Decomposition of Complex Measure into Finite Signed Measures

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then there exists unique finite signed measures $\mu_R$ and $\mu_I$ such that:

$\mu = \mu_R + i \mu_I$


Proof

Existence

For each $A \in \Sigma$ define the function $\mu_R : X \to \R$ by:

$\map {\mu_R} A = \map \Re {\map \mu A}$

Similarly, for each $A \in \Sigma$ define the function $\mu_I : X \to \R$ by:

$\map {\mu_I} A = \map \Im {\map \mu A}$

Clearly we have:

\(\ds \map \mu A\) \(=\) \(\ds \map \Re {\map \mu A} + i \map \Im {\map \mu A}\) Definition of Real Part, Definition of Imaginary Part
\(\ds \) \(=\) \(\ds \map {\mu_R} A + i \map {\mu_I} A\)

for each $A \in \Sigma$.

So:

$\mu = \mu_R + \mu_I$

It remains to show that $\mu_R$ and $\mu_I$ are finite signed measures.


We verify both of the conditions for a signed measure in turn.

We have:

$\map \mu \O = 0$

so:

$\map {\mu_R} A = \map \Re 0 = 0$

and:

$\map {\mu_I} A = \map \Im 0 = 0$

verifying $(1)$ for $\mu_R$ and $\mu_I$.

Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint $\Sigma$-measurable sets.

Then:

\(\ds \map {\mu_R} {\bigcup_{n \mathop = 1}^\infty D_n}\) \(=\) \(\ds \map \Re {\map \mu {\bigcup_{n \mathop = 1}^\infty D_n} }\)
\(\ds \) \(=\) \(\ds \map \Re {\sum_{n \mathop = 1}^\infty \map \mu {D_n} }\) since $\mu$ is countably additive
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map \Re {\map \mu {D_n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\mu_R} {D_n}\)

and:

\(\ds \map {\mu_I} {\bigcup_{n \mathop = 1}^\infty D_n}\) \(=\) \(\ds \map \Im {\map \mu {\bigcup_{n \mathop = 1}^\infty D_n} }\)
\(\ds \) \(=\) \(\ds \map \Im {\sum_{n \mathop = 1}^\infty \map \mu {D_n} }\) since $\mu$ is countably additive
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map \Im {\map \mu {D_n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map {\mu_I} {D_n}\)

verifying $(2)$ for $\mu_R$ and $\mu_I$.

So $\mu_R$ and $\mu_I$ are signed measures.


Since $\mu$ is a complex-valued function, we have:

$\cmod {\map \mu X} < \infty$

We then have, from Modulus Larger than Real Part:

$\size {\map \Re {\map \mu X} } < \infty$

so:

$\size {\map {\mu_R} X} < \infty$

Similarly, from Modulus Larger than Imaginary Part:

$\size {\map \Im {\map \mu X} } < \infty$

so:

$\size {\map {\mu_I} X} < \infty$

So:

$\mu_R$ and $\mu_I$ are finite.

$\Box$


Uniqueness

Suppose that $\mu_R^{(1)}$, $\mu_R^{(2)}$, $\mu_I^{(1)}$, $\mu_I^{(2)}$ are finite signed measures with:

$\mu = \mu_R^{(1)} + i \mu_I^{(1)} = \mu_R^{(2)} + i \mu_I^{(2)}$

Then for each $A \in \Sigma$, we have:

$\map {\mu_R^{(1)} } A - \map {\mu_R^{(2)} } A = i \paren {\map {\mu_I^{(2)} } A - \map {\mu_I^{(1)} } A}$

The left hand side is real, while the right hand side is imaginary, so:

$\map {\mu_R^{(1)} } A - \map {\mu_R^{(2)} } A = i \paren {\map {\mu_I^{(2)} } A - \map {\mu_I^{(1)} } A} = 0$

so:

$\map {\mu_R^{(1)} } A = \map {\mu_R^{(2)} } A$

and:

$\map {\mu_I^{(1)} } A = \map {\mu_I^{(2)} } A$

Since $A \in \Sigma$ was arbitrary, we have:

$\mu_R^{(1)} = \mu_R^{(2)}$

and:

$\mu_I^{(1)} = \mu_I^{(2)}$

so we obtain the desired uniqueness.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Decomposition_of_Complex_Measure_into_Finite_Signed_Measures&oldid=551497"