A measurable set from a sequence of sets












0












$begingroup$


Let (X,M, $mu$) be a measure space and $E_n$ be a sequence of measurable sets. Let k be a fixed integer and G be the set of all those points that belong to $E_n$ for at least k values of n.



a) Prove that G is measurable.



I am trying to find a way to rewrite G as a countable union of countable intersections. I have come up with the following:



For k = 2, $G_2 = (E_1 cap E_2) cup (E_1 cap E_3) cup...cup (E_2 cap E_3) cup (E_2 cap E_4) cup...cup (E_k cap E_{k+1}) cup (E_k cup E_{k+1})... $



$G_2 = bigcup_{n=2}^infty (E_1 cap E_n) cup bigcup_{n=3}^infty(E_2 cap E_n) cup...cup bigcup_{n=k+1}^infty(E_kcap E_n) cup....$



$G_2 = bigcap_{k=1}^infty bigcup_{n=k+1}^infty (E_k cap E_n)$



Since $G_2$ is a countable union of intersections it is in M and hence measurable.



I'm having trouble generalizing this approach for an arbitrary K...



b) Show $kmu(G)le sum_{n=1}^infty mu(E_n)$



One hint was to look at: $sum_{n=1}^infty int_G chi_{E_n}dmu$ however without the correct breakdown in part a) I'm not sure if I can proceed.



Edit:
Here's what I have:



$kmu(G) = int_G k dmu le int_G sum_{n=1}^infty chi_{E_n}dmu$ (not confident with the inequality here. Is this because if x is in g, it is in k or more of the En's?)



Continuing by MCT



$int_G sum_{n=1}^infty chi_{E_n}dmu = sum_{n=1}^infty int_G chi_{E_n} dmu = sum_{n=1}^infty int chi_G chi_{E_n}dmu = sum_{n=1}^infty int chi_{G cap E_n} dmu =sum_{n=1}^inftymu(Gcap E_n) le sum_{n=1}^infty mu(E_n)$



*I have not learned about Markov inequality. This is a qualifying exam question and Markov inequality is not on the syllabus.






analysis measure-theory







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let (X,M, $mu$) be a measure space and $E_n$ be a sequence of measurable sets. Let k be a fixed integer and G be the set of all those points that belong to $E_n$ for at least k values of n.



    a) Prove that G is measurable.



    I am trying to find a way to rewrite G as a countable union of countable intersections. I have come up with the following:



    For k = 2, $G_2 = (E_1 cap E_2) cup (E_1 cap E_3) cup...cup (E_2 cap E_3) cup (E_2 cap E_4) cup...cup (E_k cap E_{k+1}) cup (E_k cup E_{k+1})... $



    $G_2 = bigcup_{n=2}^infty (E_1 cap E_n) cup bigcup_{n=3}^infty(E_2 cap E_n) cup...cup bigcup_{n=k+1}^infty(E_kcap E_n) cup....$



    $G_2 = bigcap_{k=1}^infty bigcup_{n=k+1}^infty (E_k cap E_n)$



    Since $G_2$ is a countable union of intersections it is in M and hence measurable.



    I'm having trouble generalizing this approach for an arbitrary K...



    b) Show $kmu(G)le sum_{n=1}^infty mu(E_n)$



    One hint was to look at: $sum_{n=1}^infty int_G chi_{E_n}dmu$ however without the correct breakdown in part a) I'm not sure if I can proceed.



    Edit:
    Here's what I have:



    $kmu(G) = int_G k dmu le int_G sum_{n=1}^infty chi_{E_n}dmu$ (not confident with the inequality here. Is this because if x is in g, it is in k or more of the En's?)



    Continuing by MCT



    $int_G sum_{n=1}^infty chi_{E_n}dmu = sum_{n=1}^infty int_G chi_{E_n} dmu = sum_{n=1}^infty int chi_G chi_{E_n}dmu = sum_{n=1}^infty int chi_{G cap E_n} dmu =sum_{n=1}^inftymu(Gcap E_n) le sum_{n=1}^infty mu(E_n)$



    *I have not learned about Markov inequality. This is a qualifying exam question and Markov inequality is not on the syllabus.






    analysis measure-theory







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let (X,M, $mu$) be a measure space and $E_n$ be a sequence of measurable sets. Let k be a fixed integer and G be the set of all those points that belong to $E_n$ for at least k values of n.



      a) Prove that G is measurable.



      I am trying to find a way to rewrite G as a countable union of countable intersections. I have come up with the following:



      For k = 2, $G_2 = (E_1 cap E_2) cup (E_1 cap E_3) cup...cup (E_2 cap E_3) cup (E_2 cap E_4) cup...cup (E_k cap E_{k+1}) cup (E_k cup E_{k+1})... $



      $G_2 = bigcup_{n=2}^infty (E_1 cap E_n) cup bigcup_{n=3}^infty(E_2 cap E_n) cup...cup bigcup_{n=k+1}^infty(E_kcap E_n) cup....$



      $G_2 = bigcap_{k=1}^infty bigcup_{n=k+1}^infty (E_k cap E_n)$



      Since $G_2$ is a countable union of intersections it is in M and hence measurable.



      I'm having trouble generalizing this approach for an arbitrary K...



      b) Show $kmu(G)le sum_{n=1}^infty mu(E_n)$



      One hint was to look at: $sum_{n=1}^infty int_G chi_{E_n}dmu$ however without the correct breakdown in part a) I'm not sure if I can proceed.



      Edit:
      Here's what I have:



      $kmu(G) = int_G k dmu le int_G sum_{n=1}^infty chi_{E_n}dmu$ (not confident with the inequality here. Is this because if x is in g, it is in k or more of the En's?)



      Continuing by MCT



      $int_G sum_{n=1}^infty chi_{E_n}dmu = sum_{n=1}^infty int_G chi_{E_n} dmu = sum_{n=1}^infty int chi_G chi_{E_n}dmu = sum_{n=1}^infty int chi_{G cap E_n} dmu =sum_{n=1}^inftymu(Gcap E_n) le sum_{n=1}^infty mu(E_n)$



      *I have not learned about Markov inequality. This is a qualifying exam question and Markov inequality is not on the syllabus.






      analysis measure-theory







      share|cite|improve this question











      $endgroup$




      Let (X,M, $mu$) be a measure space and $E_n$ be a sequence of measurable sets. Let k be a fixed integer and G be the set of all those points that belong to $E_n$ for at least k values of n.



      a) Prove that G is measurable.



      I am trying to find a way to rewrite G as a countable union of countable intersections. I have come up with the following:



      For k = 2, $G_2 = (E_1 cap E_2) cup (E_1 cap E_3) cup...cup (E_2 cap E_3) cup (E_2 cap E_4) cup...cup (E_k cap E_{k+1}) cup (E_k cup E_{k+1})... $



      $G_2 = bigcup_{n=2}^infty (E_1 cap E_n) cup bigcup_{n=3}^infty(E_2 cap E_n) cup...cup bigcup_{n=k+1}^infty(E_kcap E_n) cup....$



      $G_2 = bigcap_{k=1}^infty bigcup_{n=k+1}^infty (E_k cap E_n)$



      Since $G_2$ is a countable union of intersections it is in M and hence measurable.



      I'm having trouble generalizing this approach for an arbitrary K...



      b) Show $kmu(G)le sum_{n=1}^infty mu(E_n)$



      One hint was to look at: $sum_{n=1}^infty int_G chi_{E_n}dmu$ however without the correct breakdown in part a) I'm not sure if I can proceed.



      Edit:
      Here's what I have:



      $kmu(G) = int_G k dmu le int_G sum_{n=1}^infty chi_{E_n}dmu$ (not confident with the inequality here. Is this because if x is in g, it is in k or more of the En's?)



      Continuing by MCT



      $int_G sum_{n=1}^infty chi_{E_n}dmu = sum_{n=1}^infty int_G chi_{E_n} dmu = sum_{n=1}^infty int chi_G chi_{E_n}dmu = sum_{n=1}^infty int chi_{G cap E_n} dmu =sum_{n=1}^inftymu(Gcap E_n) le sum_{n=1}^infty mu(E_n)$



      *I have not learned about Markov inequality. This is a qualifying exam question and Markov inequality is not on the syllabus.






      analysis measure-theory




      analysis measure-theory






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 6 at 14:36







      Math Lady

















      asked Jan 6 at 13:45









      Math LadyMath Lady

      1196




      1196






















          2 Answers
          2






          active

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          1












          $begingroup$

          Let $mathscr F$ be the set of all intersections of $k$ $E_n$'s. Then $mathscr F$ is countable and $G$ is the union of all elements of $mathscr F$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
            $endgroup$
            – Math Lady
            Jan 6 at 14:13












          • $begingroup$
            Yes, that would be correct.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 14:27



















          2












          $begingroup$

          Hint: Express $G$ as an upper level set of the measurable function
          $$
          F:xmapstosum_{n=1}^infty 1_{E_n}(x).
          $$           (What does $F(x)$ mean?) For (b), apply Markov's inequality.



          EDIT: Since $F(x)$ is the number of $n$ for which $xin E_n$, we have
          $$
          G = {x;|;F(x)ge k}
          $$           is measurable. Now, observe that $k 1_G(x) le F(x)$ since $kle F(x)$ on $G$. Integrate both sides with respect to $mu$ to get
          $$
          kmu(G) = int k1_G(x)dmu(x) le int F(x)dmu(x) = sum_{n=1}^infty mu(E_n)
          $$           by monotone convergence thereom.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
            $endgroup$
            – Math Lady
            Jan 6 at 14:37










          • $begingroup$
            Yes, it's correct. Indeed that's what exactly Markov's inequality says!
            $endgroup$
            – Song
            Jan 6 at 14:39













          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Let $mathscr F$ be the set of all intersections of $k$ $E_n$'s. Then $mathscr F$ is countable and $G$ is the union of all elements of $mathscr F$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
            $endgroup$
            – Math Lady
            Jan 6 at 14:13












          • $begingroup$
            Yes, that would be correct.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 14:27
















          1












          $begingroup$

          Let $mathscr F$ be the set of all intersections of $k$ $E_n$'s. Then $mathscr F$ is countable and $G$ is the union of all elements of $mathscr F$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
            $endgroup$
            – Math Lady
            Jan 6 at 14:13












          • $begingroup$
            Yes, that would be correct.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 14:27














          1












          1








          1





          $begingroup$

          Let $mathscr F$ be the set of all intersections of $k$ $E_n$'s. Then $mathscr F$ is countable and $G$ is the union of all elements of $mathscr F$.






          share|cite|improve this answer









          $endgroup$



          Let $mathscr F$ be the set of all intersections of $k$ $E_n$'s. Then $mathscr F$ is countable and $G$ is the union of all elements of $mathscr F$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 13:54









          José Carlos SantosJosé Carlos Santos

          156k22126227




          156k22126227












          • $begingroup$
            How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
            $endgroup$
            – Math Lady
            Jan 6 at 14:13












          • $begingroup$
            Yes, that would be correct.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 14:27


















          • $begingroup$
            How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
            $endgroup$
            – Math Lady
            Jan 6 at 14:13












          • $begingroup$
            Yes, that would be correct.
            $endgroup$
            – José Carlos Santos
            Jan 6 at 14:27
















          $begingroup$
          How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
          $endgroup$
          – Math Lady
          Jan 6 at 14:13






          $begingroup$
          How would I write this out? $F = {bigcap_I E_n : |I| = k}$ is countable since there are countable many $E_n$ so there are only a countable number of ways to pick k of them to intersect. Then $G = bigcup_{n=1}^infty F_n$ is in M because it is a union of finite intersections?
          $endgroup$
          – Math Lady
          Jan 6 at 14:13














          $begingroup$
          Yes, that would be correct.
          $endgroup$
          – José Carlos Santos
          Jan 6 at 14:27




          $begingroup$
          Yes, that would be correct.
          $endgroup$
          – José Carlos Santos
          Jan 6 at 14:27











          2












          $begingroup$

          Hint: Express $G$ as an upper level set of the measurable function
          $$
          F:xmapstosum_{n=1}^infty 1_{E_n}(x).
          $$           (What does $F(x)$ mean?) For (b), apply Markov's inequality.



          EDIT: Since $F(x)$ is the number of $n$ for which $xin E_n$, we have
          $$
          G = {x;|;F(x)ge k}
          $$           is measurable. Now, observe that $k 1_G(x) le F(x)$ since $kle F(x)$ on $G$. Integrate both sides with respect to $mu$ to get
          $$
          kmu(G) = int k1_G(x)dmu(x) le int F(x)dmu(x) = sum_{n=1}^infty mu(E_n)
          $$           by monotone convergence thereom.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
            $endgroup$
            – Math Lady
            Jan 6 at 14:37










          • $begingroup$
            Yes, it's correct. Indeed that's what exactly Markov's inequality says!
            $endgroup$
            – Song
            Jan 6 at 14:39


















          2












          $begingroup$

          Hint: Express $G$ as an upper level set of the measurable function
          $$
          F:xmapstosum_{n=1}^infty 1_{E_n}(x).
          $$           (What does $F(x)$ mean?) For (b), apply Markov's inequality.



          EDIT: Since $F(x)$ is the number of $n$ for which $xin E_n$, we have
          $$
          G = {x;|;F(x)ge k}
          $$           is measurable. Now, observe that $k 1_G(x) le F(x)$ since $kle F(x)$ on $G$. Integrate both sides with respect to $mu$ to get
          $$
          kmu(G) = int k1_G(x)dmu(x) le int F(x)dmu(x) = sum_{n=1}^infty mu(E_n)
          $$           by monotone convergence thereom.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
            $endgroup$
            – Math Lady
            Jan 6 at 14:37










          • $begingroup$
            Yes, it's correct. Indeed that's what exactly Markov's inequality says!
            $endgroup$
            – Song
            Jan 6 at 14:39
















          2












          2








          2





          $begingroup$

          Hint: Express $G$ as an upper level set of the measurable function
          $$
          F:xmapstosum_{n=1}^infty 1_{E_n}(x).
          $$           (What does $F(x)$ mean?) For (b), apply Markov's inequality.



          EDIT: Since $F(x)$ is the number of $n$ for which $xin E_n$, we have
          $$
          G = {x;|;F(x)ge k}
          $$           is measurable. Now, observe that $k 1_G(x) le F(x)$ since $kle F(x)$ on $G$. Integrate both sides with respect to $mu$ to get
          $$
          kmu(G) = int k1_G(x)dmu(x) le int F(x)dmu(x) = sum_{n=1}^infty mu(E_n)
          $$           by monotone convergence thereom.






          share|cite|improve this answer











          $endgroup$



          Hint: Express $G$ as an upper level set of the measurable function
          $$
          F:xmapstosum_{n=1}^infty 1_{E_n}(x).
          $$           (What does $F(x)$ mean?) For (b), apply Markov's inequality.



          EDIT: Since $F(x)$ is the number of $n$ for which $xin E_n$, we have
          $$
          G = {x;|;F(x)ge k}
          $$           is measurable. Now, observe that $k 1_G(x) le F(x)$ since $kle F(x)$ on $G$. Integrate both sides with respect to $mu$ to get
          $$
          kmu(G) = int k1_G(x)dmu(x) le int F(x)dmu(x) = sum_{n=1}^infty mu(E_n)
          $$           by monotone convergence thereom.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 14:44

























          answered Jan 6 at 13:53









          SongSong

          10.1k627




          10.1k627












          • $begingroup$
            Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
            $endgroup$
            – Math Lady
            Jan 6 at 14:37










          • $begingroup$
            Yes, it's correct. Indeed that's what exactly Markov's inequality says!
            $endgroup$
            – Song
            Jan 6 at 14:39




















          • $begingroup$
            Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
            $endgroup$
            – Math Lady
            Jan 6 at 14:37










          • $begingroup$
            Yes, it's correct. Indeed that's what exactly Markov's inequality says!
            $endgroup$
            – Song
            Jan 6 at 14:39


















          $begingroup$
          Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
          $endgroup$
          – Math Lady
          Jan 6 at 14:37




          $begingroup$
          Could you please look at the edit above and tell me if it is valid? My class did not cover Markov's Inequality.
          $endgroup$
          – Math Lady
          Jan 6 at 14:37












          $begingroup$
          Yes, it's correct. Indeed that's what exactly Markov's inequality says!
          $endgroup$
          – Song
          Jan 6 at 14:39






          $begingroup$
          Yes, it's correct. Indeed that's what exactly Markov's inequality says!
          $endgroup$
          – Song
          Jan 6 at 14:39




















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          Npm cannot find a required file even through it is in the searched directory

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          up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
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