Lebesgue-Radon-Nikodym Theorem shows that:

If $\mathbb{M}$ is a $\sigma$-algbra on the set X. $\mu,\lambda$ are a $\sigma$-finite positive measure and $\sigma$-finite signed measure on $\mathbb{M}$ respectively. then $\lambda$ has a decomposition.

I wander why we do not consider the case when $\mu$ is not positive.

I think maybe it is difficult to define integral with signed measure or complex-valued measure,so we just do not take the case when $\mu$ is not positive. But I am not clear about this.

Any hint will be greatly appreciated!

  • $\begingroup$ To define e.g. what the continuity $\lambda_c \ll \mu$ means you would need the total variation $|\mu|$ of $\mu$ anyway (every $|\mu|$- null set is $\lambda_c$-null). $\endgroup$ – Jochen Dec 2 '15 at 11:26
  • $\begingroup$ @Jochen Do you mean $|\mu|$ needs to be bounded? If so, then every complex-valued measure has a bounded total variation. But Lebesgue-Radon-Nikodym Theorem still do not care about the case when $\mu$ is complex-valued measure. $\endgroup$ – David Lee Dec 2 '15 at 11:35

Consider the total variation $|\mu|$ of $\mu$. Then $|\mu|$ is $\sigma$-finite and positive. Moreover $\mu \ll |\mu|$. Then, you get a function $\mu' \in L^1(|\mu|)$ (in fact $|\mu'|=1$ $|\mu|$-a.e.), such that $$\mu(E) = \int_E \mu' \, \mathrm{d}|\mu|.$$

Now, you can apply the Lebesgue-Radon-Nikodým theorem to $\lambda$ and $|\mu|$. In particular, there is $h \in L^1(|\mu|)$, such that $$\lambda_a(E) = \int_E h \, \mathrm{d}|\mu|.$$ This could also be written as $$\lambda_a(E) = \int_E \frac{h}{\mu'} \, \mathrm{d}\mu$$ if the integration w.r.t. the complex measure $\mu$ is defined accordingly. In particular, $h / \mu'$ could be considered a Radon-Nikodým derivative of $\lambda_a$ w.r.t. $\mu$.

  • $\begingroup$ I know that total variation of complex-valued measure is finite. But i do not think total variation of signed measure is $\sigma$-finite.e.g. define function f on $\mathbb{R}$ like this : $f([n,n+1/2])=1$ for $\forall n\in \mathbb{N}$, otherwise f(x) is $-\infty$. and use this function to construct another measure on $\mathbb{R}$ by integral. Then this new measure is signed measure but its total variation is not $\sigma$-finite. $\endgroup$ – David Lee Dec 2 '15 at 13:52
  • $\begingroup$ How do you define "signed measure"? $\endgroup$ – gerw Dec 2 '15 at 15:04
  • $\begingroup$ link $\mu(E)=\int_E{f}dm,\forall E\in \mathbb{M}$ $\endgroup$ – David Lee Dec 2 '15 at 15:19
  • $\begingroup$ Citation from your link: There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values. To avoid confusion, this article will call these two cases "finite signed measures" and "extended signed measures". $\endgroup$ – gerw Dec 2 '15 at 18:33
  • $\begingroup$ so what i mean signed measure is the so-called extended signed measure. and this kind of measure may not have $\sigma$-finite total variation. $\endgroup$ – David Lee Dec 3 '15 at 2:11

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