# Prove $\pi$-$\lambda$ theorem from Monotone Class Theorem

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?

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.