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April 20th, 2014, 01:40 PM   #1
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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 01:48 PM.
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April 20th, 2014, 01:57 PM   #2
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No - example: f(x) = |1/x|. Even if you exclude an interval around the origin (function is bounded), the integral still diverges.
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April 20th, 2014, 02:08 PM   #3
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Quote:
Originally Posted by mathman View Post
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})
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April 21st, 2014, 01:50 PM   #4
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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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