An extended real value function is measurable (proof verification)

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For the second proof, it says "$xin X: f_1(x) > alpha = x in X : f(x) >alpha cup B$". $f_1(x)$ cannot be equal to $-infty$. Then, why is it union of $B$?




measure-theory proof-explanation




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  • Could you explain why?
    – Sihyun Kim
    Jul 30 at 12:23










  • Consider the two cases for $f_1(x)$.
    – xbh
    Jul 30 at 12:28














up vote
0
down vote

favorite













enter image description here




For the second proof, it says "$xin X: f_1(x) > alpha = x in X : f(x) >alpha cup B$". $f_1(x)$ cannot be equal to $-infty$. Then, why is it union of $B$?




measure-theory proof-explanation




share|cite|improve this question





















  • Could you explain why?
    – Sihyun Kim
    Jul 30 at 12:23










  • Consider the two cases for $f_1(x)$.
    – xbh
    Jul 30 at 12:28












up vote
0
down vote

favorite









up vote
0
down vote

favorite












enter image description here




For the second proof, it says "$xin X: f_1(x) > alpha = x in X : f(x) >alpha cup B$". $f_1(x)$ cannot be equal to $-infty$. Then, why is it union of $B$?




measure-theory proof-explanation




share|cite|improve this question














enter image description here




For the second proof, it says "$xin X: f_1(x) > alpha = x in X : f(x) >alpha cup B$". $f_1(x)$ cannot be equal to $-infty$. Then, why is it union of $B$?





measure-theory proof-explanation





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 12:35









amWhy

189k25219431




189k25219431









asked Jul 30 at 12:17









Sihyun Kim

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701210











  • Could you explain why?
    – Sihyun Kim
    Jul 30 at 12:23










  • Consider the two cases for $f_1(x)$.
    – xbh
    Jul 30 at 12:28
















  • Could you explain why?
    – Sihyun Kim
    Jul 30 at 12:23










  • Consider the two cases for $f_1(x)$.
    – xbh
    Jul 30 at 12:28















Could you explain why?
– Sihyun Kim
Jul 30 at 12:23




Could you explain why?
– Sihyun Kim
Jul 30 at 12:23












Consider the two cases for $f_1(x)$.
– xbh
Jul 30 at 12:28




Consider the two cases for $f_1(x)$.
– xbh
Jul 30 at 12:28










3 Answers
3






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up vote
0
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accepted










For e.g. $alpha=-1$ we find that the following statements are equivalent:



  • $f_1(x)>-1$

  • $f_1(x)in(-1,infty)$

  • $f_1(x)=0vee f_1(x)in(-1,0)cup(0,infty)$

  • $f(x)in-infty,0,inftyvee f(x)in(-1,0)cup(0,infty)$

  • $f(x)in(-1,infty]vee f(x)=-infty$

  • $f(x)>-1vee xin B$





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  • What does the symbol V mean?
    – Sihyun Kim
    Jul 31 at 9:33







  • 1




    It stands for "or".
    – drhab
    Jul 31 at 9:34

















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0
down vote













If $f(x) = -infty$, then $f_1(x) = 0$.






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    If $alpha < 0$ and $f_1(x) > alpha$, then either $f(x)$ is finite and $f(x) = f_1(x) > alpha$ or $f(x) = infty$ or $f(x) = -infty$. Thus,
    $$x: f_1(x) > alpha subset x: f(x) > alpha cup A cup B.$$
    But $A subset x: f(x) > alpha$, so this reduces to
    $$x: f_1(x) > alpha subset x: f(x) > alpha cup B.$$
    Conversely, if $alpha<0$ and either $f(x)> alpha$ or $f(x) = -infty$, then, in the former case $f(x)$ is either finite or = $infty$, whence $f_1(x) = f(x)$ or $0$, respectively, whence $f_1(x)> alpha$; and in the latter case $f_1(x)=0 > alpha$. So,
    $$x: f_1(x) > alpha supset x: f(x) > alpha cup B.$$






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote



      accepted










      For e.g. $alpha=-1$ we find that the following statements are equivalent:



      • $f_1(x)>-1$

      • $f_1(x)in(-1,infty)$

      • $f_1(x)=0vee f_1(x)in(-1,0)cup(0,infty)$

      • $f(x)in-infty,0,inftyvee f(x)in(-1,0)cup(0,infty)$

      • $f(x)in(-1,infty]vee f(x)=-infty$

      • $f(x)>-1vee xin B$





      share|cite|improve this answer





















      • What does the symbol V mean?
        – Sihyun Kim
        Jul 31 at 9:33







      • 1




        It stands for "or".
        – drhab
        Jul 31 at 9:34














      up vote
      0
      down vote



      accepted










      For e.g. $alpha=-1$ we find that the following statements are equivalent:



      • $f_1(x)>-1$

      • $f_1(x)in(-1,infty)$

      • $f_1(x)=0vee f_1(x)in(-1,0)cup(0,infty)$

      • $f(x)in-infty,0,inftyvee f(x)in(-1,0)cup(0,infty)$

      • $f(x)in(-1,infty]vee f(x)=-infty$

      • $f(x)>-1vee xin B$





      share|cite|improve this answer





















      • What does the symbol V mean?
        – Sihyun Kim
        Jul 31 at 9:33







      • 1




        It stands for "or".
        – drhab
        Jul 31 at 9:34












      up vote
      0
      down vote



      accepted







      up vote
      0
      down vote



      accepted






      For e.g. $alpha=-1$ we find that the following statements are equivalent:



      • $f_1(x)>-1$

      • $f_1(x)in(-1,infty)$

      • $f_1(x)=0vee f_1(x)in(-1,0)cup(0,infty)$

      • $f(x)in-infty,0,inftyvee f(x)in(-1,0)cup(0,infty)$

      • $f(x)in(-1,infty]vee f(x)=-infty$

      • $f(x)>-1vee xin B$





      share|cite|improve this answer













      For e.g. $alpha=-1$ we find that the following statements are equivalent:



      • $f_1(x)>-1$

      • $f_1(x)in(-1,infty)$

      • $f_1(x)=0vee f_1(x)in(-1,0)cup(0,infty)$

      • $f(x)in-infty,0,inftyvee f(x)in(-1,0)cup(0,infty)$

      • $f(x)in(-1,infty]vee f(x)=-infty$

      • $f(x)>-1vee xin B$






      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Jul 30 at 12:51









      drhab

      85.9k540118




      85.9k540118











      • What does the symbol V mean?
        – Sihyun Kim
        Jul 31 at 9:33







      • 1




        It stands for "or".
        – drhab
        Jul 31 at 9:34
















      • What does the symbol V mean?
        – Sihyun Kim
        Jul 31 at 9:33







      • 1




        It stands for "or".
        – drhab
        Jul 31 at 9:34















      What does the symbol V mean?
      – Sihyun Kim
      Jul 31 at 9:33





      What does the symbol V mean?
      – Sihyun Kim
      Jul 31 at 9:33





      1




      1




      It stands for "or".
      – drhab
      Jul 31 at 9:34




      It stands for "or".
      – drhab
      Jul 31 at 9:34










      up vote
      0
      down vote













      If $f(x) = -infty$, then $f_1(x) = 0$.






      share|cite|improve this answer

























        up vote
        0
        down vote













        If $f(x) = -infty$, then $f_1(x) = 0$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          If $f(x) = -infty$, then $f_1(x) = 0$.






          share|cite|improve this answer













          If $f(x) = -infty$, then $f_1(x) = 0$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 30 at 12:23









          Hurkyl

          107k9112253




          107k9112253




















              up vote
              0
              down vote













              If $alpha < 0$ and $f_1(x) > alpha$, then either $f(x)$ is finite and $f(x) = f_1(x) > alpha$ or $f(x) = infty$ or $f(x) = -infty$. Thus,
              $$x: f_1(x) > alpha subset x: f(x) > alpha cup A cup B.$$
              But $A subset x: f(x) > alpha$, so this reduces to
              $$x: f_1(x) > alpha subset x: f(x) > alpha cup B.$$
              Conversely, if $alpha<0$ and either $f(x)> alpha$ or $f(x) = -infty$, then, in the former case $f(x)$ is either finite or = $infty$, whence $f_1(x) = f(x)$ or $0$, respectively, whence $f_1(x)> alpha$; and in the latter case $f_1(x)=0 > alpha$. So,
              $$x: f_1(x) > alpha supset x: f(x) > alpha cup B.$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                If $alpha < 0$ and $f_1(x) > alpha$, then either $f(x)$ is finite and $f(x) = f_1(x) > alpha$ or $f(x) = infty$ or $f(x) = -infty$. Thus,
                $$x: f_1(x) > alpha subset x: f(x) > alpha cup A cup B.$$
                But $A subset x: f(x) > alpha$, so this reduces to
                $$x: f_1(x) > alpha subset x: f(x) > alpha cup B.$$
                Conversely, if $alpha<0$ and either $f(x)> alpha$ or $f(x) = -infty$, then, in the former case $f(x)$ is either finite or = $infty$, whence $f_1(x) = f(x)$ or $0$, respectively, whence $f_1(x)> alpha$; and in the latter case $f_1(x)=0 > alpha$. So,
                $$x: f_1(x) > alpha supset x: f(x) > alpha cup B.$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  If $alpha < 0$ and $f_1(x) > alpha$, then either $f(x)$ is finite and $f(x) = f_1(x) > alpha$ or $f(x) = infty$ or $f(x) = -infty$. Thus,
                  $$x: f_1(x) > alpha subset x: f(x) > alpha cup A cup B.$$
                  But $A subset x: f(x) > alpha$, so this reduces to
                  $$x: f_1(x) > alpha subset x: f(x) > alpha cup B.$$
                  Conversely, if $alpha<0$ and either $f(x)> alpha$ or $f(x) = -infty$, then, in the former case $f(x)$ is either finite or = $infty$, whence $f_1(x) = f(x)$ or $0$, respectively, whence $f_1(x)> alpha$; and in the latter case $f_1(x)=0 > alpha$. So,
                  $$x: f_1(x) > alpha supset x: f(x) > alpha cup B.$$






                  share|cite|improve this answer













                  If $alpha < 0$ and $f_1(x) > alpha$, then either $f(x)$ is finite and $f(x) = f_1(x) > alpha$ or $f(x) = infty$ or $f(x) = -infty$. Thus,
                  $$x: f_1(x) > alpha subset x: f(x) > alpha cup A cup B.$$
                  But $A subset x: f(x) > alpha$, so this reduces to
                  $$x: f_1(x) > alpha subset x: f(x) > alpha cup B.$$
                  Conversely, if $alpha<0$ and either $f(x)> alpha$ or $f(x) = -infty$, then, in the former case $f(x)$ is either finite or = $infty$, whence $f_1(x) = f(x)$ or $0$, respectively, whence $f_1(x)> alpha$; and in the latter case $f_1(x)=0 > alpha$. So,
                  $$x: f_1(x) > alpha supset x: f(x) > alpha cup B.$$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 30 at 12:37









                  aduh

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