Does If f be a Measurable Function Imply Finite ∫|f|dm?

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Discussion Overview

The discussion revolves around the implications of a measurable function and whether the condition that the limit of the measure of the set where the function exceeds a threshold exists and is finite implies that the integral of the absolute value of the function is also finite. The context includes theoretical exploration and counterexamples related to Lebesgue measure.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant posits that if a measurable function f satisfies the condition that the limit of the measure of the set where f exceeds a threshold exists and is finite, it may imply that the integral of the absolute value of f is finite.
  • Another participant provides the function f(x) = 1/x for x > 0 as a counterexample to the initial claim.
  • There is a discussion about the measure of the set {x | 1/x > λ} and whether it equals 0, with participants clarifying that it does not.
  • Participants suggest rewriting the condition 1/x > λ to better understand the measure of the corresponding set.
  • One participant emphasizes the importance of visualizing the function's graph to understand the measure's behavior.
  • There is a query regarding the existence of the limit of the measure and its finiteness in relation to the integral of the function.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views regarding the implications of the measurable function and the validity of the counterexample provided.

Contextual Notes

Participants express uncertainty about the measure of specific sets and the implications of the limit condition, indicating a need for clarity on mathematical definitions and structures involved.

Funky1981
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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??
 
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f(x)=1/x, for x>0, is a counterexample.
 
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Erland said:
f(x)=1/x, for x>0, is a counterexample.

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 ??
 
Funky1981 said:
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 ??

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?
 
micromass said:
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?

i would write it into union of ##\{x~\vert~1/x>1/n}##??
 
Look at a graph. The set has a really easy structure.
 
micromass said:
Look at a graph. The set has a really easy structure.

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
 
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?
 

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