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Dynkin's π−λ theorem

The Dynkin's π-λ theorem is very useful result in a measure theory which allows to replace σ-algebra with π-system in numerous applications.
The core insight behind Dynkin's π−λ\pi-\lambda theorem is that σ\sigma-algebras are 'difficult', but π\pi-systems are 'easy', prompting us to primarily work with the latter. For example, this theorem significantly simplifies the proof of the uniqueness aspect of Carathéodory's Theorem.
π\pi-system
A class P\mathcal{P} of subsets of space Ω\Omega is called a π\pi-system if it is closed under the formation of finite intersections:

  • (π1) (\pi_1) \:

    if A,B∈P A,B \in \mathcal{P} \: then  A∩B∈P\: A \cap B \in \mathcal{P}

λ\lambda-system
A class L\mathcal{L} of subsets is a λ\lambda-system if

  • (λ1) (\lambda_1) \:

    Ω∈L\Omega \in \mathcal{L};

  • (λ2) (\lambda_2) \:

    if A,B∈LA,B \in \mathcal{L}, A⊂B A \subset B \: then  B∖A∈L \: B \setminus A \in \mathcal{L}

  • (λ3) (\lambda_3) \:

    if An∈LA_n \in \mathcal{L} and An↑AA_n \uparrow A then A∈LA \in \mathcal{L}

Because of the monotonicity condition in (λ3)(\lambda_3), the definition of λ\lambda-system is weaker than that of σ\sigma-algebra. Although a σ\sigma-algebra is a λ\lambda-system, the reverse is not true. But if the class is both a π\pi-system and a λ\lambda-system is a σ\sigma-algebra.
LEMMA
A class that is both a π\pi-system and a λ\lambda-system is a σ\sigma-algebra.
\quad Proof. The class includes Ω\Omega, as indicated by (λ1)(\lambda_1). Furthermore, it is closed under the operation of taking complements, because (λ2)(\lambda_2) implies that if A∈LA \in \mathcal{L} then Ac=Ω∖A∈LA^{c} = \Omega \setminus A \in \mathcal{L}. Moreover, the class is closed under finite unions, exemplified by the fact that for any sets AA and BB in the class, A∪BA \cup B can be expressed as (Ac∩Bc)c(A^{c} \cap B^{c})^{c} due to (π1)(\pi_1). Extending this principle, the closure under countable unions is demonstrated as ⋃i=1nAi↑⋃i=1∞Ai\bigcup_{i=1}^{n} A_i \uparrow \bigcup_{i=1}^{\infty} A_i, and the condition (λ3)(\lambda_3) ensures that ⋃i=1∞Ai\bigcup_{i=1}^{\infty} A_i is contained within the class. Therefore, the class fulfills all the criteria to be considered a σ\sigma-algebra. □\Box
The illustration of Dynkin's Theorem
Now we can formulate the main theorem of the article.
THEOREM
If P\mathcal{P} is a π\pi-system and L\mathcal{L} is λ\lambda-system that contains P\mathcal{P} then we have implication
P⊂L⟹σ(P)⊂L\begin{equation*}\mathcal{P} \subset \mathcal{L} \Longrightarrow \sigma(\mathcal{P}) \subset \mathcal{L}\end{equation*}

The foolowing proof is a combination of [Durrett2019] p. 395 and [Billingsley2012] p. 43

The proof strategy involves considering the smallest λ\lambda-system L0\mathcal{L}_0, that includes P\mathcal{P}. We will demonstrate that L0\mathcal{L}_0 is also a π\pi-system. From this, it follows by Lemma 1 that L0\mathcal{L}_0 is a σ\sigma-algebra as well. Given that σ(P)\sigma(\mathcal{P})(the smallest σ\sigma-algebra containing P\mathcal{P}) is included in L0\mathcal{L}_0 and consequently we establish the following relationship:
P⊂σ(P)⊂L0⊂L\begin{equation*}\mathcal{P} \subset \sigma(\mathcal{P}) \subset \mathcal{L}_0 \subset \mathcal{L}\end{equation*}
Thus, the critical task at hand is to verify that L0\mathcal{L}_0 is indeed a π\pi-system.
Let L0\mathcal{L}_0 denote the λ\lambda-system generated by P\mathcal{P}. By definition, L0\mathcal{L}_0 is the intersection of all λ\lambda-systems that contain P\mathcal{P}.
L0=⋂αLα⟹P⊂L0⊂L\begin{equation*}\mathcal{L}_0 = \bigcap_{\alpha} \mathcal{L}_{\alpha} \quad \Longrightarrow \quad \mathcal{P} \subset \mathcal{L}_0 \subset \mathcal{L}\end{equation*}
As such, L0\mathcal{L}_0 itself is a λ\lambda-system, incorporates P\mathcal{P}, and is simultaneously a subset of any λ\lambda-system that includes P\mathcal{P}.
Define GA={B:A∩B∈L0}G_A = \{ B : A \cap B \in \mathcal{L}_0 \} for A∈L0A \in \mathcal{L}_0 then GA G_A is a λ\lambda-system as it satisfies the following conditions:
  • Ω∈GA \Omega \in G_A since A∩Ω=A∈L0 A \cap \Omega = A \in \mathcal{L}_0, fulfilling (λ1)(\lambda_1);
  • if B,C∈GA B, C \in G_A and C⊂B C \subset B then the λ\lambda-system L0\mathcal{L}_0 contains A∩B A \cap B and A∩CA \cap C and hence contains the proper difference (A∩B)−(A∩C)=A∩(B−C)(A \cap B) - (A \cap C) = A \cap (B - C), meeting (λ2)(\lambda_2);
  • if Bn∈GAB_n \in G_A and Bn↑BB_n \uparrow B then A∩Bn↑A∩B∈L0A \cap B_n \uparrow A \cap B \in \mathcal{L}_0 since A∩Bn∈L0A \cap B_n \in \mathcal{L}_0 and L0\mathcal{L}_0 is a λ\lambda-system, so (λ3)(\lambda_3) is satisfied.
If A∈PA \in \mathcal{P}, then P⊂GA\mathcal{P} \subset G_A (since P\mathcal{P} is π\pi-system). Since L0\mathcal{L}_0 is the minimal λ\lambda-system containing P\mathcal{P}, it follows that L0⊂GA\mathcal{L}_0 \subset G_A. Hence, if A∈PA \in \mathcal{P} and B∈L0B \in \mathcal{L}_0, then A∩B∈L0A \cap B \in \mathcal{L}_0. Interchanging AA and BB in the last sentence: if A∈L0 A \in \mathcal{L}_0 and B∈PB \in \mathcal{P} then A∩B∈L0 A \cap B \in \mathcal{L}_0. But this implies that if A∈L0A \in \mathcal{L}_0 then P⊂GA\mathcal{P} \subset G_A and L0⊂GA\mathcal{L}_0 \subset G_A.
This conclusion implies that if A,B∈L0A, B \in \mathcal{L}_0, then A∩B∈L0A \cap B \in \mathcal{L}_0, thereby completing the proof that L0\mathcal{L}_0 is indeed a π\pi-system. □\Box

The foolowing lemma is taken from [Durrett2019] p. 396

To prove that the extension in the Carathéodory's Theorem is unique, we need to the following lemma.
LEMMA
Let P\mathcal{P} be a π\pi-system. If v1v_1 and v2v_2 are probability measures (on σ\sigma-algebras F1F_1 and F2F_2) that agree on P\mathcal{P} and there is a sequence An∈PA_n \in \mathcal{P} with An↑ΩA_n \uparrow \Omega, then v1v_1 and v2v_2 agree on σ(P)\sigma(\mathcal{P}).
\quad Proof. Let A∈PA \in \mathcal{P} have v1(A)=v2(A)<∞v_1(A) = v_2(A) < \infty. Let
L={B∈σ(P):  v1(A∩B)=v2(A∩B)}\begin{equation*}\mathcal{L} = \{ B \in \sigma(\mathcal{P}) : \: \: v_1(A \cap B) = v_2(A \cap B)\}\end{equation*}
We will now show that L\mathcal{L} is a λ\lambda-system. Since A∈PA \in \mathcal{P},  v1(A)=v2(A) \:v_1(A) = v_2(A) and Ω∈L\Omega \in \mathcal{L}, so (λ1)(\lambda_1) is satisfied. If B,C∈LB, C \in \mathcal{L} with C⊂BC \subset B then
v1(A∩(B−C))=v1(A∩B)−v1(A∩C)=v2(A∩B)−v2(A∩C)=v2(A∩(B−C))\begin{equation*}v_1(A \cap (B - C)) = v_1(A \cap B) - v_1(A \cap C) = v_2(A \cap B) - v_2(A \cap C) = v_2(A \cap (B - C))\end{equation*}
So (λ2)(\lambda_2) is verified. Finally, if Bn∈LB_n \in \mathcal{L} and Bn↑BB_n \uparrow B, then the continuity of probability measure implies
v1(A∩B)=lim⁡n→∞v1(A∩Bn)=lim⁡n→∞v2(A∩Bn)=v2(A∩B)\begin{equation*}v_1(A \cap B) = \lim_{n \to \infty} v_1(A \cap B_n) = \lim_{n \to \infty} v_2(A \cap B_n) = v_2(A \cap B)\end{equation*}
Since P\mathcal{P} is π\pi-system, the π−λ\pi - \lambda theorem implies σ(P)⊂L\sigma(\mathcal{P}) \subset \mathcal{L} , i.e., if A∈PA \in \mathcal{P} with v1(A)=v2(A)<∞v_1(A) = v_2(A) < \infty and B∈σ(P)B \in \sigma(\mathcal{P}) then v1(A∩B)=v2(A∩B)v_1(A \cap B) = v_2(A \cap B). Letting An∈PA_n \in \mathcal{P} with An↑ΩA_n \uparrow \Omega and using the continuity of probability measure, we have the desired conclusion. □\Box
CARATHÉODORY'S EXTENSION THEOREM
Let μ\mu be a probability measure on an algebra A\mathcal{A}. Then μ\mu has a unique extension to σ(A)\sigma(\mathcal{A}) = the smallest σ\sigma-algebra containing A\mathcal{A}.

The foolowing theorem is partially taken from [Durrett2019] p. 39

The illustration of Theorem about Independence of σ\sigma-algebras
THEOREM
Suppose A1,A2,…,An\mathcal{A}_1, \mathcal{A}_2, \ldots, \mathcal{A}_n are independent π\pi-systems. Then the minimal σ\sigma-algebras σ(A1),σ(A2),…,σ(An)\sigma(\mathcal{A}_1), \sigma(\mathcal{A}_2), \ldots, \sigma(\mathcal{A}_n) are independent.
\quad Proof. It is sufficient to show that if one of the classes i.e. A1\mathcal{A}_1 is replaced by σ(A1)\sigma(\mathcal{A}_1), then the new sequence of classes will also be independent.
Let A2,…,AnA_2, \ldots, A_n be sets with Ai∈AiA_i \in \mathcal{A}_i, let F=A2∩…∩AnF = A_2 \cap \ldots \cap A_n and let
L={A:P(A∩F)=P(A)P(F)}\begin{equation*}\mathcal{L} = \{A : \mathbb{P}(A \cap F) = \mathbb{P}(A)\mathbb{P}(F) \}\end{equation*}
is the set of all events which do not depend on A2,…,An\mathcal{A}_2, \ldots, \mathcal{A}_n. By definition, A1⊂L\mathcal{A}_1 \subset \mathcal{L} and we only need to prove that L\mathcal{L} is a λ\lambda-system in order to use Dynkin's π−λ\pi-\lambda theorem.
Since P(Ω∩F)=P(Ω)P(F)\mathbb{P}(\Omega \cap F) = \mathbb{P}(\Omega)\mathbb{P}(F), so Ω∈L\Omega \in \mathcal{L} and (λ1)(\lambda_1) is fulfilled.
To check (λ2)(\lambda_2) we note that if A,B∈LA, B \in \mathcal{L} with A⊂BA \subset B then (B−A)∩F=(B∩F)−(A∩F)(B - A) \cap F = (B \cap F) - (A \cap F):
P((B−A)∩F)=P(B∩F)−P(A∩F)=P(B)P(F)−P(A)P(F)=(P(B)−P(A))P(F)=P(B−A)P(F)\begin{equation*}\mathbb{P}((B - A) \cap F) = \mathbb{P}(B \cap F) - \mathbb{P}(A \cap F) = \mathbb{P}(B)\mathbb{P}(F) - \mathbb{P}(A)\mathbb{P}(F) = (\mathbb{P}(B) - \mathbb{P}(A))\mathbb{P}(F) = \mathbb{P}(B - A)\mathbb{P}(F)\end{equation*}
and we have B−A∈LB - A \in \mathcal{L}.
To check (λ3)(\lambda_3) let Bk∈LB_k \in \mathcal{L} with Bk↑BB_k \uparrow B and (Bk∩F)↑(B∩F)(B_k \cap F) \uparrow (B \cap F) then using the continuity property of probability measure we get:
P(B∩F)=lim⁡k→∞P(Bk∩F)=lim⁡k→∞P(Bk)P(F)=P(B)P(F)\begin{equation*}\mathbb{P}(B \cap F) = \lim_{k \to \infty} \mathbb{P}(B_k \cap F) = \lim_{k \to \infty} \mathbb{P}(B_k)\mathbb{P}(F) = \mathbb{P}(B)\mathbb{P}(F)\end{equation*}
Thus L\mathcal{L} is λ\lambda-system and with the π−λ\pi-\lambda theorem now gives σ(A1)⊂L\sigma(A_1) \subset \mathcal{L}. It follows that if A1∈σ(A1)A_1 \in \sigma(A_1) and Ai∈AiA_i \in \mathcal{A}_i for 2≤i≤n2 \leq i \leq n then
P(⋂i=1nAi)=P(A1)P(⋂i=2nAi)=∏i=1nP(Ai)\begin{equation*}\mathbb{P}\left(\bigcap_{i=1}^n A_i\right) = \mathbb{P}(A_1)\mathbb{P}\left(\bigcap_{i=2}^n A_i\right) = \prod_{i=1}^n \mathbb{P}(A_i)\end{equation*}
The theorem is thus proved. □\Box

References