A problem about total variation of complex measure [closed]

This problem is Proposition 3.13 c. ( or Exercise 18) of Folland's "real analysis: modern techniques and their applications" in page 94. I just have no idea how to prove it, can you help me? Please use notions and definitions in this book. Thanks! The following images are section 3.3 that contains this problem and related notions and definitions.

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closed as off-topic by Rory Daulton, jnh, Michael Albanese, Asaf Karagila, quidMay 16 at 22:32

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That is an also lot to ask of us to do for you, especially as you have not given us any work of your own. Do you at least understand parts $a$ and $b$ of the proposition? –  Rory Daulton May 16 at 20:14

By part (b) you have a measurable $g$ such that $d\nu = gd|\nu|$ and $|g(x)| = 1$ for all $x$. Use this and apply $|\int fd\nu| \leq \int|f|d\nu$, replacing $d\nu$ with $gd\nu$ first.
To be specific, if $f\in L^1(|\nu|)$, how to prove that $f\in L^1(\nu)$? –  zzzhhh Mar 17 '11 at 17:18
I found a method as follows: I wish to show that $\nu_r^+(E)\leq |\nu|(E)$ for all measurable set E, so I will obtain $\int fd\nu_r^+\leq\int fd|\nu|$ step by step from characteristic function to nonnegative simple function and then nonnegative measurable function. Since $\nu_r=({\rm Re}f)d\mu, |\nu_r|=|{\rm Re}f|d\mu\leq|f|d\mu=|\nu|$, but $\nu_r^+\leq|\nu_r|$, We get the desired result. But do we have better method to prove it? –  zzzhhh Mar 17 '11 at 18:06