Signed measure and complex measure integral

Question

Let $(X,\mathscr{A},\mu)$ be a measure space and $f$ an $\mathscr{A}$-measurable $\bar{\mathbb{R}}$-valued function such that $\int_{X}fd\mu$ exists (but not necessarily finite). I have show that $$\nu(E):=\int_{E}fd\mu\ \ \ \ (E\in \mathscr{A})$$ is a signed measure on $(X,\mathscr{A})$.By the Jardon decomposition, we can define two positive measures $\nu^{+}(E)$ and $\nu^{-}(E)$. Also, we can decompose $f$ into $f^+$ and $f^-$ shch that $f=f^+-f^-$ where $f^+$ is coincide with $f$ if $f$ is nonnegative and vanish at which $f$ is negative; and $f^-$ similarly. How can I show that $$\nu^+(E)=\int_{E}f^+d\mu\ \ {\rm{,and}}\ \ \nu^-(E)=\int_{E}f^-d\mu.$$

When replacing $f$ by a complex-valued integrable function $g$. I have shown that $\nu(E)$ is a complex measure on $(X,\mathscr{A})$. But how can I show that $$|\nu|(E)=\int_{E}|g|d\mu.$$ where $|\nu|(E)$ is definied by $\rm{sup}\left\{ \sum_{i=1}^{n}|\nu(A_j)| \ \mid A_1,...,A_n {\rm{\ is\ a\ partition\ of\ }}E {\rm{\ in\ }}\mathscr{A}\right\}$.

Thanks in advance.

Answer 1

Let $A=\{x~:~f(x)\geq 0\}$ and $B=\{x~:~f(x)<0\}$. Note that $$ \nu(A\cap E)=\int_{E}f^{+}~d\mu,\quad\nu(B\cap E)=\int_{E} f^{-}~d\mu, $$ so the pair $A$ and $B$ form a Hahn decomposition of $X$, and therefore $d\nu^{\pm}=f^{\pm}~d\mu$.

For your second question, consider the polar representation $d\nu=h~d\lvert\nu\rvert$. Observe $$ \lvert\nu\rvert(E)=\int_{E} \bar{h}h~d\lvert\nu\rvert=\int_{E}\bar{h}~d\nu=\int_{E}\bar{h}g~d\mu. $$ Also, the Radon-Nikodym derivative of a positive measure is positive, thus $\bar{h}g\geq 0$ almost everywhere [$\mu$], hence $\bar{h}g=\lvert{g}\rvert$ a.e. [$\mu$].