Folland complex measures total variation definition

Question

Consider the following fragments from Folland's text on real analysis:

Why can proposition 3.9 be applied to show uniqueness? I.e. why is $\rho = \mu_1 + \mu_2$ $\sigma$-finite?

Answer 1

You are right. You spotted a "typo" in Folland (there are others).

In the definition of total variation of a complex measure $\nu$, Folland is assuming that $\mu$ is a $\sigma$-finite positive measure (although he only says "$\mu$ is a positive measure"), that is why he can use Proposition 3.9 to prove the uniqueness.

Remark: You can also consider that the definition of total variation of a complex measure $\nu$ in Folland's is meant to be exacyly as written ($\mu$ being any positive measure,not necessarily $\sigma$-finite) and, to apply proposition 3.9, Folland used without mentioning the following result:

Given $\nu$ be a complex measure, $\mu$ be any positive measure (not necessarily $\sigma$-finite) and $f$ a measurable function such that $$\nu = \int f d\mu$$ (which means $\forall E \in \mathcal{M}: \nu(E) = \int_E f d \mu $). Then there is a $\sigma$-finite positive measure $\mu_f$ such that $$\nu = \int f d\mu = \int f d\mu_f $$ and $$|\nu| = \int |f| d\mu_f = \int |f| d\mu$$

Proof: Since $\nu$ is a complex measure, its a finite measure. Since $\forall E \in \mathcal{M}: \nu(E) = \int_E f d \mu $, it follows that $f \in L^1(\mu)$. So we have that $[f\neq 0]= \{x \in X : f(x) \neq 0\}$ is $\sigma$-finite. Let us define $\mu_f$ by, $\forall E \in \mathcal{M}$, $$ \mu_f(E) = \mu([f\neq 0]\cap E)$$ It is immediate that $\mu_f$ is a $\sigma$-finite positive measure and $$\nu = \int f d\mu = \int f \chi_{[f\neq 0]} d\mu =\int f d\mu_f $$ and $$|\nu| = \int |f| d\mu_f = \int |f| \chi_{[f\neq 0]} d\mu= \int |f| d\mu$$.