Sigma-Algebras with Independent Generators are Independent

Theorem

Let $\left({\Omega, \mathcal E, P}\right)$ be a probability space.

Let $\Sigma, \Sigma'$ be sub-$\sigma$-algebras of $\mathcal E$.

Suppose that $\mathcal G, \mathcal H$ are $\cap$-stable generators for $\Sigma, \Sigma'$, respectively.

Suppose that, for all $G \in \mathcal G, H \in \mathcal H$:

$(1):\quad P \left({G \cap H}\right) = P \left({G}\right) P \left({H}\right)$

Then $\Sigma$ and $\Sigma'$ are $P$-independent.

Proof

Fix $H \in \mathcal H$.

Define, for $E \in \Sigma$:

$\mu \left({E}\right) := P \left({E \cap H}\right)$
$\nu \left({E}\right) := P \left({E}\right) P \left({H}\right)$

Then by Intersection Measure is Measure and Restricted Measure is Measure, $\mu$ is a measure on $\Sigma$.

Namely, it is the intersection measure $P_H$ restricted to $\Sigma$, i.e. $P_H \restriction_\Sigma$.

Next, by Linear Combination of Measures and Restricted Measure is Measure, $\nu$ is also a measure on $\Sigma$.

Namely, it is the restricted measure $P \left({H}\right) P \restriction_\Sigma$.

Let $\mathcal{G}' := \mathcal G \cup \left\{{X}\right\}$.

It is immediate that $\mathcal{G}'$ is also a $\cap$-stable generator for $\Sigma$.

By assumption $(1)$, $\mu$ and $\nu$ coincide on $\mathcal G$ (since $P \left({X}\right) = 1$).

From Restricting Measure Preserves Finiteness, $\mu$ and $\nu$ are also finite measures.

Hence, $\mathcal G$ 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 $E' \in \Sigma'$:

$\mu'_E \left({E'}\right) := P \left({E \cap E'}\right)$
$\nu'_E \left({E'}\right) := P \left({E}\right) P \left({E'}\right)$

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 $\nu'_E$:

$\forall E \in \Sigma: \forall E' \in \Sigma': P \left({E \cap E'}\right) = P \left({E}\right) P \left({E'}\right)$

This is precisely the statement that $\Sigma$ and $\Sigma'$ are $P$-independent $\sigma$-algebras.

$\blacksquare$