Doesn't the definition of Indistinguishability of Stochastic Processes Depend on a Non-Measurable Set?

In the theory of continuous time stochastic processes, two processes $X$ and $Y$ are known as indistinguishable if $\mathbf{P}(\forall t: X_t = Y_t) = 1$. However, if $X$ and $Y$ are defined on the same sample space $\Omega$, then isn't the set

$$\{ \omega \in \Omega: [\forall t: X_t(\omega) = Y_t(\omega)] \}$$

a non-measurable set of $\mathbf{R}^{[0,\infty)}$, since it measures the values of $X$ and $Y$ at uncountably many index points $t$, and therefore cannot be a $\sigma$ cylinder? In this case, how does the definition of an indistinguishable process make sense, since the probability of a non-measurable set need not be defined?

• Well, firstly the set you give is a subset of Omega, so technically without further information we cannot say what it has to with e.g. R. Secondly, I feel that you confuse uncountability with non-measurability - e.g. clearly R is a Borel set. – Mau314 Oct 16 '17 at 6:55
• It's not uncountable sets that are a problem, it's uncountable products of uncountable sets, which are unmeasurable under the Borel topology on $\mathbf{R}^{[0,\infty)}$. – Jacob Denson Oct 16 '17 at 17:01

$X$ and $Y$ are indistinguishable if there exists a $\mathsf{P}$-null set $N$, s.t. for all $\omega\notin N$ and $t\in \mathbb{T}$, $X_t(\omega)=Y_t(\omega)$. Also the set $I:=\{\omega\in\Omega:[\forall t: X_t(\omega)=Y_t(\omega)]\}$ need not be non-measurable. For example, if $\mathbb{T}=\mathbb{R}_{+}$ and $X$ and $Y$ are real-valued stochastic processes with continuous paths, then $I=\left\{\omega\in\Omega:[\forall t\in \mathbb{Q}_{\ge 0}:X_t(\omega)=Y_t(\omega)]\right\}$ is measurable.
Here is a simple example. Let $(\Omega,\mathcal{F},\mathsf{P})=([0,1],\mathcal{B}_{[0,1]},\mu)$, where $\mu$ is the Lebesgue measure, and $\mathbb{T}=[0,1]$. Define $X_t(\omega)=1_{\{t\}\cap V}(\omega)$, where $V$ is a subset of $[0,1]$ and $Y_t(\omega)\equiv 0$. Note that $I=V$ in this case. Thus, $X$ and $Y$ are indistinguishable if $V$ is a $\mu$-null set and $I$ is non-measurable if $V\notin \mathcal{B}_{[0,1]}$.
• So this means that the set is measurable, and has measure 1 in the completion of the measure on $\mathbf{R}^{[0,\infty)}$? – Jacob Denson Oct 16 '17 at 18:32
• Nope. This set may not be measurable as well. Consider a canonical construction, i.e. $\Omega=R^{[0,\infty)}$, $\mathcal{F}$ is a cylinder $\sigma$-algebra, and $X_t(f)=f(t)$, $f\in \mathbb{R}^{[0,\infty)}$. Then, as you mentioned in the question, this set is non-measurable... – d.k.o. Oct 16 '17 at 21:24