Decomposition of Complex Measure into Finite Signed Measures

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:


 * $\mu = \mu_R + i \mu_I$

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:

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:

and:

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.

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 is real, while the  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.