If $(X_t)_{t∈I}$ is a martingale and $τ$ is a stopping time with $τ≤T$, then $X_τ=\text E[X_T\mid\mathcal F_τ]$. Why does $I$ need to be countable?

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $I\subseteq\mathbb R$ be nonempty
• $\tau:\Omega\to I\cup\sup I$ be $\mathcal A$-measurable, $$\left\{\tau\le t\right\}\in\mathcal F_t\;\;\;\text{for all }t\in I\tag1$$ and $$\tau\le T\;\;\;\operatorname P\text{-almost surely}\tag2$$ for some $T\in I$
• $X$ be an $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$

Let $$A\in\mathcal F_\tau:=\left\{A\in\mathcal A:A\cap\left\{\tau\le t\right\}\in\mathcal F_t\text{ for all }t\in I\right\}\;.$$ By definition, $$A\cap \left\{\tau=t\right\}\in\mathcal F_t\;\;\;\text{for all }t\in I\tag3\;.$$ Now, \begin{equation}\begin{split}\operatorname E[1_AX_\tau] &=\sum_{t\:\in\:I\:\cap\:(-\infty,\:T]}\operatorname E[1_{A\:\cap\:\left\{\:\tau\:=\:t\:\right\}}X_t]\\ &=\sum_{t\:\in\:I\:\cap\:(-\infty,\:T]}\operatorname E[1_{A\:\cap\:\left\{\:\tau\:=\:t\:\right\}}\operatorname E[X_T\mid\mathcal F_t]]\\ &=\sum_{t\:\in\:I\:\cap\:(-\infty,\:T]}\operatorname E[1_A1_{\left\{\:\tau\:=\:t\:\right\}}X_T]\\ &=\operatorname E[1_AX_T]\;. \end{split}\tag4\end{equation}

I didn't say that $I$ is countable. However, we can find $(4)$ in any textbook only with the assumption of countability of $I$. So, what's going wrong in $(4)$ if $I$ is not countable?

Answer 1

If $I$ is not countable, you will sum over an uncountably infinite number of terms all of which will have measure zero in general. So in the continuous-time case you have to be much more careful and make assumptions on $X$ ((right-)continuity) that will essentially allow you to work with countably many sets instead of uncountably many. There are definitely versions for uncountable $I$, e.g. here: https://almostsure.wordpress.com/2009/12/20/optional-sampling/

Answer 2

Equation $(4)$ is not true if $I\cap(-\infty,T]$ is uncountable. Specifically, your first equality is actually using countable additivity of measures, which of course requires the index set to be countable. It doesn't really have much to do with $\{\tau=t\}$ having measure zero, although it is not difficult to show that $\{t:\mathrm P[\tau=t]>0\}$ is at most countable for any random variable $\tau$.