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.