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atkrunner@hotmail.com science forum beginner
Joined: 20 Jul 2006
Posts: 5

Posted: Fri Jul 21, 2006 12:12 am Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



Okay I am obtuse, but I see it now. Your first response was right.
Just take the integral of the distribution function over [0, inf.),
which is infinite. And we know there is at least one function
with such a distribution.
I just talked with a classmate who says Tonelli is his
favorite trick in measure theory, now I see why.
Thanks both for the help.
John
Ronald Bruck wrote:
Quote:  Actually, that's why I gave the hint that f \in L^1 iff \sum_n n*m({x:n
= f(x) < n+1}) is finite. That works out to approximately
\sum_n (1/(n log n)  1/((n+1)log(n+1)))
\approx \sum_n n D_n (1/(n log n))
\approx \sum_n 1/(n (log n)^2) + 1/(n log n)
which is infinite. But I like your way much better. I didn't spend
much time on it, because it's obvious, as you say.
I guess f = g^{1} isn't explicit enough for him. I suppose we could
take something asymptotic to this which is close enough to work, and
simple in form, but I'm too lazy to try. (And it's NOT NECESSARY.)

Ron Bruck
In article <DzOvg.1090104$xm3.733246@attbi_s21>, Stephen
MontgomerySmith <stephen@math.missouri.edu> wrote:
That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if f_1=infty.
upendra0111@rediffmail.com wrote:
No, that is not what I am looking for. What I am
asking is the following:
Find a measurable function f, on [0,1] such that
1) f is not in L^1, and
2) r* m{f>=r} >0 as r >infinity.
Stephen MontgomerySmith wrote:
atkrunner@hotmail.com wrote:
My wording may have been a bit unclear.
Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) >0 as r>infinity.
But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].
Can you show that this function is not in L^1{[0,1]}?
If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)
Thanks,
John
Ronald Bruck wrote:
In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck
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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 7:39 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



atkrunner@hotmail.com wrote:
Quote:  Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} >0 in the case.
This is fine.
As far as f_1, where we have restricted f to [0,1]?,
I am stumped.
Any help toward this end will be greatly appreciated.
The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is f_1?

Or, notice that if f is restricted to [0,1]
m{f>=r} = min{ 1/{(r+1)*log(r+1)} , 1 } 

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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 7:33 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



atkrunner@hotmail.com wrote:
Quote:  Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} >0 in the case.
This is fine.
As far as f_1, where we have restricted f to [0,1]?,
I am stumped.
Any help toward this end will be greatly appreciated.
The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is f_1?

OK. Take f such that m{f>=r}=1/{(r+1)*log(r+e)}. 

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atkrunner@hotmail.com science forum beginner
Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 7:11 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} >0 in the case.
This is fine.
As far as f_1, where we have restricted f to [0,1]?,
I am stumped.
Any help toward this end will be greatly appreciated.
Quote:  The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is f_1? 


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Ronald Bruck science forum Guru
Joined: 05 Jun 2005
Posts: 356

Posted: Thu Jul 20, 2006 7:10 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



Actually, that's why I gave the hint that f \in L^1 iff \sum_n n*m({x:n
<= f(x) < n+1}) is finite. That works out to approximately
\sum_n (1/(n log n)  1/((n+1)log(n+1)))
\approx \sum_n n D_n (1/(n log n))
\approx \sum_n 1/(n (log n)^2) + 1/(n log n)
which is infinite. But I like your way much better. I didn't spend
much time on it, because it's obvious, as you say.
I guess f = g^{1} isn't explicit enough for him. I suppose we could
take something asymptotic to this which is close enough to work, and
simple in form, but I'm too lazy to try. (And it's NOT NECESSARY.)

Ron Bruck
In article <DzOvg.1090104$xm3.733246@attbi_s21>, Stephen
MontgomerySmith <stephen@math.missouri.edu> wrote:
Quote:  That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if f_1=infty.
upendra0111@rediffmail.com wrote:
No, that is not what I am looking for. What I am
asking is the following:
Find a measurable function f, on [0,1] such that
1) f is not in L^1, and
2) r* m{f>=r} >0 as r >infinity.
Stephen MontgomerySmith wrote:
atkrunner@hotmail.com wrote:
My wording may have been a bit unclear.
Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) >0 as r>infinity.
But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].
Can you show that this function is not in L^1{[0,1]}?
If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)
Thanks,
John
Ronald Bruck wrote:
In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck

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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 6:46 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



atkrunner@hotmail.com wrote:
Quote:  Stephen MontgomerySmith wrote:
That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if f_1=infty.
I know, that is just the definition of L^1.
If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)
I can show this statement also.
All of the statements you both have offered
are true, and I can prove them. I am actually
not trying to prove anything, however. I am only
looking for a specific function that has the following
properties:
Find a measurable function f, on [0,1] such that
1) f is not in L^1, and
2) r* m{f>=r} >0 as r >infinity.
Can someone find such a function? If so, please let
me know.

The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is f_1? 

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atkrunner@hotmail.com science forum beginner
Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 5:36 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



Stephen MontgomerySmith wrote:
Quote:  That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if f_1=infty.

I know, that is just the definition of L^1.
Quote:  If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)

I can show this statement also.
All of the statements you both have offered
are true, and I can prove them. I am actually
not trying to prove anything, however. I am only
looking for a specific function that has the following
properties:
Find a measurable function f, on [0,1] such that
1) f is not in L^1, and
2) r* m{f>=r} >0 as r >infinity.
Can someone find such a function? If so, please let
me know. 

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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 4:54 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if f_1=infty.
upendra0111@rediffmail.com wrote:
Quote:  No, that is not what I am looking for. What I am
asking is the following:
Find a measurable function f, on [0,1] such that
1) f is not in L^1, and
2) r* m{f>=r} >0 as r >infinity.
Stephen MontgomerySmith wrote:
atkrunner@hotmail.com wrote:
My wording may have been a bit unclear.
Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) >0 as r>infinity.
But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].
Can you show that this function is not in L^1{[0,1]}?
If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)
Thanks,
John
Ronald Bruck wrote:
In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck



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upendra0111@rediffmail.co science forum beginner
Joined: 20 Jul 2006
Posts: 1

Posted: Thu Jul 20, 2006 4:48 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



No, that is not what I am looking for. What I am
asking is the following:
Find a measurable function f, on [0,1] such that
1) f is not in L^1, and
2) r* m{f>=r} >0 as r >infinity.
Stephen MontgomerySmith wrote:
Quote:  atkrunner@hotmail.com wrote:
My wording may have been a bit unclear.
Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) >0 as r>infinity.
But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].
Can you show that this function is not in L^1{[0,1]}?
If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)
Thanks,
John
Ronald Bruck wrote:
In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck



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Stephen MontgomerySmith1 science forum Guru
Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 4:26 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



atkrunner@hotmail.com wrote:
Quote:  My wording may have been a bit unclear.
Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) >0 as r>infinity.
But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].
Can you show that this function is not in L^1{[0,1]}?

If f is positive, then f_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)
Quote: 
Thanks,
John
Ronald Bruck wrote:
In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck



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atkrunner@hotmail.com science forum beginner
Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 4:43 am Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



My wording may have been a bit unclear.
Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) >0 as r>infinity.
But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].
Can you show that this function is not in L^1{[0,1]}?
Thanks,
John
Ronald Bruck wrote:
Quote:  In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck 


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Ronald Bruck science forum Guru
Joined: 05 Jun 2005
Posts: 356

Posted: Thu Jul 20, 2006 3:26 am Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?



In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
<atkrunner@hotmail.com> wrote:
Quote:  Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?
First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,
g(r) = m({x : f(x) >= r}).
I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.
g(r) = 1/((r+1)log(r+1)),
then choose an f which has g as its distribution function. We have
r g(r) > 0 as r > infinity, f >= 0, and f is not in L^1.
Am I missing something?
If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And
\int f dm = \int_0^\infty r dg(r),
where dg(r) is the LebesgueStieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

Ron Bruck 

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atkrunner@hotmail.com science forum beginner
Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 12:46 am Post subject:
Function not in L_1 {[0,1]}, but satisfies ...?



Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that
r* m( f >= r) > 0 as r > infinity?
Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that
N* f^1(N) >0 as N > inf.
If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.
thanks... 

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