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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. Tags measure, theory Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jeh Real Analysis 0 July 19th, 2012 09:13 PM stephanie_unizh Real Analysis 1 February 2nd, 2012 12:48 PM ANANYA Real Analysis 10 June 25th, 2010 01:57 PM Real Analysis 1 April 14th, 2009 04:00 PM sriram Real Analysis 0 June 16th, 2007 05:17 AM

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