L^1 versus level set measure

Shadow · **Posted:** Mon, 3 Oct 2011 23:36:48 UTC

I'm asked to show, for a positive, measurable function

, that

is of class $L^1(\Omega)$ --where $(\Omega, \Sigma,\mu)$ 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 $f\in L^1$ implies $\mu(\{f\ge n\})=o({1\over n})$ as $n\to\infty$ , 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 $\mu(\Omega)<\infty$ 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 L^1

, I require that

be unbounded on $\Omega$ by finiteness of $\mu(\Omega)$ . At the same time, I can WLOG assume $f<\infty$ off a set of measure 0, since otherwise it's clear that $f\not\in L^1(\Omega)$ .

I'd like to write $$g_n=\sum_{k=1}^{n2^n} {k-1\over 2^n}1_{\{(k-1)2^{-n}\le f<k2^{-n}\}}+n1_{\{f\ge n\}}$ and $h_n=g_n+1_{\{f\ge n\}}$

so that I can write the sum of the measures of the set as $$\int h_n-g_n$ and use some kind of convergence theorem to conclude the integral converges iff $f\in L^1(\Omega)$ , 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 $\mu(\Omega)<\infty$ , that's definitely a flaw.

Any help appreciated.

outermeasure · **Posted:** Tue, 4 Oct 2011 05:34:55 UTC

Shadow wrote:

I'm asked to show, for a positive, measurable function

, that

is of class $L^1(\Omega)$ --where $(\Omega, \Sigma,\mu)$ 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 $f\in L^1$ implies $\mu(\{f\ge n\})=o({1\over n})$ as $n\to\infty$ , 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 $\mu(\Omega)<\infty$ 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 L^1

, I require that

be unbounded on $\Omega$ by finiteness of $\mu(\Omega)$ . At the same time, I can WLOG assume $f<\infty$ off a set of measure 0, since otherwise it's clear that $f\not\in L^1(\Omega)$ .

I'd like to write $$g_n=\sum_{k=1}^{n2^n} {k-1\over 2^n}1_{\{(k-1)2^{-n}\le f<k2^{-n}\}}+n1_{\{f\ge n\}}$ and $h_n=g_n+1_{\{f\ge n\}}$

so that I can write the sum of the measures of the set as $$\int h_n-g_n$ and use some kind of convergence theorem to conclude the integral converges iff $f\in L^1(\Omega)$ , 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 $\mu(\Omega)<\infty$ , that's definitely a flaw.

Any help appreciated.

Dominate your f by $\lim_{\epsilon\downarrow 0}\lceil f+\epsilon\rceil=1+\lfloor f\rfloor$ , and of course f dominates $\lfloor f\rfloor$

Shadow · **Posted:** Tue, 4 Oct 2011 06:06:41 UTC

I think I understand, when I integrate that step I get exactly my sum, and the integral of 1+f

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?

outermeasure · **Posted:** Tue, 4 Oct 2011 06:19:13 UTC

Shadow wrote:

I think I understand, when I integrate that step I get exactly my sum, and the integral of 1+f

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|.

Shadow · **Posted:** Tue, 4 Oct 2011 08:07:15 UTC

outermeasure wrote:

Shadow wrote:

I think I understand, when I integrate that step I get exactly my sum, and the integral of 1+f

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 $$\lfloor f\rfloor\,$ is the sum of $n\mu(\{n\le f<n+1\})$ , so I think I still don't see how this relates, because it's not the full set of $f\ge n$ and if it were, then it the sum in question would be $$\sum n\mu(\{f\ge n\})$ which is not (obviously) equivalent to the convergence of the sum $$\sum \mu(\{f\ge n\})$ , although one direction is trivial.

Edit: $\mu(\{n\le f <n+1\})$ , not $\mu(\{n\le f <n\})$

outermeasure · **Posted:** Tue, 4 Oct 2011 10:54:27 UTC

Shadow wrote:

outermeasure wrote:

Shadow wrote:

I think I understand, when I integrate that step I get exactly my sum, and the integral of 1+f

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 $$\lfloor f\rfloor\,$ is the sum of $n\mu(\{n\le f<n\})$ , so I think I still don't see how this relates, because it's not the full set of $f\ge n$ and if it were, then it the sum in question would be $$\sum n\mu(\{f\ge n\})$ which is not (obviously) equivalent to the convergence of the sum $$\sum \mu(\{f\ge n\})$ , 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 $1+\lfloor f\rfloor$ instead.

$\begin{aligned} \sum_{n=0}^\infty \mu\{f\geq n\} &=\sum_{n=0}^\infty\sum_{r=n}^\infty\mu\{r\leq f<r+1\}\\ &=\sum_{r=0}^\infty\sum_{n=0}^r \mu\{r\leq f<r+1\}\\ &=\sum_{r=0}^\infty (r+1)\mu\{\lfloor f\rfloor=r\} \end{aligned}$

Edit: Correct typo.

Shadow · **Posted:** Tue, 4 Oct 2011 18:10:45 UTC

outermeasure wrote:

Shadow wrote:

outermeasure wrote:

Shadow wrote:

I think I understand, when I integrate that step I get exactly my sum, and the integral of 1+f

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 $$\lfloor f\rfloor\,$ is the sum of $n\mu(\{n\le f<n\})$ , so I think I still don't see how this relates, because it's not the full set of $f\ge n$ and if it were, then it the sum in question would be $$\sum n\mu(\{f\ge n\})$ which is not (obviously) equivalent to the convergence of the sum $$\sum \mu(\{f\ge n\})$ , 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 $1+\lfloor f\rfloor$ instead.

$\begin{aligned} \sum_{n=0}^\infty \mu\{f\geq n\} &=\sum_{n=0}^\infty\sum_{r=n}^\infty\mu\{r\leq f<r+1\}\\ &=\sum_{r=0}^\infty\sum_{n=0}^r \mu\{r\leq f<r+1\}\\ &=\sum_{r=0}^\infty (r+1)\mu\{\lfloor\floor r\rfloor=r\} \end{aligned}$

I suppose that last sum should be: $$\sum_{r=0}^\infty (r+1) \mu(\{\lfloor f\rfloor = r\})$ ?

outermeasure · **Posted:** Tue, 4 Oct 2011 19:05:56 UTC

Shadow wrote:

I suppose that last sum should be: $$\sum_{r=0}^\infty (r+1) \mu(\{\lfloor f\rfloor = r\})$ ?

Yes.

Shadow · **Posted:** Tue, 4 Oct 2011 19:08:47 UTC

Thanks outermeasure!

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L^1 versus level set measure

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