Induction proof of $\pi-\lambda$ theorem

Question

A $\pi$-system on a set $X$ is a collection $\mathcal P$ of subsets of $X$ closed under finite intersections. (For our purposes $\varnothing, X \in\mathcal P$.)

A $\lambda$-system on $X$ is a collection $\mathcal L$ of subsets of $X$ such that $X \in \mathcal L$ and $\mathcal L$ is closed under proper differences (i.e., if $A, B \in \mathcal L$ and $A \subseteq B,$ then $B\setminus A \in \mathcal L$) and countable unions of disjoint sets. Observe that $\lambda$-systems are also closed under countable increasing unions.

Let $\mathcal P$ and $\mathcal L$ be an arbitrary $\pi$-system and $\lambda$-system, respectively.

Theorem. Whenever $\mathcal P \subseteq \mathcal L,$ then $\sigma(\mathcal P) \subseteq \mathcal L,$ where $\sigma(\mathcal P)$ is the smallest $\sigma$-algebra that contains $\mathcal P.$

I already know a proof of this statement, but I am interested in proofs that use transfinite induction. Does anyone know where I can find such a proof?

In any case, below is my attempt to prove the theorem. I'll be thankful to whoever points out any mistakes or suggests ways to improve the proof.

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Proof. First note that, whenever $A, B, A\cap B$ are all in $\mathcal L,$ their union $$ A \cup B = (A \cap B) \cup (A \setminus (A \cap B)) \cup (B \setminus (A \cap B))$$ is in $\mathcal L.$ Let $\mathcal P'$ be any $\pi$-system contained in $\mathcal L.$ Let's prove by induction that for any $A_1, \dots, A_n \in \mathcal P',$ their union is in $\mathcal L.$ We have just proved the case $n = 2.$ Assume the result holds for a certain $n,$ and let $A_1, \dots, A_{n+1} \in \mathcal P'.$ Then by induction hypothesis both $A_1 \cup \cdots \cup A_n$ and $$(A_1 \cup \cdots \cup A_n) \cap A_{n+1} = (A_1 \cap A_{n+1}) \cup \cdots \cup (A_n \cap A_{n+1})$$ are in $\mathcal L.$ The claim is proved by applying the case $n =2.$

Let $\mathcal B_0 = \mathcal P.$ For $\alpha \ge 1,$ define inductively $\mathcal A_\alpha$ to be the class consisting of countable unions of sets, each from some previous $\mathcal B_\beta$. Define $\mathcal B_\alpha$ to be the class consisting of all the differences $A \setminus B$ for $A, B \in \mathcal A_\alpha.$

Let $\alpha \ge 1$ and suppose that $\mathcal B_\beta$ is a $\pi$-system contained in $\mathcal L$ for every $\beta < \alpha.$ This is certainly the case for $\alpha = 1.$ Note that $$\mathcal A_{<\alpha} = \bigcup_{\beta < \alpha} \mathcal B_{\beta}$$ is a $\pi$-system contained in $\mathcal L.$ Hence, given a sequence $A_n \in \mathcal A_{<\alpha}$ for each $n \in \mathbb N,$ we can express its union as a monotone one: $$\bigcup_n A_n = \bigcup_n (A_1 \cup \cdots \cup A_n),$$ and so $\mathcal A_\alpha \subseteq \mathcal L.$ It is also clear that $\mathcal A_\alpha$ is still a $\pi$-system, and also closed under finite unions.

Let $A,B,C,D \in \mathcal A_\alpha.$ Because $A \setminus B = A \setminus (A \cap B),$ it is immediate that $\mathcal B_\alpha \subseteq \mathcal L.$ Also, $(A \setminus B) \cap (C \setminus D) = (A \cap C) \setminus (B \cup D),$ and so $\mathcal B_{\alpha}$ is a $\pi$-system.

To finish, notice that $\mathcal A_{<\omega_1}$ is closed under complements ($A \in \mathcal B_{\alpha} \subseteq \mathcal A_{\alpha+1} \implies X \setminus A \in \mathcal B_{\alpha+1}$) and under countable unions (if $A_n \in \mathcal B_{\alpha_n}$ for each $n \in \mathbb N$ and $\alpha_n < \omega_1$ then for $\alpha = \sup_n \alpha_n < \omega_1$ we have $\bigcup_n A_n \in \mathcal B_{\alpha+1}$). So $\mathcal A_{< \omega_1}$ is a $\sigma$-algebra, and by induction it is contained in $\mathcal L.$

$\square$

Answer 1

Let's me sketch a simpler proof

1. Note that if a $\lambda$-system is closed under finite intersection, then it's a $\sigma$-algebra.

2. So, it suffices to prove the generated lambda system $L = \lambda(\mathcal{P})$ is closed under bi-intersection.

3. Build the lambda system $L$ bottom-up in the following way. Define a double-indexed collections: $P_{\alpha, \beta}$ such that: $$ P_{\alpha, \beta + 1} = \{U \setminus V | U, V \in P_{\alpha, \beta} \} \\ P_{\alpha^{'} , \beta} = \operatorname{CountableDisjointUnion}({\bigcup_{\alpha < \alpha^{'}} P_{\alpha, \beta}}) \\ L = P_{\omega_1 , \omega} $$

4. Now it suffice to prove $$\forall \alpha \beta \gamma \theta \forall (A, B) \in P_{\alpha, \beta} \times P_{\gamma, \theta} \exists (\eta, \mu, C \in P_{\eta, \mu}) \therefore C = A \cap B $$

5. Denote this proposition $Q(\alpha, \beta, \gamma, \theta)$, apply induction, we need to prove: $$ \forall \gamma \theta \therefore Q(0, 0, \gamma, \theta) \\ \forall \gamma \theta \therefore \forall \beta \therefore ( \forall \alpha < \alpha' \therefore Q(\alpha, \beta, \gamma, \theta) ) → Q(\alpha', \beta, \gamma, \theta) \\ \forall \gamma \theta \therefore \forall \alpha, \beta \therefore Q(\alpha, \beta, \gamma, \theta) → Q(\alpha , \beta + 1, \gamma, \theta) \\ $$ The second and third statement is obvious to prove, apply induction to the first statement again, now it suffices to prove: $$ Q(0,0,0,0) \\ \forall \beta \therefore ( \forall \alpha < \alpha' \therefore Q(0, 0, \alpha, \beta) ) → Q(0, 0, \alpha', \beta) \\ \forall \alpha, \beta \therefore Q(0, 0, \alpha, \beta) → Q(0, 0, \alpha , \beta + 1) \\ $$ All of which is obvious. This is actually the approach many textbooks use in a disguised way.