Show that $lim_{ntoinfty}int_{mathbb{R}}f_ndmu$ exists, where $mu=sum_{k=1}^M delta_{y_k}$












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I want to show that $lim_{ntoinfty}int_{mathbb{R}}f_ndmu$ exists and calculate it, with $mu=sum_{k=1}^M delta_{y_k}$ and $delta_{y_k}$ being the dirac measure.
Additionally, $f_n$ is defined from $mathbb{R}tomathbb{R}$ where $$x to arctan(n(x-a))-arctan(n(x-b))$$ with $a,binmathbb{R}, alt b, ninmathbb{N}$ and $y_kinmathbb{R}$ is given.
What I tried:
$$lim_{ntoinfty}int_{mathbb{R}}f_ndmu=lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)$$ Is this correct? Then, $$lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)=lim_{ntoinfty}left(sum_{k=1}^Mf_n(y_k)right)=sum_{k=1}^Mleft(lim_{ntoinfty}f_n(y_k)right)$$ If I can now show that $f_n$ converges pointwise to a limit function $f$, I should be done. However, I struggle with that and am completely unsure if this is anyhow correct. Thanks for help!






real-analysis measure-theory







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    $begingroup$


    I want to show that $lim_{ntoinfty}int_{mathbb{R}}f_ndmu$ exists and calculate it, with $mu=sum_{k=1}^M delta_{y_k}$ and $delta_{y_k}$ being the dirac measure.
    Additionally, $f_n$ is defined from $mathbb{R}tomathbb{R}$ where $$x to arctan(n(x-a))-arctan(n(x-b))$$ with $a,binmathbb{R}, alt b, ninmathbb{N}$ and $y_kinmathbb{R}$ is given.
    What I tried:
    $$lim_{ntoinfty}int_{mathbb{R}}f_ndmu=lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)$$ Is this correct? Then, $$lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)=lim_{ntoinfty}left(sum_{k=1}^Mf_n(y_k)right)=sum_{k=1}^Mleft(lim_{ntoinfty}f_n(y_k)right)$$ If I can now show that $f_n$ converges pointwise to a limit function $f$, I should be done. However, I struggle with that and am completely unsure if this is anyhow correct. Thanks for help!






    real-analysis measure-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I want to show that $lim_{ntoinfty}int_{mathbb{R}}f_ndmu$ exists and calculate it, with $mu=sum_{k=1}^M delta_{y_k}$ and $delta_{y_k}$ being the dirac measure.
      Additionally, $f_n$ is defined from $mathbb{R}tomathbb{R}$ where $$x to arctan(n(x-a))-arctan(n(x-b))$$ with $a,binmathbb{R}, alt b, ninmathbb{N}$ and $y_kinmathbb{R}$ is given.
      What I tried:
      $$lim_{ntoinfty}int_{mathbb{R}}f_ndmu=lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)$$ Is this correct? Then, $$lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)=lim_{ntoinfty}left(sum_{k=1}^Mf_n(y_k)right)=sum_{k=1}^Mleft(lim_{ntoinfty}f_n(y_k)right)$$ If I can now show that $f_n$ converges pointwise to a limit function $f$, I should be done. However, I struggle with that and am completely unsure if this is anyhow correct. Thanks for help!






      real-analysis measure-theory







      share|cite|improve this question











      $endgroup$




      I want to show that $lim_{ntoinfty}int_{mathbb{R}}f_ndmu$ exists and calculate it, with $mu=sum_{k=1}^M delta_{y_k}$ and $delta_{y_k}$ being the dirac measure.
      Additionally, $f_n$ is defined from $mathbb{R}tomathbb{R}$ where $$x to arctan(n(x-a))-arctan(n(x-b))$$ with $a,binmathbb{R}, alt b, ninmathbb{N}$ and $y_kinmathbb{R}$ is given.
      What I tried:
      $$lim_{ntoinfty}int_{mathbb{R}}f_ndmu=lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)$$ Is this correct? Then, $$lim_{ntoinfty}left(sum_{k=1}^Mint_{mathbb{R}}f_ndelta_{y_k}right)=lim_{ntoinfty}left(sum_{k=1}^Mf_n(y_k)right)=sum_{k=1}^Mleft(lim_{ntoinfty}f_n(y_k)right)$$ If I can now show that $f_n$ converges pointwise to a limit function $f$, I should be done. However, I struggle with that and am completely unsure if this is anyhow correct. Thanks for help!






      real-analysis measure-theory




      real-analysis measure-theory






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      edited Jan 8 at 13:21









      Christian Blatter

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      asked Jan 8 at 12:53









      Michael MaierMichael Maier

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          1 Answer
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          $begingroup$

          The only thing you have to know on the $arctan$ function is that
          $$
          lim_{xto -infty}arctan x=-frac{pi}2mbox{ and }lim_{xto +infty}arctan x= frac{pi}2.
          $$
          Therefore, the value of $lim_{nto +infty}f_nleft(y_kright)$ depends whether $y_klt a$, $y_k=a$, $alt y_klt b$, $y_k=b$ or $y_kgt b$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
            $endgroup$
            – Michael Maier
            Jan 8 at 15:24












          • $begingroup$
            And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
            $endgroup$
            – Michael Maier
            Jan 8 at 15:30











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          1 Answer
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          active

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          $begingroup$

          The only thing you have to know on the $arctan$ function is that
          $$
          lim_{xto -infty}arctan x=-frac{pi}2mbox{ and }lim_{xto +infty}arctan x= frac{pi}2.
          $$
          Therefore, the value of $lim_{nto +infty}f_nleft(y_kright)$ depends whether $y_klt a$, $y_k=a$, $alt y_klt b$, $y_k=b$ or $y_kgt b$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
            $endgroup$
            – Michael Maier
            Jan 8 at 15:24












          • $begingroup$
            And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
            $endgroup$
            – Michael Maier
            Jan 8 at 15:30
















          1












          $begingroup$

          The only thing you have to know on the $arctan$ function is that
          $$
          lim_{xto -infty}arctan x=-frac{pi}2mbox{ and }lim_{xto +infty}arctan x= frac{pi}2.
          $$
          Therefore, the value of $lim_{nto +infty}f_nleft(y_kright)$ depends whether $y_klt a$, $y_k=a$, $alt y_klt b$, $y_k=b$ or $y_kgt b$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
            $endgroup$
            – Michael Maier
            Jan 8 at 15:24












          • $begingroup$
            And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
            $endgroup$
            – Michael Maier
            Jan 8 at 15:30














          1












          1








          1





          $begingroup$

          The only thing you have to know on the $arctan$ function is that
          $$
          lim_{xto -infty}arctan x=-frac{pi}2mbox{ and }lim_{xto +infty}arctan x= frac{pi}2.
          $$
          Therefore, the value of $lim_{nto +infty}f_nleft(y_kright)$ depends whether $y_klt a$, $y_k=a$, $alt y_klt b$, $y_k=b$ or $y_kgt b$.






          share|cite|improve this answer









          $endgroup$



          The only thing you have to know on the $arctan$ function is that
          $$
          lim_{xto -infty}arctan x=-frac{pi}2mbox{ and }lim_{xto +infty}arctan x= frac{pi}2.
          $$
          Therefore, the value of $lim_{nto +infty}f_nleft(y_kright)$ depends whether $y_klt a$, $y_k=a$, $alt y_klt b$, $y_k=b$ or $y_kgt b$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 15:13









          Davide GiraudoDavide Giraudo

          126k16150261




          126k16150261












          • $begingroup$
            Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
            $endgroup$
            – Michael Maier
            Jan 8 at 15:24












          • $begingroup$
            And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
            $endgroup$
            – Michael Maier
            Jan 8 at 15:30


















          • $begingroup$
            Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
            $endgroup$
            – Michael Maier
            Jan 8 at 15:24












          • $begingroup$
            And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
            $endgroup$
            – Michael Maier
            Jan 8 at 15:30
















          $begingroup$
          Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
          $endgroup$
          – Michael Maier
          Jan 8 at 15:24






          $begingroup$
          Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that?
          $endgroup$
          – Michael Maier
          Jan 8 at 15:24














          $begingroup$
          And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
          $endgroup$
          – Michael Maier
          Jan 8 at 15:30




          $begingroup$
          And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again!
          $endgroup$
          – Michael Maier
          Jan 8 at 15:30


















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