# 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 $\lambda$.

$\blacksquare$