Show that if $\nu(E) = \int_E f d \mu$, then $|\nu|(E) = \int_E |f| d\mu$

Question

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.

Answer 1

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:

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 $$|\nu| = \int |f| d\mu_f = \int |f| \chi_{[f\neq 0]} d\mu= \int |f| d\mu$$.

Answer 2

I would say the following things (but I'm not used with non $\sigma$-finite measures, so there might be some mistakes...)

For simplify let's assume that we only want to go from positive measures to real measures.

I don't know how he defines $|\nu|$, but I guess he shows that a $\sigma$-finite real measure $\nu$ can be decomposed as $\nu_+-\nu_-$ for some positive mutually singular measures $\nu_+,\nu_-$. Assume that such a decomposition holds without the $\sigma$-finite assumption (which I guess you can prove. Otherwise, your $|\nu|$ is not even well-defined, right?).

Then, there is a set $A$ such that for any $E$, $\nu_+(E)=\nu(E\cap A)$ and $\nu_-(E)=-\nu(E\setminus A)$.

Then, if $\nu=f\mu$, $\nu_+(E)=\nu(E\cap A)=\int_{E\cap A} f d \mu$ and $\nu_-(E)=-\nu(E\setminus A)=\int_{E\setminus A} - f d \mu$.

If you know that $f\geq0 $ on $A$ (in the almost everywhere sense) and $f\leq 0$ on its complementary, you are done.

$f$ being positive $\mu$-almost everywhere is equivalent to $\int_E f d \mu\geq 0$ for any $E$ . If $E$ is a subset of $A$, then $\int_E f d \mu=\nu(E)=\nu_+(E)\geq 0$, so $f$ is indeed positive almost everywhere on $A$. ''$f$ is negative almost everywhere on the complemetary of $A$'' is similar.