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I am trying to solve this exercise from Billingsley's Probability and Measure

Exercise 3.12 Deduce the $\pi-\lambda$ theorem from the monotone class theorem by showing directly that, if a $\lambda$-system $\mathscr{L}$ contains a $\pi$-system $\mathscr{P}$, then $\mathscr{L}$ also contains the field generated by $\mathscr{P}$.

Here is what I have come up with:

The monotone class theorem applies to an (algebra/field). Therefore, we need some field to apply it to and not just a $\pi$-system. The natural choice would be the field (not $\sigma$-field) generated by $\mathscr{P}$. We also need some monotone class. We claim that $\mathscr{L}$ is a monotone class. Using the alternate axiom $(\lambda'_2)$ which says that $A,B \in \mathscr{L}$ and $A\subset B$ imply $B\setminus A = \mathscr{L}$, we can easily show that $\mathscr{L}$ is closed under monotone unions and intersections. To see this, fix $A_1,A_2,\dots,\in \mathscr{L}$ where $A_n \uparrow A$. Then define the sets $B_1=A_1$ $B_2=A_2\setminus B_1$ , $B_3=A_3\setminus B_1 \cup B_2$ and so on which are elements of $\mathscr{L}$ by $(\lambda_2')$. Then $B_1\cup B_2 \cup \dots$ are all disjoint so by $(\lambda_3)$ which says that $A_1,A_2,\dots, \in \mathscr{L}$ and $A_n\cap A_m=\emptyset$ for $m\neq n$ imply $\bigcup_n \in \mathscr{L}$, we conclude that $\mathscr{L}$ is closed under increasing unions. The proof for decreasing unions is similar. Now to use the monotone class theorem, we finally need to show that the field generated by $\mathscr{P}$ is a subset of $\mathscr{L}$. This holds by minimality of the field generated by $\mathscr{P}$ because $\mathscr{L}$ is a field. To see this, it is closed under complements as it is a $\lambda$-system, it contains $\Omega$ as it is a $\lambda$-system, but I am not sure how to show $A\cup B\in \mathscr{L}$. What can I do from here to show this fact? Is it even true?

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1 Answer 1

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If I get things wrong, let me know, please. Exercise 2.5(b) is useful in this exercise.

Exercise 2.5(b). Let $f(\mathscr{A})$ be the field generated by a class $\mathscr{A}$ in $\Omega$. For nonempty $\mathscr{A}$, $f(\mathscr{A})$ is the class of sets of the form $$\bigcup_{i=1}^m\bigcap_{j=1}^{n_i}A_{ij},$$ where for each $i$ and $j$ either $A_{ij}\in\mathscr{A}$ or $A_{ij}^c\in\mathscr{A}$, and where the $m$ sets $\bigcap_{j=1}^{n_i}A_{ij}$, $1\leq i\le m$, are disjoint.

Now we prove Exercise 3.12. As was noticed by OP, it suffices to prove that $f(\mathscr{P})\subset \mathscr{L}$. Since $\mathscr{L}$ is closed under the disjoint unions, it suffices to prove that $\bigcap_{j=1}^{n}P_{j}\in\mathscr{L}$, where for each $j$ either $P_j\in\mathscr{P}$ or $P_j^c\in\mathscr{P}$. Without loss of generality, we assume that $P_1^c,\cdots P_k^c\in \mathscr{P}$ and $P_{k+1},\cdots, P_n\in\mathscr{P}$. Let $P=P_{k+1}\cap\cdots\cap P_n\in\mathscr{P}$ and $A_j=P_j^c\in\mathscr{P}$ for $1\leq j\le k$, then \begin{align*} \bigcap_{j=1}^{n}P_{j}&=P\cap A_1^c\cap A_2^c\cap\cdots\cap A_k^c=P\bigcap\left(A_1\cup A_2\cup \cdots \cup A_k\right)^c\\ &=P\setminus \left(P\cap\left(A_1\cup A_2\cup \cdots \cup A_k\right)\right)\\ &=P\setminus\left((P\cap A_1)\cup(P\cap A_2)\cup\cdots\cup(P\cap A_k)\right). \end{align*} Since $\mathscr{L}$ is closed under the formation of proper differences, it suffices to show that $$(P\cap A_1)\cup(P\cap A_2)\cup\cdots\cup(P\cap A_k)\in\mathscr{L}.\tag{1}$$ The key point is to write the above expression as a dijoint union of sets in $\mathscr{L}$. Indeed, we have \begin{align*} &(P\cap A_1)\cup(P\cap A_2)\cup\cdots\cup(P\cap A_k)\\&=(P\cap A_1)\cup((P\cap A_2)\setminus(P\cap A_1\cap A_2))\cup\cdots\cup((P\cap A_k)\setminus(P\cap A_1\cap A_2\cap\cdots\cap A_k)). \end{align*} Now since $\mathscr{P}$ is closed under finite intersections, $\mathscr{L}$ is closed under the formation of proper differences and the disjoint unions, we get $(1)$. Therefore, $\bigcap_{j=1}^{n}P_{j}\in\mathscr{L}$ and the proof is completed.

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  • $\begingroup$ The last expression attempting to express a finite union as a disjoint finite union is wrong. For example, A_3 \ (A1 u A2) is what was desired, but A3 \ (A1 n A2 n A3) may intersect A1 or A2 or both. The correct expression of a union in terms of a disjoint union, proper difference, and intersection looks a lot like the inclusion-exclusion principle. $\endgroup$ Commented Mar 10 at 3:35

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