Properties of Random and Stopping Sets Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $S$ be a countable set, and we define a family of sub-$\sigma$-algebras $(F_A)_{A \subseteq S}$ such that $A \subseteq B \Rightarrow \mathcal{F}_A \subseteq \mathcal{F}_B$ for all $A, B \subseteq S$. We define a random subset $\zeta$ of $S$ to be a mapping $\zeta : \Omega \to \mathcal{P}(S)$ such that $\{\zeta \subseteq A\} \in \mathcal{F}$ for all $A \subseteq S$. Finally, we say that $S$ is a stopping set if $\{\zeta \subseteq A\} \in \mathcal{F_A}$ for all $A \subseteq S$. I have two problems:


*

*Show that $\zeta$ is a stopping set $\Leftrightarrow$ $\{\zeta = A\} \in \mathcal{F}_A$ for all $A \subseteq S$.

*For two stopping sets $\zeta$ and $\eta$, are $\zeta \cup \eta$ and $\zeta \cap \eta$ necessarily stopping sets as well?



For the first one, the forward direction is straightforward as $S$ is countable, so $A$ is countable. We can simply observe that $\{\zeta = A\} = \{\zeta \subseteq A\} \setminus \bigcup_{a \in A} \{\zeta \subseteq A \setminus \{a\}\} \in \mathcal{F}_A$. The converse is however much less straightforward, as the number of subsets of $A$ may be uncountable (e.g. $A = S$). Since $\sigma$-algebra only guarantees closure for countable unions, I'm unsure how to proceed.
For second, the first part is also straightforward, with the observation that $\{\zeta \cup \eta \subseteq A\} = \{\zeta \subseteq A\} \cap \{\eta \subseteq A\}$. I know the answer for the second part is false, but I'm not sure how to construct an explicit counterexample.
Thanks in advance.
 A: $\newcommand{\si}{\sigma}
\newcommand{\om}{\omega}
\newcommand{\Om}{\Omega}
\newcommand{\F}{\mathcal{F}}$
The answer to your first question is negative. E.g., let $\Omega:=2^S=\mathcal P(S)$, with $\mathcal F:=2^\Omega$, and let $\zeta$ be the identity map of $\Omega$, so that $\zeta(A)=A$ for all $A\subseteq S$. For each $A\subseteq S$, let 
\begin{equation}
 \F_A:=\si(\big\{\{C\}\colon C\subseteq A\big\})
 =\si(\big\{\{C\}\colon C\in2^A\big\})=\si(\big\{\{C\}\colon\{C\}\subseteq 2^A\big\}),
\end{equation}
where (for any set $\mathcal E\subseteq2^\Om$) $\si(\mathcal E)$ denotes, as usual, the smallest sigma-algebra $\mathcal G$ over $\Om$ such that $\mathcal G\supseteq\mathcal E$. So, here $\mathcal F_A$ is the smallest sigma-algebra containing/generated by the set of all singleton sets of the form $\{C\}$ with $C\subseteq A$. 
It is easy to see that 
\begin{equation}
 \F_A=\{\Phi\colon\Phi\subseteq2^A\text{ and $\Phi$ is countable}\}
 \cup
\{\Phi\colon\Om\setminus\Phi\subseteq2^A\text{ and $\Om\setminus\Phi$ is countable}\}. 
\end{equation}
Moreover, $\mathcal F_A\subseteq\mathcal F_B\subseteq\mathcal F$ for all subsets $A$ and $B$ of $S$ such that $A\subseteq B$.  
Furthermore, what you denote informally by $\{\zeta=A\}$ is $\zeta^{-1}(\{A\})=\{A\}\in\mathcal F_A$ for all $A\subseteq S$. 
However, what you denote informally by $\{\zeta\subseteq A\}$ is $\zeta^{-1}(2^A)=2^A\notin\mathcal F_A$ for any infinite $A\subsetneq S$, because then neither $2^A$ nor $\Omega\setminus2^A$ is countable. So, $\zeta$ is not a stopping set. 

A counterexample for your second question is as follows. Let $S:=U\times V$, where $U$ and $V$ are any sets of cardinalities $\ge2$ (you may assume that $U$ and $V$ are denumerable and hence $S$ is also denumerable). Let $\Omega:=2^S$, with $\mathcal F:=2^\Om$. For any $\om\in\Om$, let 
\begin{equation}
 \zeta(\om):=p_1(\om)\times V,\quad\eta(\om):=U\times p_2(\om), 
\end{equation}
where 
\begin{equation}
 p_1(\om):=\{u\in U\colon\exists v\in V\ (u,v)\in\om\}\quad\text{and}\quad
 p_2(\om):=\{v\in V\colon\exists u\in U\ (u,v)\in\om\}, 
\end{equation}
the projections of the set $\om$ onto $U$ and $V$. Then 
\begin{equation}
 (\zeta\cap\eta)(\om)=\zeta(\om)\cap\eta(\om)=p_1(\om)\times p_2(\om)
\end{equation}
for all $\om\in\Om$. 
For any $A\subseteq S$, let 
\begin{equation}
 \F_A:=\si(\{\zeta^{-1}(2^C)\colon C\subseteq A\}
 \cup \{\eta^{-1}(2^C)\colon C\subseteq A\}). 
\end{equation}
Then $\mathcal F_A\subseteq\mathcal F_B\subseteq\mathcal F$ for all subsets $A$ and $B$ of $S$ such that $A\subseteq B$. Also, obviously $\zeta^{-1}(2^A)\in\F_A$ and $\eta^{-1}(2^A)\in\F_A$, for any $A\subseteq S$, so that $\zeta$ and $\eta$ are stopping sets. 
Take now any $(x,y)\in S=U\times V$ and let $A$ be the singleton set $\{(x,y)\}$. Then for any $C\subseteq A$
\begin{equation}
 \zeta^{-1}(2^C)=\{\om\in\Om\colon p_1(\om)\times V\subseteq C\}=\{\emptyset\}
\end{equation}
and similarly $\eta^{-1}(2^C)=\{\emptyset\}$, whence
\begin{equation}
 \F_A=\si\big(\big\{\{\emptyset\}\big\}\big)
 =\big\{\emptyset,\{\emptyset\},\Om,\Om\setminus\{\emptyset\}\big\}. 
\end{equation}
On the other hand, the set 
\begin{equation}
 (\zeta\cap\eta)^{-1}(2^A)
 =\{\om\in\Om\colon p_1(\om)\times p_2(\om)\subseteq\{(x,y)\}\}
 =\big\{\emptyset,\{(x,y)\}\big\}  
\end{equation}
is of cardinality $2$ and hence is not in $\F_A$, because none of the four members of $\F_A$ is of cardinality $2$. Thus, $\zeta\cap\eta$ is not a stopping set. 
