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 April 20th, 2014, 12:40 PM #1 Newbie   Joined: Oct 2013 Posts: 25 Thanks: 0 measure theory Let f be a measurable function. Assume that lim λm({x|f(x)>λ}) exists and is finite as λ tends to infinite Here m is the Lebesgue measure in R Does this imply that ∫|f|dm is finite?? Last edited by James1973; April 20th, 2014 at 12:48 PM.
 April 20th, 2014, 12:57 PM #2 Global Moderator   Joined: May 2007 Posts: 6,766 Thanks: 698 No - example: f(x) = |1/x|. Even if you exclude an interval around the origin (function is bounded), the integral still diverges. Thanks from James1973
April 20th, 2014, 01:08 PM   #3
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 Originally Posted by mathman No - example: f(x) = |1/x|. Even if you exclude an interval around the origin (function is bounded), the integral still diverges.
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({x|1/x>λ}) and limit of λ m({x|1/x>λ})

 April 21st, 2014, 12:50 PM #4 Global Moderator   Joined: May 2007 Posts: 6,766 Thanks: 698 Consider interval (0,∞) - sufficient because of symmetry. Let λ = 1/y. x < y => 1/x > 1/y = λ. Therefore λ m({x|1/x>λ}) = λy = 1. If you exclude an interval around the origin, then 1/x is bounded and m({x|1/x>λ}) = 0 for large enough λ.

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