Proving existence of dense subset $D subseteq mathbb R$, so that ${f geq alpha} in mathcal{A}$, $forall alpha...











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Background to the question is that I have to show the equivalance of the following three statements:
$(X, mathcal{A})$ a measure space and $f: X to overline{mathbb R}$



Show the equivalence of the three following statements:



$i)$ $f$ is $mathcal{A}-mathcal{B}(overline{mathbb R})-$measureable



$ii)$There exists a dense subset $D subseteq mathbb R$, so that ${f geq alpha}in mathcal{A}$, $forall alpha in D$.



$iii)$ $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A}$



I thought I took the "easy route" when I showed $ii) Rightarrow i)$ and then $i) Rightarrow iii)$. Now I am having trouble proving that $iii) Rightarrow ii)$, which leads me to think that it may not be possible directly and so I'll have to redo my other implications. Is there a way to directly solve:



If $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A} Rightarrow$ I can find $D subseteq mathbb R$ such that ${f geq alpha} in mathcal{A}$, $forall alpha in D$.










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    up vote
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    down vote

    favorite












    Background to the question is that I have to show the equivalance of the following three statements:
    $(X, mathcal{A})$ a measure space and $f: X to overline{mathbb R}$



    Show the equivalence of the three following statements:



    $i)$ $f$ is $mathcal{A}-mathcal{B}(overline{mathbb R})-$measureable



    $ii)$There exists a dense subset $D subseteq mathbb R$, so that ${f geq alpha}in mathcal{A}$, $forall alpha in D$.



    $iii)$ $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A}$



    I thought I took the "easy route" when I showed $ii) Rightarrow i)$ and then $i) Rightarrow iii)$. Now I am having trouble proving that $iii) Rightarrow ii)$, which leads me to think that it may not be possible directly and so I'll have to redo my other implications. Is there a way to directly solve:



    If $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A} Rightarrow$ I can find $D subseteq mathbb R$ such that ${f geq alpha} in mathcal{A}$, $forall alpha in D$.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Background to the question is that I have to show the equivalance of the following three statements:
      $(X, mathcal{A})$ a measure space and $f: X to overline{mathbb R}$



      Show the equivalence of the three following statements:



      $i)$ $f$ is $mathcal{A}-mathcal{B}(overline{mathbb R})-$measureable



      $ii)$There exists a dense subset $D subseteq mathbb R$, so that ${f geq alpha}in mathcal{A}$, $forall alpha in D$.



      $iii)$ $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A}$



      I thought I took the "easy route" when I showed $ii) Rightarrow i)$ and then $i) Rightarrow iii)$. Now I am having trouble proving that $iii) Rightarrow ii)$, which leads me to think that it may not be possible directly and so I'll have to redo my other implications. Is there a way to directly solve:



      If $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A} Rightarrow$ I can find $D subseteq mathbb R$ such that ${f geq alpha} in mathcal{A}$, $forall alpha in D$.










      share|cite|improve this question















      Background to the question is that I have to show the equivalance of the following three statements:
      $(X, mathcal{A})$ a measure space and $f: X to overline{mathbb R}$



      Show the equivalence of the three following statements:



      $i)$ $f$ is $mathcal{A}-mathcal{B}(overline{mathbb R})-$measureable



      $ii)$There exists a dense subset $D subseteq mathbb R$, so that ${f geq alpha}in mathcal{A}$, $forall alpha in D$.



      $iii)$ $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A}$



      I thought I took the "easy route" when I showed $ii) Rightarrow i)$ and then $i) Rightarrow iii)$. Now I am having trouble proving that $iii) Rightarrow ii)$, which leads me to think that it may not be possible directly and so I'll have to redo my other implications. Is there a way to directly solve:



      If $f^{-1}({+infty}) in mathcal{A}$ and $f^{-1}(mathcal{B}(mathbb R)) subseteq mathcal{A} Rightarrow$ I can find $D subseteq mathbb R$ such that ${f geq alpha} in mathcal{A}$, $forall alpha in D$.







      real-analysis measure-theory borel-sets






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      edited Nov 19 at 21:13

























      asked Nov 18 at 18:55









      SABOY

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      532211






















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          Hint: $f^{-1}([alpha,infty]) = f^{-1}([alpha,infty))cup f^{-1}({infty}).$






          share|cite|improve this answer





















          • How can I use that to find my dense subset $D$?
            – SABOY
            Nov 19 at 21:15






          • 1




            Your dense subset is all of $mathbb R.$
            – zhw.
            Nov 19 at 21:20










          • got it, thanks!
            – SABOY
            Nov 19 at 21:25











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          1 Answer
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          active

          oldest

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          active

          oldest

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          up vote
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          down vote



          accepted










          Hint: $f^{-1}([alpha,infty]) = f^{-1}([alpha,infty))cup f^{-1}({infty}).$






          share|cite|improve this answer





















          • How can I use that to find my dense subset $D$?
            – SABOY
            Nov 19 at 21:15






          • 1




            Your dense subset is all of $mathbb R.$
            – zhw.
            Nov 19 at 21:20










          • got it, thanks!
            – SABOY
            Nov 19 at 21:25















          up vote
          1
          down vote



          accepted










          Hint: $f^{-1}([alpha,infty]) = f^{-1}([alpha,infty))cup f^{-1}({infty}).$






          share|cite|improve this answer





















          • How can I use that to find my dense subset $D$?
            – SABOY
            Nov 19 at 21:15






          • 1




            Your dense subset is all of $mathbb R.$
            – zhw.
            Nov 19 at 21:20










          • got it, thanks!
            – SABOY
            Nov 19 at 21:25













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hint: $f^{-1}([alpha,infty]) = f^{-1}([alpha,infty))cup f^{-1}({infty}).$






          share|cite|improve this answer












          Hint: $f^{-1}([alpha,infty]) = f^{-1}([alpha,infty))cup f^{-1}({infty}).$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 21:08









          zhw.

          70.4k42975




          70.4k42975












          • How can I use that to find my dense subset $D$?
            – SABOY
            Nov 19 at 21:15






          • 1




            Your dense subset is all of $mathbb R.$
            – zhw.
            Nov 19 at 21:20










          • got it, thanks!
            – SABOY
            Nov 19 at 21:25


















          • How can I use that to find my dense subset $D$?
            – SABOY
            Nov 19 at 21:15






          • 1




            Your dense subset is all of $mathbb R.$
            – zhw.
            Nov 19 at 21:20










          • got it, thanks!
            – SABOY
            Nov 19 at 21:25
















          How can I use that to find my dense subset $D$?
          – SABOY
          Nov 19 at 21:15




          How can I use that to find my dense subset $D$?
          – SABOY
          Nov 19 at 21:15




          1




          1




          Your dense subset is all of $mathbb R.$
          – zhw.
          Nov 19 at 21:20




          Your dense subset is all of $mathbb R.$
          – zhw.
          Nov 19 at 21:20












          got it, thanks!
          – SABOY
          Nov 19 at 21:25




          got it, thanks!
          – SABOY
          Nov 19 at 21:25


















           

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