Lebesgue integral simple function query
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I'm trying to show that $int_{-infty}^{infty}Big(sum_{n=1}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}Big)dmu = infty$ where $mu$ is the Lebesgue measure.
Approach:
I swap the sum and the integral (justified by the fact that we have a non-negative function).
Then I am slightly confused.
To evaluate this integral I need to be able to write it as a simple function. In the examples I've seen the function has always been written as a finite linear combination of indicator functions.
i.e. $$f(x)=sum_{n=1}^{N}b_{n}chi_{B_{n}}(x)$$ where $B_{n}$ is a measurable set and $b_{n} in mathbb{R}$ and $int f dmu:= sum_{n=1}^{N} b_{n}mu(B_{n}).$
Here it seems to be the case that $N=infty$ and therefore it would be an infinite linear combination. Is this allowed?
measure-theory lebesgue-integral
asked Nov 17 at 8:39
mathlearner
6
|
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up vote
0
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I'm trying to show that $int_{-infty}^{infty}Big(sum_{n=1}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}Big)dmu = infty$ where $mu$ is the Lebesgue measure.
Approach:
I swap the sum and the integral (justified by the fact that we have a non-negative function).
Then I am slightly confused.
To evaluate this integral I need to be able to write it as a simple function. In the examples I've seen the function has always been written as a finite linear combination of indicator functions.
i.e. $$f(x)=sum_{n=1}^{N}b_{n}chi_{B_{n}}(x)$$ where $B_{n}$ is a measurable set and $b_{n} in mathbb{R}$ and $int f dmu:= sum_{n=1}^{N} b_{n}mu(B_{n}).$
Here it seems to be the case that $N=infty$ and therefore it would be an infinite linear combination. Is this allowed?
measure-theory lebesgue-integral
asked Nov 17 at 8:39
mathlearner
6
You say you have swapped the integral and sum. Doesn't that mean you get the integral in from of a single simple function? Namely $$int nchi_{(frac 1{n-1}, frac 1 n]} , dmu$$
– Shashi
Nov 17 at 8:55
@Shashi If i swap them I have $sum_{n=1}^{infty}int_{-infty}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}.$ But then this a sum of an integral, but I wasn't sure whether I can evaluate this as $n mu((frac{1}{n+1}, frac{1}{n}])$ because the $n$ that is multiplying the measure could be infinite and I thought it had to be a finite linear combination?
– mathlearner
Nov 17 at 8:57
Yes. You can do the integration first, right?
– Shashi
Nov 17 at 8:58
Notice that it’s not hard to compute $int_{-infty}^infty nchi_{(frac1{n+1}, frac1n]} dmu$
– Alonso Delfín
Nov 17 at 9:00
@AlonsoDelfín so is it okay for there to be an infinite number of linear combinations of indicator functions then. My confusion is that the index in the sum goes to $infty$ and therefore we have an infinite linear combination (potentally). I know that $n mu((frac{1}{n+1},frac{1}{n}]) = frac{1}{n+1}$ if I can do that.
– mathlearner
Nov 17 at 9:04
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to show that $int_{-infty}^{infty}Big(sum_{n=1}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}Big)dmu = infty$ where $mu$ is the Lebesgue measure.
Approach:
I swap the sum and the integral (justified by the fact that we have a non-negative function).
Then I am slightly confused.
To evaluate this integral I need to be able to write it as a simple function. In the examples I've seen the function has always been written as a finite linear combination of indicator functions.
i.e. $$f(x)=sum_{n=1}^{N}b_{n}chi_{B_{n}}(x)$$ where $B_{n}$ is a measurable set and $b_{n} in mathbb{R}$ and $int f dmu:= sum_{n=1}^{N} b_{n}mu(B_{n}).$
Here it seems to be the case that $N=infty$ and therefore it would be an infinite linear combination. Is this allowed?
measure-theory lebesgue-integral
asked Nov 17 at 8:39
mathlearner
6
I'm trying to show that $int_{-infty}^{infty}Big(sum_{n=1}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}Big)dmu = infty$ where $mu$ is the Lebesgue measure.
Approach:
I swap the sum and the integral (justified by the fact that we have a non-negative function).
Then I am slightly confused.
To evaluate this integral I need to be able to write it as a simple function. In the examples I've seen the function has always been written as a finite linear combination of indicator functions.
i.e. $$f(x)=sum_{n=1}^{N}b_{n}chi_{B_{n}}(x)$$ where $B_{n}$ is a measurable set and $b_{n} in mathbb{R}$ and $int f dmu:= sum_{n=1}^{N} b_{n}mu(B_{n}).$
Here it seems to be the case that $N=infty$ and therefore it would be an infinite linear combination. Is this allowed?
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Nov 17 at 8:39
mathlearner
6
asked Nov 17 at 8:39
mathlearner
6
asked Nov 17 at 8:39
mathlearner
6
asked Nov 17 at 8:39
mathlearner
6
asked Nov 17 at 8:39
mathlearner
6
6
You say you have swapped the integral and sum. Doesn't that mean you get the integral in from of a single simple function? Namely $$int nchi_{(frac 1{n-1}, frac 1 n]} , dmu$$
– Shashi
Nov 17 at 8:55
@Shashi If i swap them I have $sum_{n=1}^{infty}int_{-infty}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}.$ But then this a sum of an integral, but I wasn't sure whether I can evaluate this as $n mu((frac{1}{n+1}, frac{1}{n}])$ because the $n$ that is multiplying the measure could be infinite and I thought it had to be a finite linear combination?
– mathlearner
Nov 17 at 8:57
Yes. You can do the integration first, right?
– Shashi
Nov 17 at 8:58
Notice that it’s not hard to compute $int_{-infty}^infty nchi_{(frac1{n+1}, frac1n]} dmu$
– Alonso Delfín
Nov 17 at 9:00
@AlonsoDelfín so is it okay for there to be an infinite number of linear combinations of indicator functions then. My confusion is that the index in the sum goes to $infty$ and therefore we have an infinite linear combination (potentally). I know that $n mu((frac{1}{n+1},frac{1}{n}]) = frac{1}{n+1}$ if I can do that.
– mathlearner
Nov 17 at 9:04
|
show 1 more comment
You say you have swapped the integral and sum. Doesn't that mean you get the integral in from of a single simple function? Namely $$int nchi_{(frac 1{n-1}, frac 1 n]} , dmu$$
– Shashi
Nov 17 at 8:55
@Shashi If i swap them I have $sum_{n=1}^{infty}int_{-infty}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}.$ But then this a sum of an integral, but I wasn't sure whether I can evaluate this as $n mu((frac{1}{n+1}, frac{1}{n}])$ because the $n$ that is multiplying the measure could be infinite and I thought it had to be a finite linear combination?
– mathlearner
Nov 17 at 8:57
Yes. You can do the integration first, right?
– Shashi
Nov 17 at 8:58
Notice that it’s not hard to compute $int_{-infty}^infty nchi_{(frac1{n+1}, frac1n]} dmu$
– Alonso Delfín
Nov 17 at 9:00
@AlonsoDelfín so is it okay for there to be an infinite number of linear combinations of indicator functions then. My confusion is that the index in the sum goes to $infty$ and therefore we have an infinite linear combination (potentally). I know that $n mu((frac{1}{n+1},frac{1}{n}]) = frac{1}{n+1}$ if I can do that.
– mathlearner
Nov 17 at 9:04
You say you have swapped the integral and sum. Doesn't that mean you get the integral in from of a single simple function? Namely $$int nchi_{(frac 1{n-1}, frac 1 n]} , dmu$$
– Shashi
Nov 17 at 8:55
You say you have swapped the integral and sum. Doesn't that mean you get the integral in from of a single simple function? Namely $$int nchi_{(frac 1{n-1}, frac 1 n]} , dmu$$
– Shashi
Nov 17 at 8:55
@Shashi If i swap them I have $sum_{n=1}^{infty}int_{-infty}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}.$ But then this a sum of an integral, but I wasn't sure whether I can evaluate this as $n mu((frac{1}{n+1}, frac{1}{n}])$ because the $n$ that is multiplying the measure could be infinite and I thought it had to be a finite linear combination?
– mathlearner
Nov 17 at 8:57
@Shashi If i swap them I have $sum_{n=1}^{infty}int_{-infty}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}.$ But then this a sum of an integral, but I wasn't sure whether I can evaluate this as $n mu((frac{1}{n+1}, frac{1}{n}])$ because the $n$ that is multiplying the measure could be infinite and I thought it had to be a finite linear combination?
– mathlearner
Nov 17 at 8:57
Yes. You can do the integration first, right?
– Shashi
Nov 17 at 8:58
Yes. You can do the integration first, right?
– Shashi
Nov 17 at 8:58
Notice that it’s not hard to compute $int_{-infty}^infty nchi_{(frac1{n+1}, frac1n]} dmu$
– Alonso Delfín
Nov 17 at 9:00
Notice that it’s not hard to compute $int_{-infty}^infty nchi_{(frac1{n+1}, frac1n]} dmu$
– Alonso Delfín
Nov 17 at 9:00
@AlonsoDelfín so is it okay for there to be an infinite number of linear combinations of indicator functions then. My confusion is that the index in the sum goes to $infty$ and therefore we have an infinite linear combination (potentally). I know that $n mu((frac{1}{n+1},frac{1}{n}]) = frac{1}{n+1}$ if I can do that.
– mathlearner
Nov 17 at 9:04
@AlonsoDelfín so is it okay for there to be an infinite number of linear combinations of indicator functions then. My confusion is that the index in the sum goes to $infty$ and therefore we have an infinite linear combination (potentally). I know that $n mu((frac{1}{n+1},frac{1}{n}]) = frac{1}{n+1}$ if I can do that.
– mathlearner
Nov 17 at 9:04
|
show 1 more comment
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You say you have swapped the integral and sum. Doesn't that mean you get the integral in from of a single simple function? Namely $$int nchi_{(frac 1{n-1}, frac 1 n]} , dmu$$
– Shashi
Nov 17 at 8:55
@Shashi If i swap them I have $sum_{n=1}^{infty}int_{-infty}^{infty}nchi_{(frac{1}{n+1},frac{1}{n}]}.$ But then this a sum of an integral, but I wasn't sure whether I can evaluate this as $n mu((frac{1}{n+1}, frac{1}{n}])$ because the $n$ that is multiplying the measure could be infinite and I thought it had to be a finite linear combination?
– mathlearner
Nov 17 at 8:57
Yes. You can do the integration first, right?
– Shashi
Nov 17 at 8:58
Notice that it’s not hard to compute $int_{-infty}^infty nchi_{(frac1{n+1}, frac1n]} dmu$
– Alonso Delfín
Nov 17 at 9:00
@AlonsoDelfín so is it okay for there to be an infinite number of linear combinations of indicator functions then. My confusion is that the index in the sum goes to $infty$ and therefore we have an infinite linear combination (potentally). I know that $n mu((frac{1}{n+1},frac{1}{n}]) = frac{1}{n+1}$ if I can do that.
– mathlearner
Nov 17 at 9:04