Complex Measures: Integrability

Question

Problem

On the one hand, a complex measure decomposes into: $$\mu=\Re_+\mu-\Re_-\mu+i\Im_+\mu-i\Im_-\mu=:\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha$$

This gives rise to the integrability condition: $$f\in L(\mu)\iff f\in L(\mu_\alpha)\quad(\alpha=1,\ldots,3)$$

On the other hand, a complex measure admits a derivative: $$\mu(E)=\int_Eu\mathrm{d}|\mu|\quad(|u|=1)$$

This gives rise to the integrability condition: $$f\in L(\mu):\iff fu\in L(|\mu|)\iff f\in L(|\mu|)$$

So the question arises wether these approaches coincide: $$f\in L(|\mu|)\iff f\in L(\mu_\alpha)\quad(\alpha=1,\ldots,3)$$

Attempt

So far by construction it is: $$\mu_\alpha(E)=\int_Eu_\alpha\mathrm{d}|\mu|\quad(0\leq u_\alpha\leq1)$$ which gives for positive functions: $$\int|f|\mathrm{d}\mu_\alpha=\int|f|u_\alpha\mathrm{d}|\mu|\leq\int |f|\mathrm{d}|\mu|$$ That proves the inclusion: $$f\in L(|\mu|)\implies f\in L(\mu_\alpha)\quad(\alpha=1,\ldots,3)$$ But what about the converse?

Related

For a formal treatment see: Complex Measures: Integration

For a related recapitulation see: Complex Measures: Variation

For a similar problem see: Complex Functions: Integrability

For a general problem see: Radon-Nikodym: Integrability?

Answer 1

In fact, the very very heart of the whole story here are the two inequalities: $$|\mu(E)|\leq|\mu|(E)<\infty\quad(E\in\Sigma)$$

For positive measures given as derivative: $$\kappa(E)=\int_Eh\mathrm{d}\lambda\quad(h\geq0)$$ the positive integrals agree: $$\int f\mathrm{d}\kappa=\int fh\mathrm{d}\lambda\quad(f\geq0)$$ (Note, these are allowed to be infinite!)

Now, consider the representation by the Radon-Nikodym derivative: $$\mu_\alpha=\int_Eu_\alpha\mathrm{d}|\mu|\quad(u_\alpha\geq0,|u|=1)$$ Exploiting the bounds: $$u_\alpha\leq|u|$$ $$|u|\leq\sum_{\alpha=0\ldots3}u_\alpha$$ one has the estimates: $$\int|f|\mathrm{d}\mu_\alpha=\int|f|\cdot u_\alpha\mathrm{d}|\mu|\leq\int|f|\cdot|u|\mathrm{d}|\mu|=\int|f|\mathrm{d}|\mu|$$ $$\int|f|\mathrm{d}|\mu|=\int|f|\cdot|u|\mathrm{d}|\mu|\leq\sum_{\alpha=0\ldots3}\int|f|\cdot u_\alpha\mathrm{d}|\mu|=\sum_{\alpha=0\ldots3}\int|f|\mathrm{d}\mu_\alpha$$ and therefore: $$f\in L(|\mu|)\iff f\in L(\mu_\alpha)\quad(\alpha=0\ldots3)$$

Answer 2

If we have a signed measure $\mu$ on a measurable space $(\Omega,\Sigma)$, then $\mu$ may be allowed to assume values of $\pm \infty$, but not both. A complex measure $\mu$ is not allowed to assume any type of infinite value. For either type of measure $\mu$ which is not allowed to assume infinite values, the variation measure $|\mu|$ is a finite measure, where $$ |\mu|E = \sup_{\pi}\sum_{S \in \pi}|\mu S| $$ is taken over all finite partitions $\pi$ of measurable subsets of $E$. Because we want to deal with a complex measure, then we assume without loss of generality that all measures related to this discussion are finite.

If $\mu$ is a signed measure, then we can decompose $\mu$ into positive meausres $\mu_{+}$, $\mu_{-}$ according to the (Hahn) Jordan decomposition. This decomposition is the same as a variation measure approach: $$ |\mu|=\mu_{+}+\mu_{-},\;\;\; \mu = \mu_{+}-\mu_{-}. $$ This extends to the complex case as well $$ \mu = (\Re\mu)_{+}-(\Re\mu)_{-}+i\{(\Im\mu)_{+}-(\Im\mu)_{-}\} $$ with $$ |\mu| \le |\Re\mu| + |\Im\mu| = |(\Re\mu)_{-}|+|(\Re\mu)_{+}|+|(\Im\mu)_{-}|+|(\Im\mu)_{+}|. $$ But we also have $$ |\Re\mu| \le |\mu|,\;\;\; |\Im\mu| \le |\mu|. $$ So integrability of a function $f$ with respect to $|\mu|$ is equivalent to simultaneous integrability with respect to $|(\Re\mu)_{-}|,|(\Re\mu)_{+}|,|(\Im\mu)_{-}|,|(\Im\mu)_{+}|$.