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The problem says

For a non-empty set $\Omega$, recall that a $\pi$-system $\mathcal{P}$ is a collection of subsets of $\Omega$ that is closed under finite intersections. Let $\mathcal{A}$ be the smallest algebra over the $\pi$-system $\mathcal{P}$. Show that probability measures agreeing on $\mathcal{P}$ must also agree on $\mathcal{A}$.

Solution: It says that I should use inclusion-exclusion principle. To be honest, I don't know exactly how I can begin. When it says probability measures agreeing on $\mathcal{P}$ must also agree on $\mathcal{A}$, I don't know how to write out as a set. For sure, $\mathcal{P} \subset \mathcal{A}$, so I should worry about subsets that are not in $\mathcal{D}$ but are in $\mathcal{A}$.

Thanks for any help/extra hint.

You meant something like this: let $\mathcal{P}$ be a $\pi$-system of subsets of $\Omega$ such that $\mathcal{G}:=\mathcal{A}(\mathcal{P})$ (the smallest algebra containing $\mathcal{P}$). Following what you said, we should prove that $$ \mathcal{L}=\{A \in \mathcal{G}:\mathbb{P}_1(A)=\mathbb{P}_2(A)\} \quad \textrm{where }\mathbb{P}_1, \mathbb{P}_2 \textrm{ are two probability measures on }\mathcal{A} $$ is a $\lambda$-system...

Charles Seife
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  • If you're familiar with Dynkin-($\lambda$-)systems, you can show that (1) the set of all sets where the probability measures agree, is a Dynkin-system; (2) the Dynkin-system generated by $P$ and the $\sigma$-algebra generated by $P$ are equal, provided $P$ is intersection-stable. – Juno Kim Oct 12 '19 at 19:13
  • Yes, actually the $\lambda$-system part is easy: the cases of empty set, complement and countable union follow trivially from properties of probabilty measures. (2), Dynkin's $\pi - \lambda$ theorem, is trickier. There is an elegant proof using more Dynkin-systems, but I can't remember it off the top of my head... – Juno Kim Oct 12 '19 at 20:16
  • I was reading a theorem in probability-1 by Albert Shiryaev pag. 171 which says if $\mathcal{E}$ is a $\pi$-system, $\mathcal{L}$ is a $\lambda$-system and $\mathcal{E} \subset \mathcal{L}$. Then $\sigma(\mathcal{E})\subset \mathcal{L}.$ – Charles Seife Oct 12 '19 at 20:34
  • Well then there you have it. If you put $\mathcal{L}$ as the smallest $\lambda$-system containing $\mathcal{P}$, the inclusion in the other direction holds trivially since $\mathcal{L}$ is the intersection of all $\lambda$-systems containing $\mathcal{P}$, which automatically includes all $\sigma$-algebras containing $\mathcal{P}$ (every $\sigma$-algebra is a $\lambda$-system). – Juno Kim Oct 13 '19 at 02:29

1 Answers1

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Let $\mu, \nu$ be two probability measures that agree on $\mathcal P.$ Without loss of generality, assume $\varnothing, \Omega \in \mathcal P.$ Let $\mathcal B$ be the set of all finite unions $A_1 \cup \cdots \cup A_n$ for $A_1,\dots, A_n \in \mathcal P,\ n \ge 1.$ Let's show that both measures agree on $\mathcal B.$

$$\mu(A_1 \cup \cdots \cup A_n) = \sum_{1 \le i_1 < \cdots < i_k \le n \\ k = 1,\dots,n} (-1)^{k-1} \mu(A_{i_1} \cap \cdots \cap A_{i_k}) = \sum_{1 \le i_1 < \cdots < i_k \le n \\ k = 1,\dots,n} (-1)^{k-1} \nu(A_{i_1} \cap \cdots \cap A_{i_k})= \nu(A_1 \cup \cdots \cup A_n). $$

An overcomplicated formula for a simple fact: because $\mathcal P$ is a $\pi$-system, the two measures coincide on the finite intersections $A_{i_1} \cap \cdots \cap A_{i_k},$ and the measure of the union $A_1 \cup \cdots \cup A_n$ is determined by those (by the principle of inclusion-exclusion). Hence the measures agree on $\mathcal B.$

Note that $\mathcal B$ is a lattice. That is, the union of two sets in $\mathcal B$ is in $\mathcal B$ (which is obvious from the definition) and the intersection of two elements in $\mathcal B$ is in $\mathcal B$ (because intersection distributes over union and $\mathcal P$ is a $\pi$-system).

Now, I am going to proceed using induction, because sets in $\mathcal A$ can be written by applying finitely many set operations to the sets in $\mathcal P.$ However, if possible I recommend a more structured approach, where we directly specify what the sets in $\mathcal A$ look like (I haven't found a simple expression).

Let $\mathcal P'$ consist of all the differences $A \setminus B$ for $A, B \in \mathcal B.$ Then $\mathcal P'$ is a $\pi$-system. For, whenever we are given two of its elements as $A \setminus B,\ C \setminus D,$ where $A,B,C,D \in \mathcal B,$ then $$(A \setminus B) \cap (C \setminus D) = (A \cap C) \setminus (B \cup D)$$ is in $\mathcal P'$ (because $\mathcal B$ is a lattice). Furthermore, $$ \mu(A\setminus B) = \mu(A) - \mu(B \cap A) = \nu(A) - \nu(B \cap A) = \nu(A \setminus B).$$ That is, both measures agree on $\mathcal P'.$ It is also clear by construction that $$\mathcal P \subseteq \mathcal B \subseteq \mathcal P' \subseteq \mathcal A.$$ So we can apply the exact same process to $\mathcal P',$ obtaining a lattice and a larger $\pi$-system on which the measures agree. Going on like this, we obtain a sequence $$\mathcal P = \mathcal P_0 \subseteq \mathcal B_0 \subseteq \mathcal P_1 \subseteq \mathcal B_1 \subseteq \cdots \subseteq \mathcal A.$$ The measures agree on each $\mathcal P_n$ and on every $\mathcal B_n.$ It suffices to show that the limit $$\bigcup_{n \in \mathbb N} \mathcal P_n = \bigcup_{n \in \mathbb N}\mathcal B_n,$$ in which the measures clearly agree, is an algebra (and because it is contained in $\mathcal A,$ which is the smallest algebra containing $\mathcal P,$ necessarily it is equal to $\mathcal A$). It's good to prove this statement if one does not find it obvious.

(This is the main idea of my attempt to prove the full Dynkin's theorem.)

Keplerto
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  • I think a slight simplification would be to take $\mathcal P' = { \Omega \setminus A \ |\ A \in \mathcal B }.$ Then checking that $\mathcal P'$ is a $\pi$-system is easier (you only need that $\mathcal B$ is closed for unions). However, one then has to be more careful to specify which inclusions hold. – Keplerto Aug 28 '24 at 14:21