Is this generating family of a measurable space of point measures a pi-system?

Question

I'm learning some probability and measure theory and working my way through the first few paragraphs of [1]. My question is perhaps too basic for Math Overflow, but I hope it is welcome here.

Point Measures: Let $X = (E, \mathscr{E})$ be a measurable space, and define a point measure on $X$ to be a measure $\omega$ determined by (or determining) a finite or countable subset $D$ of $E$ such that $\ \omega(A) = |A \cap D|$.

Let $M$ be a set of point measures over $X$. For each $A \in \mathscr{E}$ let $A_k = \{\omega \in M : \omega(A) \leq k\}$. Let $T = \{A_k : A \in \mathscr{E}, 0 \leq k \leq \infty\}$, and $T_0 = \{A_0 : A \in \mathscr{E}\}$.

Question: It turns out that $T_0$ is a $\pi$-system. Is $T$ also a $\pi$-system?

What I have figured out so far: Let $A, B \in \mathscr{E}$, and $k < j$. It is straightforward to show:

1. $A_k \subset A_j$.
2. If $A \subset B$ then $B_k \subset A_k$.
3. If $A_k \cup B_k \subset (A \triangle B)_0$ then $A_k = B_k$.
4. $A_0 \cap B_0 = (A \cup B)_0$.
5. $A_k \cap B_j \subset (A \cup B)_{k+j}$.
   • In general we don't have equality between these two sets. We have $A_0 \cap B_{j+k} \subset (A \cup B)_{k+j}$, but there are models for which $A_0 \cap B_{j+k} \subset A_j \cap B_k$ doesn't hold.
6. Let $0 \leq i \leq j - k$. Then $A_i \cap B_{j+k-i} \subset (A \cup B)_{j+k}$.

None of this stuff seems to help much. I have a suspicion that $T$ isn't a $\pi$-system. I would love to see a counterexample which confirms this, and would be surprised and delighted if it does turn out to be a $\pi$-system. I also like the $A_k$ subsets of $M$, and any background on these would be neat. It turns out that given a measurable function $f: X \rightarrow \mathbb{R}$ the function

$$ \omega \mapsto \int_E fd\omega = \sum_{x \in D} f(x)$$

is a measurable function from $(M, \sigma(T))$ to $\mathbb{R}$, a result which seems to depend on the countable nature of point measures, which I hope to verify soon.

Thanks, all!

Reference: [1] Finkelstein, Tucker, Veeh. "Point Processes Without Topology", Statistics, probability and game theory: Papers in honor of David Blackwell, 1996.

Answer 1

$T$ is not a $\pi$-system.

For example, take $E = \{a,b,c,d\}$ with $\mathcal{E} = \mathcal{P}(E)$, and let $A = \{a,b\}$, $B = \{c,d\}$. If we identify a point measure with the set on which it is supported, then $$A_1 \cap B_1 = \{\emptyset, \{a\}, \{b\}, \{c\}, \{d\}, \{a,c\}, \{a,d\}, \{b,c\}, \{b,d\}\}.$$ This is not of the form $C_k$ for any $C \subset E$, $k \ge 0$. I wrote a short program to verify this by brute force (there are only 64 possibilities or so) but you could also do it by hand with some case analysis.

Answer 2

It's not a $\pi $-system. For a counter example let $ A=\{0, 1\} $, $ B=\{2, 3\}$, $ u=v=1$. Show there is no $ C $ and $ w $ with $$ A_u\cap B_v= C_w $$ by first finding out what $ w $ would have to be.