Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. enter image description here

enter image description here

share|improve this question

1 Answer 1

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.

share|improve this answer
I can not get it. could you please explain in more detail? –  zzzhhh Mar 17 '11 at 16:19
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
@zzzhhh do you still need a proof ? –  Airbag Dec 13 '14 at 19:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.