Let $\mu$ be a complex Radon measure (it's real and imaginary parts are signed radon measures) on a locally compact Hausdorff space $X$. Is its total variation, $|\mu|$, Radon?

I define $|\mu|$ in the followin way, write $d\mu=fd\alpha$, with $\alpha$ any $\sigma$finite positive measure, the $d|\mu|=|f|d\alpha$ defines $|\mu|$.

  • $\begingroup$ You should probably explain how you define $|\mu|$. Some people would write $\mu = (\mu_1 - \mu_2) + i (\mu_3 - \mu_4)$ with $\mu_k \geq 0$ and define $|\mu| = \mu_1 + \mu_2 + \mu_3 + \mu_4$ which is of course Radon. $\endgroup$ – Martin Jun 12 '13 at 2:46
  • $\begingroup$ Thanks, I edited it and added the definition i'm working with. Is the definition you wrote equivalent to the one I wrote? $\endgroup$ – Bill Jun 12 '13 at 2:54
  • $\begingroup$ Yes it is (provided the decomposition of the signed measures are the Jordan decompositions). First treat the case where $\mu$ is a signed measure. Then show that real and imaginary part of $\mu$ have real and imaginary part of $f$ as Radon-Nikodym derivative with respect to $\alpha$. $\endgroup$ – Martin Jun 12 '13 at 3:02
  • $\begingroup$ Jus to check that i'm followin you: Isn't there a constant somewhere, given that $|z|$ isn't $|Re(z)|+|Im(z)|$ but $\sqrt{Re(z)^2 + Im(z)^2}$ @Martin $\endgroup$ – Bill Jun 12 '13 at 3:26
  • $\begingroup$ I'm sorry but I don't follow hoy can you take a decomposition like that. For example if $\mu_{1}(X)$ is to be zero, $\mu_1$ is a positive measure, then $\mu_{1}=0$. @Martin $\endgroup$ – Bill Jun 12 '13 at 3:39

Ok, following @Martin hint, here's why $|\mu|$ is Radon.

  1. Write $\mu=(\mu_1 - \mu_2) +i (\mu_3-\mu_4)$ Using Jordan's decomposition.

  2. Show that there are positive constants $c_1,c_2$ such that

$c_1 (\mu_1+\mu_2+\mu_3+\mu_4)\leq |\mu|\leq c_2(\mu_1+\mu_2+\mu_3+\mu_4)$

  1. Show that when $\mu=\mu_1-\mu_2$ is real, $|\mu|=\mu_1+\mu_2$

  2. Show that for a finite positive borel measure $\mu$, $\mu$ is Radon iff for every borelian $E$, $\epsilon >0$ there are $V,K$, $V$ open, $K$ compact, $K\subseteq E\subseteq V$ such that $\mu(V\setminus K)<\epsilon$.

This actually shows that $|\mu|$ Radon implies $\mu$ Radon

  • 1
    $\begingroup$ My apologies for all this silliness in the comments. You are absolutely right the "identification" of $|\mu|$ with $\mu_1 + \mu_2+\mu_3+\mu_4$ is sheer nonsense. Maybe another way of looking at this: if $\alpha$ is a Radon measure and $f$ is a (locally) integrable function then $f d\alpha$ is a Radon measure. If $f$ is integrable then $|f|$ is integrable, so $|\mu| = |f|\,d\alpha$ is a Radon measure. $\endgroup$ – Martin Jun 12 '13 at 11:02

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