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   Author  Topic: measure theory ....  (Read 2270 times)
trusure
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measure theory ....  
« on: Oct 12th, 2009, 7:08pm »
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I just proved the following result:
if  
(1)  f_n is a sequence which converges to f a.e.,  
(2)  f_n>=0, and lebesgue measurable on R.  
(3)  integral(f)  is finite, and  
(4)  integral (f_n)   converges to integral (f)  then:
 
    integral (f_n) converges to integral (f)  over any measurable set A.
 
(note: the integration is Lebesgue integral)
 
Now, I'm looking for a counterexample that this result is not true if integral (f) is infinity.
 
so.... any one can help me  ....
 
 
thanks for helping in advance
 
 
 
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Obob
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Re: measure theory ....  
« Reply #1 on: Oct 12th, 2009, 9:30pm »
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Let fn(x) = max (1/x2,1/n) on the whole real line.
« Last Edit: Oct 12th, 2009, 9:30pm by Obob » IP Logged
trusure
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Re: measure theory ....  
« Reply #2 on: Oct 12th, 2009, 10:29pm »
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ok, so we have f_n ---> f=1/x^2,
but integral of f over R is finite =0, ??!!
 
moreover, what about the set A we will use to get a contradiction ?
 
« Last Edit: Oct 12th, 2009, 10:30pm by trusure » IP Logged
Obob
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Re: measure theory ....  
« Reply #3 on: Oct 13th, 2009, 5:18am »
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I'm guessing this is homework, so you need to fill in some of the details.  And it is definitely not true that the integral of 1/x^2 over R is finite (let alone zero).  How on earth could it be zero?  It is a function that is strictly positive everywhere.
 
There's a couple other examples that fail for the same reason:
 
Let your measure space X consist of two points p & q, both with infinite measure.  Put f_n(p) = 1, f_n(q) = 1/n.
 
Or let your measure space X be a disjoint union of two smaller spaces A and B, where A has infinite measure and B has positive measure.  Let g be a function on B with infinite integral, and define f_n(a) = 1/n for a in A, f_n(b) = g(b) for b in B.
 
The first example I gave can be seen as being essentially equivalent to the second construction here.
« Last Edit: Oct 13th, 2009, 5:52am by Obob » IP Logged
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