Folland Theorem 3.22 Proof Explanation

Question

At the first step of Theorem 3.22 of Folland, he mentioned that $d\nu = d\lambda + f\,dm$ implies $d|\nu| = d|\lambda| + |f| dm$:

I am aware of the similar questions has been asked before on this site, but I am not entirely convinced of the answers provided. In particular, I have been trying to show the claim for the case where $\nu$ is a signed measure:

Since the total variation $|\lambda| \perp m$ by the Radon Nikodym representation of $\nu$, we can find set $N \in \mathcal{B}_{\mathbb{R}^n}$ such that $|\lambda|(N) = m(N^c) = 0$. Write $\nu(A) = \lambda(A) + \int_A f \,d\mu$ for all $A \in \mathcal{B}_{\mathbb{R}^n}$. Moreover, we have $$ |\nu|(A) = \nu^+(A) + \nu^-(A). $$ I then thought it would be helpful if I could identity what $\nu^+$ and $\nu^-$ are, but haven't been able to do so. Any suggestions?

Answer 1

What you are trying to prove follows pretty quickly from the Hahn decomposition. You have a decomposition $S_+\cup S_-= \mathbb R^n$ with $\nu^+=\nu\llcorner_{S_+}$, $\nu^-=-\nu\llcorner_{S_-}$, $S_+\cap S_-=\emptyset$.

Since $\nu\llcorner_{S_+}$ is a positive measure, it follows (by considering $S_+\cap N$ and $S_+\cap N^c$ separately) that $f\geq 0$ $m$-a.e. on $S_+$ and $\lambda\llcorner_{S_+}$ is a positive measure, and similar statements for the negative variations hold for $S_-$, so we have

$$|\nu|(A\cap S_+)=\nu(A\cap S_+)=\lambda(A\cap S_+)+\int_{A\cap S_+}f\,dm =|\lambda|(A\cap S_+) + \int_{A\cap S_+}|f|\,dm \text{,}$$ and similarly $$|\nu|(A\cap S_-)=-\nu(A\cap S_-)=-\lambda(A\cap S_-)-\int_{A\cap S_-}f\,dm =|\lambda|(A\cap S_-) + \int_{A\cap S_-}|f|\,dm \text{,}$$ and the result follows.

Answer 2

Let $\mu_1$ and $\mu_2$ be signed measures.

The Jordan decomposition theorem implies there exist unique positive measures $\mu_1^+$ and $\mu_1^-$ such that $\mu_1 = \mu_1^+ - \mu_1^-$. Similarly there are positive measures $\mu_2^+$ and $\mu_2^-$ such that $\mu_2 = \mu_2^+ - \mu_2^-$. The total variation of $\mu_1$ is defined as $|\mu_1| = \mu_1^+ + \mu_1^-$; similarly $|\mu_2| = \mu_2^+ + \mu_2^-$.

So if $\nu = \mu_1 + \mu_2$, it is easy to verify that $\nu = (\mu_1^+ + \mu_2^+) - (\mu_1^- + \mu_2^-)$ is the Jordan decomposition of the signed measure $\nu$, and thus $|\nu| = \mu_1^+ + \mu_2^+ + \mu_1^- + \mu_2^- = |\mu_1| + |\mu_2|$.