User:Caliburn/s/mt/Definition:Lebesgue Decomposition

Definition

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

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

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

Then from the Lebesgue Decomposition Theorem for Finite Signed Measures there exists unique 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

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

We say that $\struct {\nu_a, \nu_s}$ is the Lebesgue decomposition of $\nu$ (with respect to $\mu$).

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

Then from the Lebesgue Decomposition Theorem for Complex Measures there exists unique 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

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

We say that $\struct {\nu_a, \nu_s}$ is the Lebesgue decomposition of $\nu$ (with respect to $\mu$).

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

Then from the Lebesgue Decomposition Theorem for Sigma-Finite Measures there exists unique $\sigma$-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

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

We say that $\struct {\nu_a, \nu_s}$ is the Lebesgue decomposition of $\nu$ (with respect to $\mu$).