Sigma-Algebras with Independent Generators are Independent
Theorem
Let $\struct {\Omega, \EE, \Pr}$ be a probability space.
Let $\Sigma, \Sigma'$ be sub-$\sigma$-algebras of $\EE$.
Suppose that $\GG, \HH$ are $\cap$-stable generators for $\Sigma, \Sigma'$, respectively.
Suppose that, for all $G \in \GG, H \in \HH$:
- $(1): \quad \map \Pr {G \cap H} = \map \Pr G \map \Pr H$
Then $\Sigma$ and $\Sigma'$ are $\Pr$-independent.
Proof
Fix $H \in \HH$.
Define, for $E \in \Sigma$:
- $\map \mu E := \map \Pr {E \cap H}$
- $\map \nu E := \map \Pr E \map \Pr H$
Then by Intersection Measure is Measure and Restricted Measure is Measure, $\mu$ is a measure on $\Sigma$.
Namely, it is the intersection measure $\Pr_H$ restricted to $\Sigma$, that is $\Pr_H \restriction_\Sigma$.
Next, by Linear Combination of Measures and Restricted Measure is Measure, $\nu$ is also a measure on $\Sigma$.
Namely, it is the restricted measure $\map \Pr H \Pr \restriction_\Sigma$.
Let $\GG' := \GG \cup \set X$.
It is immediate that $\GG'$ is also a $\cap$-stable generator for $\Sigma$.
By assumption $(1)$, $\mu$ and $\nu$ coincide on $\GG$ (since $\map \Pr X = 1$).
From Restricting Measure Preserves Finiteness, $\mu$ and $\nu$ are also finite measures.
Hence, $\GG$ contains the exhausting sequence of which every term equals $X$.
Having verified all conditions, Uniqueness of Measures applies to yield $\mu = \nu$.
Now fix $E \in \Sigma$ and define, for $E' \in \Sigma'$:
- $\map {\mu'_E} {E'} := \map \Pr {E \cap E'}$
- $\map {\nu'_E} {E'} := \map \Pr E \map \Pr {E'}$
Mutatis mutandis, above consideration applies again, and we conclude by Uniqueness of Measures:
- $\mu'_E = \nu'_E$
for all $E \in \Sigma$.
That is, expanding the definition of the measures $\mu'_E$ and $\nu'_E$:
- $\forall E \in \Sigma: \forall E' \in \Sigma': \map \Pr {E \cap E'} = \map \Pr E \map \Pr {E'}$
This is precisely the statement that $\Sigma$ and $\Sigma'$ are $\Pr$-independent $\sigma$-algebras.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 5$: Problem $10$