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

[Funky1981]

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??

 

[Erland]

f(x)=1/x, for x>0, is a counterexample.

 

[micromass]

Consider $f(x) = 1/x$.

Also check http://en.wikipedia.org/wiki/Marcinkiewicz_interpolation_theorem#Preliminaries

 

[Funky1981]

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 ??

 

[micromass]

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?

 

[Funky1981]

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 $\{x~\vert~1/x>1/n}$??

 

[micromass]

Look at a graph. The set has a really easy structure.

 

[Funky1981]

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

 

[Erland]

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?