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?

  • $\begingroup$ 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. $\endgroup$ – Mau314 Oct 16 '17 at 6:55
  • $\begingroup$ It's not uncountable sets that are a problem, it's uncountable products of uncountable sets, which are unmeasurable under the Borel topology on $\mathbf{R}^{[0,\infty)}$. $\endgroup$ – 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 $V\notin \mathcal{B}_{[0,1]}$.

  • $\begingroup$ So this means that the set is measurable, and has measure 1 in the completion of the measure on $\mathbf{R}^{[0,\infty)}$? $\endgroup$ – Jacob Denson Oct 16 '17 at 18:32
  • $\begingroup$ @JacobDenson What set are talking about? $\endgroup$ – d.k.o. Oct 16 '17 at 20:51
  • $\begingroup$ The set of sample points where the functions are equal at all time points. $\endgroup$ – Jacob Denson Oct 16 '17 at 21:01
  • $\begingroup$ 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... $\endgroup$ – d.k.o. Oct 16 '17 at 21:24
  • $\begingroup$ But what you say amounts to saying the set has inner probability measure 1, and therefore trivially has outer probability measure 1, and therefore the set is measurable in the completion of the sigma algebra with respect to the induced probability measure? $\endgroup$ – Jacob Denson Oct 17 '17 at 18:55

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