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!

• 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). – Jochen Dec 2 '15 at 11:26
• @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. – 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. $\mu$.
• 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. – David Lee Dec 2 '15 at 13:52
• link $\mu(E)=\int_E{f}dm,\forall E\in \mathbb{M}$ – David Lee Dec 2 '15 at 15:19
• 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. – David Lee Dec 3 '15 at 2:11