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?










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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















up vote
0
down vote

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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?










share|cite|improve this question






















  • 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













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?










share|cite|improve this question













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






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share|cite|improve this question











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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


















  • 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















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