$f$ measurable iff $f^2$ measurable and ${f > 0}$ measurable












1














I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:



$$ {f > 0} text{ measurable }$$



play into making $f$ automatically measurable?






measure-theory lebesgue-measure borel-sets borel-measures







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    1














    I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:



    $$ {f > 0} text{ measurable }$$



    play into making $f$ automatically measurable?






    measure-theory lebesgue-measure borel-sets borel-measures







    share|cite|improve this question









    New contributor




    sasha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1












      1








      1


      1





      I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:



      $$ {f > 0} text{ measurable }$$



      play into making $f$ automatically measurable?






      measure-theory lebesgue-measure borel-sets borel-measures







      share|cite|improve this question









      New contributor




      sasha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:



      $$ {f > 0} text{ measurable }$$



      play into making $f$ automatically measurable?






      measure-theory lebesgue-measure borel-sets borel-measures




      measure-theory lebesgue-measure borel-sets borel-measures






      share|cite|improve this question









      New contributor




      sasha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      sasha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 19:24









      Umberto P.

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      38.5k13064






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      asked Jan 4 at 19:09









      sashasasha

      133




      133




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          2 Answers
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          1














          Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.






          share|cite|improve this answer





















          • Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
            – sasha
            Jan 4 at 19:25



















          1














          Another Hint: It holds
          $$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
          $$






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
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            active

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            active

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            1














            Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.






            share|cite|improve this answer





















            • Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
              – sasha
              Jan 4 at 19:25
















            1














            Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.






            share|cite|improve this answer





















            • Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
              – sasha
              Jan 4 at 19:25














            1












            1








            1






            Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.






            share|cite|improve this answer












            Hint: For $a>0$, we have ${f>a} ={f^2 >a^2} cap{f>0}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 19:13









            BerciBerci

            59.8k23672




            59.8k23672












            • Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
              – sasha
              Jan 4 at 19:25


















            • Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
              – sasha
              Jan 4 at 19:25
















            Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
            – sasha
            Jan 4 at 19:25




            Ohhh, I knew it had to do with a way of rewriting the sets in some sort of way.
            – sasha
            Jan 4 at 19:25











            1














            Another Hint: It holds
            $$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
            $$






            share|cite|improve this answer


























              1














              Another Hint: It holds
              $$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
              $$






              share|cite|improve this answer
























                1












                1








                1






                Another Hint: It holds
                $$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
                $$






                share|cite|improve this answer












                Another Hint: It holds
                $$f =sqrt{f^2}1_{{f>0}}-sqrt{f^2}1_{{f>0}^c}.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 19:17









                SongSong

                6,705319




                6,705319






















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