $$\forall t \in T, \hspace{0.1cm} X_t = Y_t \hspace{0.1cm} P-a.s$$
What is true is that $$X,Y$$ being indistinguishable implies that $$X$$ is a modification of $$Y$$ since $$\forall t \in T \{ X_t \ne Y_t \} \subseteq \{\exists t \in T : X_t \ne Y_t\}$$.
That being said the converse is not true, i.e modification doesn't imply indistinguishable, since you have the conterexample with $$\Omega = ([0,1], \mathcal{B}(\mathbb{R}),\mathcal{L})$$ and $$X(x,t) = \begin{cases} 1 & x = t \\ 0 \end{cases}$$ and $$Y(x,t) \equiv 0$$