Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measure theory

  1. Apr 20, 2014 #1
    If f be a measurable function. Assume that

    lim λm({x|f(x)>λ}) exists and is finite as λ tends to infinite

    Does this imply that ∫|f|dm is finite?

    Here m is the Lebesgue measure in R

    If not can anyone give me an example??
  2. jcsd
  3. Apr 20, 2014 #2


    User Avatar
    Science Advisor

    f(x)=1/x, for x>0, is a counterexample.
  4. Apr 20, 2014 #3
  5. Apr 20, 2014 #4
    Thanks for your example. But I cannot convince myself to understand the measure of your case here. Is the measure m({x|1/x>λ}) equal to 0 ??
  6. Apr 20, 2014 #5
    No, it's not.

    Can you write ##\{x~\vert~1/x>\lambda\}## in a more convenient way that allows you to see easily what the Lebesgue measure is?
  7. Apr 20, 2014 #6
    i would write it into union of ##\{x~\vert~1/x>1/n}##??
  8. Apr 20, 2014 #7
    Look at a graph. The set has a really easy structure.
  9. Apr 20, 2014 #8
    1/x> 0 for all x >0 then so the measure should be infinte. but now why lim λm({x|f(x)>λ}) exists and is finite
  10. Apr 22, 2014 #9


    User Avatar
    Science Advisor

    As Micromass wrote: try to rewrite the condition 1/x > λ in a way such that the measure of the corresponding set is easily seen:

    If 1/x > λ, what can you then say about x, in terms of λ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook