User:Caliburn/s/mt/Lebesgue Decomposition Theorem

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a measure on $\struct {X, \Sigma}$.

Finite Measure

Let $\nu$ be a finite measure on $\struct {X, \Sigma}$.

Then there exists finite measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that:

$(1) \quad$ $\nu_a$ is absolutely continuous with respect to $\mu$

$(2) \quad$ $\nu_s$ and $\mu$ are mutually singular

$(3) \quad$ $\nu = \nu_a + \nu_s$.

Complex Measure

Let $\nu$ be a complex measure on $\struct {X, \Sigma}$.

Then there exists complex measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that:

$(1) \quad$ $\nu_a$ is absolutely continuous with respect to $\mu$

$(2) \quad$ $\nu_s$ and $\mu$ are mutually singular

$(3) \quad$ $\nu = \nu_a + \nu_s$.

Finite Signed Measure

Let $\nu$ be a finite signed measure on $\struct {X, \Sigma}$.

Then there exists finite signed measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that:

$(1) \quad$ $\nu_a$ is absolutely continuous with respect to $\mu$

$(2) \quad$ $\nu_s$ and $\mu$ are mutually singular

$(3) \quad$ $\nu = \nu_a + \nu_s$.

$\sigma$-Finite Measure

Let $\nu$ be a $\sigma$-finite measure on $\struct {X, \Sigma}$.

Then there exists $\sigma$-finite measures $\nu_a$ and $\nu_s$ on $\struct {X, \Sigma}$ such that:

$(1) \quad$ $\nu_a$ is absolutely continuous with respect to $\mu$

$(2) \quad$ $\nu_s$ and $\mu$ are mutually singular

$(3) \quad$ $\nu = \nu_a + \nu_s$.