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 Post subject: L^1 versus level set measurePosted: Mon, 3 Oct 2011 23:36:48 UTC
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I'm asked to show, for a positive, measurable function , that is of class --where is a finite measure space-- is equivalent to [tex$\sum_{n=0}^\infty\mu(\{f\ge n\})<\infty[/tex]. However, I don't know if I believe this. I know by construction of the approximation to by simple functions that implies as , but I don't know how to conclude that the decay is fast enough for the measure sum to go to 0. Then again, I didn't use the assumption that yet, so I'm certain that if it's true it's that hypothesis that I am not using effectively, since the simple function approximation is where I use the assumption f>0. Some ideas: it is evident that for NOT to be in , I require that be unbounded on by finiteness of . At the same time, I can WLOG assume off a set of measure 0, since otherwise it's clear that . I'd like to write and so that I can write the sum of the measures of the set as and use some kind of convergence theorem to conclude the integral converges iff , which seems like it would be the LDCT, but I cannot see what the dominating function ought to be. . . perhaps or something like that? And I'm still not seeing how to use , that's definitely a flaw. Any help appreciated. _________________ (\ /) (O.o) (> <) This is Bunny. Copy Bunny into your signature to help him on his way to world domination Top  Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 05:34:55 UTC  Moderator Joined: Mon, 29 Dec 2008 17:49:32 UTC Posts: 6066 Location: 127.0.0.1, ::1 (avatar courtesy of UDN) Shadow wrote: I'm asked to show, for a positive, measurable function , that is of class --where is a finite measure space-- is equivalent to [tex$\sum_{n=0}^\infty\mu(\{f\ge n\})<\infty[/tex]. However, I don't know if I believe this. I know by construction of the approximation to by simple functions that implies as , but I don't know how to conclude that the decay is fast enough for the measure sum to go to 0. Then again, I didn't use the assumption that yet, so I'm certain that if it's true it's that hypothesis that I am not using effectively, since the simple function approximation is where I use the assumption f>0.

Some ideas: it is evident that for NOT to be in , I require that be unbounded on by finiteness of . At the same time, I can WLOG assume off a set of measure 0, since otherwise it's clear that .

I'd like to write and

so that I can write the sum of the measures of the set as and use some kind of convergence theorem to conclude the integral converges iff , which seems like it would be the LDCT, but I cannot see what the dominating function ought to be. . . perhaps or something like that? And I'm still not seeing how to use , that's definitely a flaw.

Any help appreciated.

Dominate your f by , and of course f dominates

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 06:06:41 UTC
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I think I understand, when I integrate that step I get exactly my sum, and the integral of is easy to calculate, and this gives me one direction, Perhaps I do the ceiling function to get the other bound, since I think that is just a shifted sum?

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 06:19:13 UTC
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I think I understand, when I integrate that step I get exactly my sum, and the integral of is easy to calculate, and this gives me one direction, Perhaps I do the ceiling function to get the other bound, since I think that is just a shifted sum?

Essentially, the step and step+1 giving you control over f from below and above, which you use to get the lower and upper bounds on the integral of f --- the sum and sum+|mu|.

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 08:07:15 UTC
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outermeasure wrote:
I think I understand, when I integrate that step I get exactly my sum, and the integral of is easy to calculate, and this gives me one direction, Perhaps I do the ceiling function to get the other bound, since I think that is just a shifted sum?

Essentially, the step and step+1 giving you control over f from below and above, which you use to get the lower and upper bounds on the integral of f --- the sum and sum+|mu|.

Ah wait a second though, the integral of is the sum of , so I think I still don't see how this relates, because it's not the full set of and if it were, then it the sum in question would be which is not (obviously) equivalent to the convergence of the sum , although one direction is trivial.

Edit: , not

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 10:54:27 UTC
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outermeasure wrote:
I think I understand, when I integrate that step I get exactly my sum, and the integral of is easy to calculate, and this gives me one direction, Perhaps I do the ceiling function to get the other bound, since I think that is just a shifted sum?

Essentially, the step and step+1 giving you control over f from below and above, which you use to get the lower and upper bounds on the integral of f --- the sum and sum+|mu|.

Ah wait a second though, the integral of is the sum of , so I think I still don't see how this relates, because it's not the full set of and if it were, then it the sum in question would be which is not (obviously) equivalent to the convergence of the sum , although one direction is trivial.

Oops, I thought the sum was from n=1. The effect of starting at n=0 is that it is the integral of instead.

Edit: Correct typo.

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 18:10:45 UTC
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outermeasure wrote:
outermeasure wrote:
I think I understand, when I integrate that step I get exactly my sum, and the integral of is easy to calculate, and this gives me one direction, Perhaps I do the ceiling function to get the other bound, since I think that is just a shifted sum?

Essentially, the step and step+1 giving you control over f from below and above, which you use to get the lower and upper bounds on the integral of f --- the sum and sum+|mu|.

Ah wait a second though, the integral of is the sum of , so I think I still don't see how this relates, because it's not the full set of and if it were, then it the sum in question would be which is not (obviously) equivalent to the convergence of the sum , although one direction is trivial.

Oops, I thought the sum was from n=1. The effect of starting at n=0 is that it is the integral of instead.

I suppose that last sum should be: ?

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 19:05:56 UTC
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I suppose that last sum should be: ?

Yes.

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 Post subject: Re: L^1 versus level set measurePosted: Tue, 4 Oct 2011 19:08:47 UTC
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Thanks outermeasure!

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