$f:X rightarrow [0,infty]$, $A = {x:f(x)>0}$. If $A$ uncountable, $sum_{xin X}f(x) = infty$












1












$begingroup$


This comes from Chapter 0, Proposition 0.20, Real Analysis, by Folland




$f:X rightarrow [0,infty]$, $A = {x:f(x)>0}$. If $A$ uncountable, $sum_{xin X}f(x) = infty$




I am confused about the proof:




Let $A = bigcup_1^infty A_n$ where $A_n = {x:f(x)>1/n}$. If A is uncountable, then some $A_n$ must be uncountable, and $sum_{xin F} f(x) > mbox{card}(F)/n$ for $F$ a finite subset of $A_n$; it follows that $sum_{xin X}f(x) = infty$.




My questions are:




  1. How to get "$sum_{xin F} f(x) > mbox{card}(F)/n$"?

  2. How to get "$sum_{xin X}f(x) = infty$" based on previous result?


Note that $$sum_{xin X}f(x) = sup left{sum_{xin F}f(x): Fsubset X, F text{ finite } right}$$



Could anyone please provide me with detailed steps? Thanks!






real-analysis functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Masacroso It should be $sum_{xin X}$ from Folland.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 10:59










  • $begingroup$
    @Masacroso $X$ should be the domain of the function. $X$ is an arbitrary set.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 11:00
















1












$begingroup$


This comes from Chapter 0, Proposition 0.20, Real Analysis, by Folland




$f:X rightarrow [0,infty]$, $A = {x:f(x)>0}$. If $A$ uncountable, $sum_{xin X}f(x) = infty$




I am confused about the proof:




Let $A = bigcup_1^infty A_n$ where $A_n = {x:f(x)>1/n}$. If A is uncountable, then some $A_n$ must be uncountable, and $sum_{xin F} f(x) > mbox{card}(F)/n$ for $F$ a finite subset of $A_n$; it follows that $sum_{xin X}f(x) = infty$.




My questions are:




  1. How to get "$sum_{xin F} f(x) > mbox{card}(F)/n$"?

  2. How to get "$sum_{xin X}f(x) = infty$" based on previous result?


Note that $$sum_{xin X}f(x) = sup left{sum_{xin F}f(x): Fsubset X, F text{ finite } right}$$



Could anyone please provide me with detailed steps? Thanks!






real-analysis functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Masacroso It should be $sum_{xin X}$ from Folland.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 10:59










  • $begingroup$
    @Masacroso $X$ should be the domain of the function. $X$ is an arbitrary set.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 11:00














1












1








1





$begingroup$


This comes from Chapter 0, Proposition 0.20, Real Analysis, by Folland




$f:X rightarrow [0,infty]$, $A = {x:f(x)>0}$. If $A$ uncountable, $sum_{xin X}f(x) = infty$




I am confused about the proof:




Let $A = bigcup_1^infty A_n$ where $A_n = {x:f(x)>1/n}$. If A is uncountable, then some $A_n$ must be uncountable, and $sum_{xin F} f(x) > mbox{card}(F)/n$ for $F$ a finite subset of $A_n$; it follows that $sum_{xin X}f(x) = infty$.




My questions are:




  1. How to get "$sum_{xin F} f(x) > mbox{card}(F)/n$"?

  2. How to get "$sum_{xin X}f(x) = infty$" based on previous result?


Note that $$sum_{xin X}f(x) = sup left{sum_{xin F}f(x): Fsubset X, F text{ finite } right}$$



Could anyone please provide me with detailed steps? Thanks!






real-analysis functions







share|cite|improve this question











$endgroup$




This comes from Chapter 0, Proposition 0.20, Real Analysis, by Folland




$f:X rightarrow [0,infty]$, $A = {x:f(x)>0}$. If $A$ uncountable, $sum_{xin X}f(x) = infty$




I am confused about the proof:




Let $A = bigcup_1^infty A_n$ where $A_n = {x:f(x)>1/n}$. If A is uncountable, then some $A_n$ must be uncountable, and $sum_{xin F} f(x) > mbox{card}(F)/n$ for $F$ a finite subset of $A_n$; it follows that $sum_{xin X}f(x) = infty$.




My questions are:




  1. How to get "$sum_{xin F} f(x) > mbox{card}(F)/n$"?

  2. How to get "$sum_{xin X}f(x) = infty$" based on previous result?


Note that $$sum_{xin X}f(x) = sup left{sum_{xin F}f(x): Fsubset X, F text{ finite } right}$$



Could anyone please provide me with detailed steps? Thanks!






real-analysis functions




real-analysis functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 10:56









Masacroso

13.2k41749




13.2k41749










asked Dec 30 '18 at 10:39









sleeve chensleeve chen

3,23942256




3,23942256








  • 1




    $begingroup$
    @Masacroso It should be $sum_{xin X}$ from Folland.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 10:59










  • $begingroup$
    @Masacroso $X$ should be the domain of the function. $X$ is an arbitrary set.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 11:00














  • 1




    $begingroup$
    @Masacroso It should be $sum_{xin X}$ from Folland.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 10:59










  • $begingroup$
    @Masacroso $X$ should be the domain of the function. $X$ is an arbitrary set.
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 11:00








1




1




$begingroup$
@Masacroso It should be $sum_{xin X}$ from Folland.
$endgroup$
– sleeve chen
Dec 30 '18 at 10:59




$begingroup$
@Masacroso It should be $sum_{xin X}$ from Folland.
$endgroup$
– sleeve chen
Dec 30 '18 at 10:59












$begingroup$
@Masacroso $X$ should be the domain of the function. $X$ is an arbitrary set.
$endgroup$
– sleeve chen
Dec 30 '18 at 11:00




$begingroup$
@Masacroso $X$ should be the domain of the function. $X$ is an arbitrary set.
$endgroup$
– sleeve chen
Dec 30 '18 at 11:00










3 Answers
3






active

oldest

votes


















2












$begingroup$

If $xin F$, then $xin A_n$, and $f(x)>1/n$. Let $|F|=m$, and write $F={x_1,ldots,x_m}$. Then
$$sum_{xin F}f(x)=f(x_1)+cdots+f(x_n)ge 1/n+cdots +1/n=m/n.$$



If $|A_n|=infty$, then $A_n$ has arbitrarily large finite subsets $F$,
and then
$$sum_{xin A_n}f(x)ge sum_{xin F}f(x)gefrac{|F|}{n}$$
which can be arbitrarily large.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 10:51












  • $begingroup$
    As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
    $endgroup$
    – Lord Shark the Unknown
    Dec 30 '18 at 10:55










  • $begingroup$
    I get it, thanks!
    $endgroup$
    – sleeve chen
    Dec 30 '18 at 10:57



















2












$begingroup$

Since $A = bigcup_1^infty A_n$ is uncountable, then there is some positive integer $n$ such that $A_n$ is infinite.
Hence for any positive integer $m$ there is a finite set $F_msubset A_n$ such that $mbox{card}(F_m)>m$, and it follows
$$sum_{xin A_n} f(x)geq sum_{xin F_m} f(x) >
sum_{xin F_m} frac{1}{n}=frac{mbox{card}(F_m)}{n}>frac{m}{n}$$
which is arbitrary large. We may conclude that
$$sum_{xin A} f(x)geq sum_{xin A_n} f(x)=+infty.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Well, the first part is clear: $sum_{xin F}f(x)>sum_{xin F} frac{1}{n} = |F|/n$.






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      If $xin F$, then $xin A_n$, and $f(x)>1/n$. Let $|F|=m$, and write $F={x_1,ldots,x_m}$. Then
      $$sum_{xin F}f(x)=f(x_1)+cdots+f(x_n)ge 1/n+cdots +1/n=m/n.$$



      If $|A_n|=infty$, then $A_n$ has arbitrarily large finite subsets $F$,
      and then
      $$sum_{xin A_n}f(x)ge sum_{xin F}f(x)gefrac{|F|}{n}$$
      which can be arbitrarily large.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:51












      • $begingroup$
        As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
        $endgroup$
        – Lord Shark the Unknown
        Dec 30 '18 at 10:55










      • $begingroup$
        I get it, thanks!
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:57
















      2












      $begingroup$

      If $xin F$, then $xin A_n$, and $f(x)>1/n$. Let $|F|=m$, and write $F={x_1,ldots,x_m}$. Then
      $$sum_{xin F}f(x)=f(x_1)+cdots+f(x_n)ge 1/n+cdots +1/n=m/n.$$



      If $|A_n|=infty$, then $A_n$ has arbitrarily large finite subsets $F$,
      and then
      $$sum_{xin A_n}f(x)ge sum_{xin F}f(x)gefrac{|F|}{n}$$
      which can be arbitrarily large.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:51












      • $begingroup$
        As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
        $endgroup$
        – Lord Shark the Unknown
        Dec 30 '18 at 10:55










      • $begingroup$
        I get it, thanks!
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:57














      2












      2








      2





      $begingroup$

      If $xin F$, then $xin A_n$, and $f(x)>1/n$. Let $|F|=m$, and write $F={x_1,ldots,x_m}$. Then
      $$sum_{xin F}f(x)=f(x_1)+cdots+f(x_n)ge 1/n+cdots +1/n=m/n.$$



      If $|A_n|=infty$, then $A_n$ has arbitrarily large finite subsets $F$,
      and then
      $$sum_{xin A_n}f(x)ge sum_{xin F}f(x)gefrac{|F|}{n}$$
      which can be arbitrarily large.






      share|cite|improve this answer









      $endgroup$



      If $xin F$, then $xin A_n$, and $f(x)>1/n$. Let $|F|=m$, and write $F={x_1,ldots,x_m}$. Then
      $$sum_{xin F}f(x)=f(x_1)+cdots+f(x_n)ge 1/n+cdots +1/n=m/n.$$



      If $|A_n|=infty$, then $A_n$ has arbitrarily large finite subsets $F$,
      and then
      $$sum_{xin A_n}f(x)ge sum_{xin F}f(x)gefrac{|F|}{n}$$
      which can be arbitrarily large.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 30 '18 at 10:42









      Lord Shark the UnknownLord Shark the Unknown

      109k1163136




      109k1163136












      • $begingroup$
        I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:51












      • $begingroup$
        As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
        $endgroup$
        – Lord Shark the Unknown
        Dec 30 '18 at 10:55










      • $begingroup$
        I get it, thanks!
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:57


















      • $begingroup$
        I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:51












      • $begingroup$
        As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
        $endgroup$
        – Lord Shark the Unknown
        Dec 30 '18 at 10:55










      • $begingroup$
        I get it, thanks!
        $endgroup$
        – sleeve chen
        Dec 30 '18 at 10:57
















      $begingroup$
      I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
      $endgroup$
      – sleeve chen
      Dec 30 '18 at 10:51






      $begingroup$
      I am still a bit confused on "if $xin F$ then $xin A_n$". How to claim this? I still miss something..
      $endgroup$
      – sleeve chen
      Dec 30 '18 at 10:51














      $begingroup$
      As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
      $endgroup$
      – Lord Shark the Unknown
      Dec 30 '18 at 10:55




      $begingroup$
      As you wrote, "...for $F$ a finite subset of $A_n$...." @sleevechen
      $endgroup$
      – Lord Shark the Unknown
      Dec 30 '18 at 10:55












      $begingroup$
      I get it, thanks!
      $endgroup$
      – sleeve chen
      Dec 30 '18 at 10:57




      $begingroup$
      I get it, thanks!
      $endgroup$
      – sleeve chen
      Dec 30 '18 at 10:57











      2












      $begingroup$

      Since $A = bigcup_1^infty A_n$ is uncountable, then there is some positive integer $n$ such that $A_n$ is infinite.
      Hence for any positive integer $m$ there is a finite set $F_msubset A_n$ such that $mbox{card}(F_m)>m$, and it follows
      $$sum_{xin A_n} f(x)geq sum_{xin F_m} f(x) >
      sum_{xin F_m} frac{1}{n}=frac{mbox{card}(F_m)}{n}>frac{m}{n}$$
      which is arbitrary large. We may conclude that
      $$sum_{xin A} f(x)geq sum_{xin A_n} f(x)=+infty.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Since $A = bigcup_1^infty A_n$ is uncountable, then there is some positive integer $n$ such that $A_n$ is infinite.
        Hence for any positive integer $m$ there is a finite set $F_msubset A_n$ such that $mbox{card}(F_m)>m$, and it follows
        $$sum_{xin A_n} f(x)geq sum_{xin F_m} f(x) >
        sum_{xin F_m} frac{1}{n}=frac{mbox{card}(F_m)}{n}>frac{m}{n}$$
        which is arbitrary large. We may conclude that
        $$sum_{xin A} f(x)geq sum_{xin A_n} f(x)=+infty.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Since $A = bigcup_1^infty A_n$ is uncountable, then there is some positive integer $n$ such that $A_n$ is infinite.
          Hence for any positive integer $m$ there is a finite set $F_msubset A_n$ such that $mbox{card}(F_m)>m$, and it follows
          $$sum_{xin A_n} f(x)geq sum_{xin F_m} f(x) >
          sum_{xin F_m} frac{1}{n}=frac{mbox{card}(F_m)}{n}>frac{m}{n}$$
          which is arbitrary large. We may conclude that
          $$sum_{xin A} f(x)geq sum_{xin A_n} f(x)=+infty.$$






          share|cite|improve this answer











          $endgroup$



          Since $A = bigcup_1^infty A_n$ is uncountable, then there is some positive integer $n$ such that $A_n$ is infinite.
          Hence for any positive integer $m$ there is a finite set $F_msubset A_n$ such that $mbox{card}(F_m)>m$, and it follows
          $$sum_{xin A_n} f(x)geq sum_{xin F_m} f(x) >
          sum_{xin F_m} frac{1}{n}=frac{mbox{card}(F_m)}{n}>frac{m}{n}$$
          which is arbitrary large. We may conclude that
          $$sum_{xin A} f(x)geq sum_{xin A_n} f(x)=+infty.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 10:53

























          answered Dec 30 '18 at 10:42









          Robert ZRobert Z

          102k1072145




          102k1072145























              1












              $begingroup$

              Well, the first part is clear: $sum_{xin F}f(x)>sum_{xin F} frac{1}{n} = |F|/n$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Well, the first part is clear: $sum_{xin F}f(x)>sum_{xin F} frac{1}{n} = |F|/n$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Well, the first part is clear: $sum_{xin F}f(x)>sum_{xin F} frac{1}{n} = |F|/n$.






                  share|cite|improve this answer









                  $endgroup$



                  Well, the first part is clear: $sum_{xin F}f(x)>sum_{xin F} frac{1}{n} = |F|/n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 10:42









                  WuestenfuxWuestenfux

                  5,5911513




                  5,5911513






























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