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.

I attempted some easy multiple-choice and T/F questinos to test if I am entirely clear about the topic before I do any proofy works. It's essential to be clear on these basic concepts and ideas. Let me know whether I am right or wrong and I'll appreciate if you briefly explain:) Thanks for helping! I want to check quickly if my answer is right.

You may just put the answers,or if you may, some explain would be great!

enter image description here

  1. a. Yes. Because f(x) is always positive b. Yes. because of the term " x^2"

  2. a. no,because charactertistic function of the rational are not measure zero. therefore not Riemann integrable?? b. Probably no. Because again, charactertistic function of the rational are very "discrete".

3 a.Yes, Cantor set is measure zero. And one function is Riemann integrable as long as its discontinuity forms measure zero. b. Dounno. Honestly, I have no clue. Probably no again. where should I start?

share|improve this question

1 Answer 1

up vote 2 down vote accepted
  1. a) YES: $f$ strictly increases if $f'>0$ everywhere, b) NO: $f$ would be convex if $f''\ge 0$ was everywhere, but now $f''$ takes strictly negative as well.
  2. $\Bbb Q$ is Lebesgue measurable, has measure $0$ as a countable union of points. So, a) YES, b) YES only because that limit (the differential) would be $0$ anyway.
  3. a) YES, as you said, and b) NO, because now that limit would be $1$ for $[0,1]\setminus C$.
share|improve this answer

Your Answer

 
discard

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