Consider the following fragment from Folland's book on real analysis:
I'm trying to show that if $$\forall E \in \mathcal{M}: \nu(E) = \int_E f d \mu \quad \quad(d \nu = f d \mu)$$
where $f \in L^1(\mu)$ and $\mu$ is a positive measure, then
$$\forall E \in \mathcal{M}:|\nu|(E) = \int_E |f| d \mu \quad \quad (|\nu| = |f|d \mu)$$
You may ask: Isn't this immediate from Folland's definition? It turns out the answer is no: in Folland's definition $\mu$ is taken to be $\sigma$-finite (see here: Folland complex measures total variation definition)
I have however not a clue how I can reduce this to the $\sigma$-finite case.
Best Answer
Let us go step by step.
Step 1. Given $\nu$ a complex measure, let $\Re(\nu)$ be the real part of $\nu$ and $\Im(\nu)$ be the imaginary part of $\nu$. It is easy to see that $\Re(\nu)$ and $\Im(\nu)$ are finite signed measure and
$$ \nu = \Re(\nu) +i\Im(\nu) $$ Let us then consider, initially, finite signed measures
Step 2. Given $\nu$ a finite signed measure, using Jordan decomposition, we have two finite positive measures: $\nu^+$ and $\nu^-$ . We have that $$ \nu = \nu^+ - \nu^- $$
Step 3. Let us prove (using Folland's definition of $|\nu|$) that $$ |\nu| = \nu^+ + \nu^- $$ Proof: Since a finite signed measure is a special case of a complex measure, we can use Folland's definition of total variation for complex mesures.
Let $\mu$ be any $\sigma$-finite positive measure 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 $).
Applying Jordan decomposition to the measure $\int f d\mu$, we have :
$$ \nu^+ = \left ( \int f d\mu \right)^+ = \int f^+ d\mu $$ and $$ \nu^- = \left ( \int f d\mu \right)^- = \int f^- d\mu $$
According to Folland's definition, we have $$ |\nu| = \int |f| d\mu = \int f^+ d\mu + \int f^- d\mu = \nu^+ + \nu^- $$
Step 4. Given $\nu$ be a finite signed 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 $$ |\nu| = \int |f| d\mu $$ Proof: Let $\nu$ be a finite signed measure, $\mu$ be any positive measure (not necessarily $\sigma$-finite) and $f$ a measurable function such that $$\nu = \int f d\mu$$
Applying Jordan decomposition, we have (as in step 3): $$ \nu^+ = \left ( \int f d\mu \right)^+ = \int f^+ d\mu $$ and $$ \nu^- = \left ( \int f d\mu \right)^- = \int f^- d\mu $$ Now, using step 3, we know $$|\nu|= \nu^+ + \nu^- = \int f^+ d\mu + \int f^- d\mu = \int |f| d\mu $$ So we have proved $$ |\nu| = \int |f| d\mu $$
Step 5. Given $\nu$ be 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 $$ |\nu| = \int |f| d\mu $$ Proof: Apply step 4 to the real and imaginary parts of $\nu$ and $f$, and combine the results.
Important Remark: There is a shorter (and more elegant) way to prove your result. It is using the following result:
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 $$|\nu| = \int |f| d\mu_f = \int |f| \chi_{[f\neq 0]} d\mu= \int |f| d\mu$$.