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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, jnh, Michael Albanese, Asaf Karagila, quid
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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

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