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Topic: measure theory .... (Read 2271 times) |
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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.
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| « Last Edit: Oct 12th, 2009, 9:30pm by Obob » |
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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 ?
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| « Last Edit: Oct 12th, 2009, 10:30pm by trusure » |
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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.
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| « Last Edit: Oct 13th, 2009, 5:52am by Obob » |
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