Prove that $lim_{nrightarrow infty} int_{[-n,n]} f,dlambda= int f,dlambda.$












1














I am working on the following exercise:




Let $lambda$ denote Lebesgue measure on $mathbf{R}$. Suppose
$f:mathbf{R}rightarrow mathbf{R}$ is a Borel measurable function
such that $int|f|<infty$. Prove that $$lim_{nrightarrow infty}
int_{[-n,n]} f,dlambda= int f,dlambda.$$




Now, I have some ideas as to things that could help me here, but I can't really put any of it together. Here is what I have so far:





  1. I know that, given a measure space $(X,mathcal{S},mu)$, a set $Ein mathcal{S}$, and an $mathcal{S}$-measurable function $f$, $$int_E f,dmu=int fchi_{E},dmu$$ if the RHS is defined. In this case, I believe it is, since $int |f|<infty$.


  2. Then this question got me thinking: what if I write $[-n,n]$ as the limit of an increasing sequence of sets? That is, can I write $E=[-n,n]=bigcup_{n=1}^infty E_n$ where $E_n=[-n,n]$ for $ninmathbf{N}$?


  3. If (2) is true, then I can define $f_n=fchi_{E_n}$ and $$lim_{nrightarrowinfty}f_n=fchi_E.$$ Then I think I would have my answer by the Dominated Convergence Theorem.




Am I on the right track here?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question


















  • 1




    Yes, and you're pretty much done.
    – T. Bongers
    Nov 19 at 0:54










  • @T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $bigcup_{n=1}^infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $bigcup_{n=1}^infty E_n=(-infty,infty)$?
    – Thy Art is Math
    Nov 19 at 0:59








  • 1




    It is $(-infty, infty)$, but don't you want it to be? You're integrating over $mathbb{R}$.
    – T. Bongers
    Nov 19 at 1:28










  • Ah, that makes sense! Thanks!
    – Thy Art is Math
    Nov 19 at 1:40
















1














I am working on the following exercise:




Let $lambda$ denote Lebesgue measure on $mathbf{R}$. Suppose
$f:mathbf{R}rightarrow mathbf{R}$ is a Borel measurable function
such that $int|f|<infty$. Prove that $$lim_{nrightarrow infty}
int_{[-n,n]} f,dlambda= int f,dlambda.$$




Now, I have some ideas as to things that could help me here, but I can't really put any of it together. Here is what I have so far:





  1. I know that, given a measure space $(X,mathcal{S},mu)$, a set $Ein mathcal{S}$, and an $mathcal{S}$-measurable function $f$, $$int_E f,dmu=int fchi_{E},dmu$$ if the RHS is defined. In this case, I believe it is, since $int |f|<infty$.


  2. Then this question got me thinking: what if I write $[-n,n]$ as the limit of an increasing sequence of sets? That is, can I write $E=[-n,n]=bigcup_{n=1}^infty E_n$ where $E_n=[-n,n]$ for $ninmathbf{N}$?


  3. If (2) is true, then I can define $f_n=fchi_{E_n}$ and $$lim_{nrightarrowinfty}f_n=fchi_E.$$ Then I think I would have my answer by the Dominated Convergence Theorem.




Am I on the right track here?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question


















  • 1




    Yes, and you're pretty much done.
    – T. Bongers
    Nov 19 at 0:54










  • @T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $bigcup_{n=1}^infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $bigcup_{n=1}^infty E_n=(-infty,infty)$?
    – Thy Art is Math
    Nov 19 at 0:59








  • 1




    It is $(-infty, infty)$, but don't you want it to be? You're integrating over $mathbb{R}$.
    – T. Bongers
    Nov 19 at 1:28










  • Ah, that makes sense! Thanks!
    – Thy Art is Math
    Nov 19 at 1:40














1












1








1







I am working on the following exercise:




Let $lambda$ denote Lebesgue measure on $mathbf{R}$. Suppose
$f:mathbf{R}rightarrow mathbf{R}$ is a Borel measurable function
such that $int|f|<infty$. Prove that $$lim_{nrightarrow infty}
int_{[-n,n]} f,dlambda= int f,dlambda.$$




Now, I have some ideas as to things that could help me here, but I can't really put any of it together. Here is what I have so far:





  1. I know that, given a measure space $(X,mathcal{S},mu)$, a set $Ein mathcal{S}$, and an $mathcal{S}$-measurable function $f$, $$int_E f,dmu=int fchi_{E},dmu$$ if the RHS is defined. In this case, I believe it is, since $int |f|<infty$.


  2. Then this question got me thinking: what if I write $[-n,n]$ as the limit of an increasing sequence of sets? That is, can I write $E=[-n,n]=bigcup_{n=1}^infty E_n$ where $E_n=[-n,n]$ for $ninmathbf{N}$?


  3. If (2) is true, then I can define $f_n=fchi_{E_n}$ and $$lim_{nrightarrowinfty}f_n=fchi_E.$$ Then I think I would have my answer by the Dominated Convergence Theorem.




Am I on the right track here?






real-analysis measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question













I am working on the following exercise:




Let $lambda$ denote Lebesgue measure on $mathbf{R}$. Suppose
$f:mathbf{R}rightarrow mathbf{R}$ is a Borel measurable function
such that $int|f|<infty$. Prove that $$lim_{nrightarrow infty}
int_{[-n,n]} f,dlambda= int f,dlambda.$$




Now, I have some ideas as to things that could help me here, but I can't really put any of it together. Here is what I have so far:





  1. I know that, given a measure space $(X,mathcal{S},mu)$, a set $Ein mathcal{S}$, and an $mathcal{S}$-measurable function $f$, $$int_E f,dmu=int fchi_{E},dmu$$ if the RHS is defined. In this case, I believe it is, since $int |f|<infty$.


  2. Then this question got me thinking: what if I write $[-n,n]$ as the limit of an increasing sequence of sets? That is, can I write $E=[-n,n]=bigcup_{n=1}^infty E_n$ where $E_n=[-n,n]$ for $ninmathbf{N}$?


  3. If (2) is true, then I can define $f_n=fchi_{E_n}$ and $$lim_{nrightarrowinfty}f_n=fchi_E.$$ Then I think I would have my answer by the Dominated Convergence Theorem.




Am I on the right track here?






real-analysis measure-theory lebesgue-integral lebesgue-measure




real-analysis measure-theory lebesgue-integral lebesgue-measure






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 19 at 0:52









Thy Art is Math

484211




484211








  • 1




    Yes, and you're pretty much done.
    – T. Bongers
    Nov 19 at 0:54










  • @T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $bigcup_{n=1}^infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $bigcup_{n=1}^infty E_n=(-infty,infty)$?
    – Thy Art is Math
    Nov 19 at 0:59








  • 1




    It is $(-infty, infty)$, but don't you want it to be? You're integrating over $mathbb{R}$.
    – T. Bongers
    Nov 19 at 1:28










  • Ah, that makes sense! Thanks!
    – Thy Art is Math
    Nov 19 at 1:40














  • 1




    Yes, and you're pretty much done.
    – T. Bongers
    Nov 19 at 0:54










  • @T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $bigcup_{n=1}^infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $bigcup_{n=1}^infty E_n=(-infty,infty)$?
    – Thy Art is Math
    Nov 19 at 0:59








  • 1




    It is $(-infty, infty)$, but don't you want it to be? You're integrating over $mathbb{R}$.
    – T. Bongers
    Nov 19 at 1:28










  • Ah, that makes sense! Thanks!
    – Thy Art is Math
    Nov 19 at 1:40








1




1




Yes, and you're pretty much done.
– T. Bongers
Nov 19 at 0:54




Yes, and you're pretty much done.
– T. Bongers
Nov 19 at 0:54












@T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $bigcup_{n=1}^infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $bigcup_{n=1}^infty E_n=(-infty,infty)$?
– Thy Art is Math
Nov 19 at 0:59






@T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $bigcup_{n=1}^infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $bigcup_{n=1}^infty E_n=(-infty,infty)$?
– Thy Art is Math
Nov 19 at 0:59






1




1




It is $(-infty, infty)$, but don't you want it to be? You're integrating over $mathbb{R}$.
– T. Bongers
Nov 19 at 1:28




It is $(-infty, infty)$, but don't you want it to be? You're integrating over $mathbb{R}$.
– T. Bongers
Nov 19 at 1:28












Ah, that makes sense! Thanks!
– Thy Art is Math
Nov 19 at 1:40




Ah, that makes sense! Thanks!
– Thy Art is Math
Nov 19 at 1:40










1 Answer
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Your proof is fine as-is.




For an alternative proof, you could apply the monotone convergence theorem to the sequences $f^+ chi_{E_n}$ and $f^{-} chi_{E_n}$ separately.






share|cite|improve this answer























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    1 Answer
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    1 Answer
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    Your proof is fine as-is.




    For an alternative proof, you could apply the monotone convergence theorem to the sequences $f^+ chi_{E_n}$ and $f^{-} chi_{E_n}$ separately.






    share|cite|improve this answer




























      1














      Your proof is fine as-is.




      For an alternative proof, you could apply the monotone convergence theorem to the sequences $f^+ chi_{E_n}$ and $f^{-} chi_{E_n}$ separately.






      share|cite|improve this answer


























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        1








        1






        Your proof is fine as-is.




        For an alternative proof, you could apply the monotone convergence theorem to the sequences $f^+ chi_{E_n}$ and $f^{-} chi_{E_n}$ separately.






        share|cite|improve this answer














        Your proof is fine as-is.




        For an alternative proof, you could apply the monotone convergence theorem to the sequences $f^+ chi_{E_n}$ and $f^{-} chi_{E_n}$ separately.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Nov 19 at 0:54


























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