April 20th, 2014, 01:40 PM  #1 
Newbie Joined: Oct 2013 Posts: 25 Thanks: 0  measure theory
Let f be a measurable function. Assume that lim λm({xf(x)>λ}) exists and is finite as λ tends to infinite Here m is the Lebesgue measure in R Does this imply that ∫fdm is finite?? Last edited by James1973; April 20th, 2014 at 01:48 PM. 
April 20th, 2014, 01:57 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625 
No  example: f(x) = 1/x. Even if you exclude an interval around the origin (function is bounded), the integral still diverges.

April 20th, 2014, 02:08 PM  #3 
Newbie Joined: Oct 2013 Posts: 25 Thanks: 0  Thanks for your example. But I cannot convince myself to understand the measure of your case here. I mean how to consider the measure m({x1/x>λ}) and limit of λ m({x1/x>λ})

April 21st, 2014, 01:50 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,641 Thanks: 625 
Consider interval (0,∞)  sufficient because of symmetry. Let λ = 1/y. x < y => 1/x > 1/y = λ. Therefore λ m({x1/x>λ}) = λy = 1. If you exclude an interval around the origin, then 1/x is bounded and m({x1/x>λ}) = 0 for large enough λ. 

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