Extension: Uniqueness

The uniqueness follows from the fact that $\cF_0$ is a $\pi$-system (closed under finite intersections) and $\Pr$ is a probability measure on $\cF_0$. By the uniqueness theorem for measures, the extension to $\sigma(\cF_0)$ is unique. More rigorous proof below.

π-system

Definition: π-system

A collection of sets $\cP$ of $\Omega$ is called a π-system if it is closed under finite intersections, i.e., for any $A, B \in \cP$, $A \cap B \in \cP$.

λ-system

Definition: λ-system

A collection of sets $\cL$ of $\Omega$ is called a λ-system if it satisfies the following three properties:

1. $\Omega \in \cL$
2. $A \in \cL \implies A^c \in \cL$
3. For disjoint sets $A_1, A_2, \ldots$, $$A_1, A_2, \ldots \in \cL \implies \bigcup_{i=1}^{\infty} A_i \in \cL$$

Facts about π-systems and λ-systems

Fact (a): π-system + λ-system = σ-field

Statement

A collection of sets that that is both a $\pi$-system and a $\lambda$-system is a $\sigma$-field.

Proof

Note that a $\lambda$-system is slightly weaker (less restrictive) than a $\sigma$-field, which requires closure under countable unions for any sets rather than just disjoint sets.

Essentially, the condition of a $\pi$-system to be closed under finite intersections compensates for the weaker condition of a $\lambda$-system to be closed under countable disjoint unions.

Let $\cL$ be both a $\lambda$-system and a $\pi$-system. Note that the first two properties of a $\lambda$-system are the same as that of a $\sigma$-field. So we just need to show that if $\cL$ is also a $\pi$-system, then it is closed under countable unions of sets that are not necessarily disjoint.

Let $A_1, A_2, \ldots \in \cL$ be arbitrary sets. Define:

$$B_i := A_i \bigcap_{j=1}^{i-1} A_j^c.$$

See that $B_i$ are disjoint sets. Also see that since $\cL$ is a $\pi$-system, $B_i \in \cL$ for all $i$. Therefore, $\bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty B_i \in \cL$, satisfying the third property of a $\sigma$-field.

Fact (b): Complement in a λ-system

Statement

If a $\lambda$-system $\cL$ contains both $A$ and $A\cap B$, then it also contains $A\cap B^c$.

Proof

$$\begin{aligned} A\in \cL &\implies A^c \in \cL & \text{ (first property of a }\lambda\text{-system)}\\ A\cap B \in \cL &\implies A^c \cup (A\cap B) \in \cL & \text{ (third property of a }\lambda\text{-system)}\\ &\implies (A^c \cup (A\cap B))^c \in \cL & \text{ (first property of a }\lambda\text{-system)}\\ \end{aligned}$$

Now see that

$$(A^c \cup (A\cap B))^c = A\cap (A \cap B)^c = A \cap (A^c \cup B^c) = A\cap B^c.$$

Hence, $A\cap B^c \in \cL$.

π-λ Theorem

Here’s a theorem that relates $\pi$-systems and $\lambda$-systems.

π-λ Theorem

Statement

If $\cP$ is a $\pi$-system and $\cL$ is a $\lambda$-system such that $\cP \subseteq \cL$, then $\sigma(\cP) \subseteq \cL$.

Proof

1. Let $L(\cP)$ be the smallest $\lambda$-system containing $\cP$ (in other words, the intersection of all $\lambda$-systems containing $\cP$). It is easy to check that $L(\cP)$ is also a $\lambda$-system. Hence, let $L(\cP)$ be the smallest $\lambda$-system containing $\cP$.

2. Now define

   $$\cG_A := \{B \in \cL : A \cap B \in L(\cP)\},$$

   for any $A \in \cP$. We claim that $\cG_A$ is a $\lambda$-system. To see this, let’s verify the three conditions of a $\lambda$-system:

   1. $\Omega \in \cG_A$ since $A \cap \Omega = A \in L(\cP)$.

   2. If $B \in \cG_A$, then both $A\in L(\cP)$ (assumption) and $A \cap B \in L(\cP)$ (defintion of $\cG_A$). This means $A \cap B^c \in L(\cP)$ (by Fact (b) above).

      By definition of $\cG_A$, we thus have $B^c \in \cG_A$.

   3. If $B_1, B_2, \ldots \in \cG_A$ (all disjoint sets),

   then $A\cap B_1, A\cap B_2, \ldots$ are disjoint sets in $L(\cP)$ (by definition of $\cG_A$). Since $L(\cP)$ is a $\lambda$-system, we have $\bigcup_{i=1}^\infty (A\cap B_i) \in L(\cP)$.

   $\implies A \bigcap \left(\bigcup_{i=1}^\infty B_i\right) \in L(\cP)$ (disjoint union property)

   $\implies \bigcup_{i=1}^\infty B_i \in \cG_A$ (by definition of $\cG_A$).

   Therefore, $\cG_A$ is a $\lambda$-system.

3. Now, if $A\in \cP$, then for any $B\in \cP$, $A\cap B\in \cP\subseteq L(\cP)$. Thus, $B\in \cG_A$ for all $B\in \cP$, implying that $\cP\subseteq \cG_A$.

   This means that $L(\cP)\subseteq \cG_A$, since $\cG_A$ is a $\lambda$-system that contains $\cP$ and $L(\cP)$ is the smallest $\lambda$-system containing $\cP$.

4. Now assume that $A\in \cP$ and $B\in L(\cP)$. Then $B\in \cG_A$ because $L(\cP)\subseteq \cG_A$. By definition of $\cG_A$, we have $A\cap B\in L(\cP)$. This also means $A\in \cG_B$. This holds for any $A\in \cP$. We thus have

   $$B\in L(\cP) \implies \cP \subseteq \cG_B \implies L(\cP) \subseteq \cG_B.$$

   Therefore, for any $A,B\in L(\cP)$, we have from $L(\cP)\subseteq \cG_B$ that $A\in \cG_B$

   $$\begin{aligned} &\implies A\cap B\in L(\cP)\\ &\implies L(\cP) \text{ is a } \pi\text{-system}\\ &\implies L(\cP) \text{ is both a } \pi\text{-system and a } \lambda\text{-system}\\ &\implies L(\cP) \text{ is a } \sigma\text{-field (by Fact (a))}. \end{aligned}$$

   Hence, $\cP \subseteq \sigma(\cP) \subseteq L(\cP) \subseteq \cL$. In words,

   • $\cP$ is a subset of $\sigma(\cP)$, the smallest $\sigma$-field containing $\cP$.
   • $\sigma(\cP)$ is a subset of $L(\cP)$, the smallest $\lambda$-system containing $\cP$ (the intersection of all $\lambda$-systems containing $\cP$).
   • $L(\cP)$ is a subset of $\cL$ which is some arbitrary $\lambda$-system containing $\cP$.

This completes the proof.

Equivalence of Probability Measures

Corollary

Statement

Let $P_1$ and $P_2$ be two probability measures that agree on the $\pi$-system $\cP$, i.e., $P_1(A) = P_2(A)$ for all $A\in \cP$. Then they also agree on $\sigma(\cP)$, i.e., $P_1(A) = P_2(A)$ for all $A\in \sigma(\cP)$.

Proof

The key idea is to define the set on which $P_1$ and $P_2$ agree and show that it is a $\lambda$-system. Then given that it is also a $\pi$-system, we can use the $\pi$-$\lambda$ theorem to show that it is a $\sigma$-field. Define $\cL := \{A \in \sigma(\cP) : P_1(A) = P_2(A)\}$. Check that $\cL$ is a $\lambda$-system.

1. $P_1(\Omega) = P_2(\Omega) = 1$.
2. We have that $$\begin{aligned} A\in \cL &\implies P_1(A) = P_2(A)\\ &\implies P_1(A^c) = 1 - P_1(A) = 1 - P_2(A) = P_2(A^c)\\ &\implies A^c \in \cL. \end{aligned}$$
3. By countable additivity of $P_1$ and $P_2$, we have for $A_1, A_2, \ldots \in \cL$ disjoint sets that $$\begin{aligned} &\implies P_1(\bigcup_i A_i) = \sum_i P_1(A_i) = \sum_i P_2(A_i) = P_2(\bigcup_i A_i).\\ &\implies \bigcup_i A_i \in \cL. \end{aligned}$$

Therefore, $\cL$ is a $\lambda$-system containing $\cP$ (because $\cP$ is the original set where $P_1$ and $P_2$ agree so $\cP \subseteq \cL$). Then by the $\pi$-$\lambda$ theorem, $\sigma(\cP) \subseteq \cL$. But $\cL \subseteq \sigma(\cP)$ by definition. Hence, $\cL = \sigma(\cP)$, implying that $P_1(A) = P_2(A)$ for all $A\in \sigma(\cP)$.

Uniqueness of Extension

Theorem: Uniqueness of Extension

Statement

The probability measure $\Qv$ shown to exist earlier is unique.

Proof

The uniqueness of the extension of a probability measure from a field to a $\sigma$-field follows immediately from the corollary above, combined with the fact that a field is a $\pi$-system.

To see why, note that if the field $\cF$ is a $\pi$-system, then if two measures $\Qv_1, \Qv_2$ agree on $\cF$, then they also agree on $\sigma(\cF)$ by the corollary above. So there can only be a single extension $\Qv$ from $\cF$ to $\sigma(\cF)$.