Absolute Continuity of Measures is Transitive Relation
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Theorem
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$, $\nu$ and $\lambda$ be measures on $\struct {X, \Sigma}$ such that:
- $\mu$ is absolutely continuous with respect to $\nu$
and:
- $\nu$ is absolutely continuous with respect to $\lambda$.
That is:
- $\mu \ll \nu$
and:
- $\nu \ll \lambda$
Then:
- $\mu$ is absolutely continuous with respect to $\lambda$.
That is:
- $\mu \ll \lambda$
In other words:
- $\ll$ is a transitive relation.
Proof
Let $A \in \Sigma$ be such that $\map \lambda A = 0$.
Then, since $\nu \ll \lambda$ we have $\map \nu A = 0$ from the definition of absolute continuity.
Since $\mu \ll \nu$ we similarly have $\map \mu A = 0$ again applying the definition of absolute continuity.
So, whenever $A \in \Sigma$ is such that $\map \lambda A = 0$ we have $\map \mu A = 0$.
So $\mu$ is absolutely continuous with respect to $\lambda$.
$\blacksquare$