Let $f: mathbb{R} rightarrow mathbb{R}$ be a measurable function, is $f circ sin$ measurable?












4












$begingroup$


If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44
















4












$begingroup$


If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44














4












4








4





$begingroup$


If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




If $f circ sin$ is measurable, then we need to show that $(f circ sin)^{-1}(B) = sin^{-1}(f^{-1}(B))$ is measurable for every Borel set $B$, it is sufficient to show that $sin^{-1}$ takes a Lebesgue set into a Lebesgue set. We have also tried to work with the fact that $arcsin$ is locally Lipschitz, so if this property is sufficient to show that the image of a Lebesgue set is Lebesgue, then $sin^{-1}(f^{-1}(B)) = cup_{n in mathbb{Z}} arcsin(f^{-1}(B)) + 2npi $ would measurable.



It could happen that $fcirc sin$ is not measurable, but the task of finding a set $B$ such that $sin^{-1}(f^{-1}(B))$ is not measurable seems very difficult (maybe impossible?).






measure-theory lebesgue-measure




measure-theory lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 4:42







Non

















asked Dec 9 '18 at 2:00









NonNon

538




538








  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44














  • 3




    $begingroup$
    the composition of measurable functions is measurable.
    $endgroup$
    – Tsemo Aristide
    Dec 9 '18 at 2:01






  • 2




    $begingroup$
    @TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
    $endgroup$
    – Will M.
    Dec 9 '18 at 2:05








  • 2




    $begingroup$
    Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
    $endgroup$
    – Shaun
    Dec 9 '18 at 2:20








  • 5




    $begingroup$
    @Tsemo actually that’s not true in general
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:42






  • 4




    $begingroup$
    @Tsemo see this post for instance
    $endgroup$
    – Omnomnomnom
    Dec 9 '18 at 2:44








3




3




$begingroup$
the composition of measurable functions is measurable.
$endgroup$
– Tsemo Aristide
Dec 9 '18 at 2:01




$begingroup$
the composition of measurable functions is measurable.
$endgroup$
– Tsemo Aristide
Dec 9 '18 at 2:01




2




2




$begingroup$
@TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
$endgroup$
– Will M.
Dec 9 '18 at 2:05






$begingroup$
@TsemoAristide He is considering $f$ to be measurable $Lambda/mathscr{B},$ so he wants to prove $sin$ is $Lambda/Lambda$ measurable, I guess.
$endgroup$
– Will M.
Dec 9 '18 at 2:05






2




2




$begingroup$
Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
$endgroup$
– Shaun
Dec 9 '18 at 2:20






$begingroup$
Use $sin$ for $sin$, rather than $sin$ for $sin$, since the latter typically denotes the product of $s, i, n$. The same goes for using $arcsin$ for $arcsin$, and other trigonometric functions.
$endgroup$
– Shaun
Dec 9 '18 at 2:20






5




5




$begingroup$
@Tsemo actually that’s not true in general
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:42




$begingroup$
@Tsemo actually that’s not true in general
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:42




4




4




$begingroup$
@Tsemo see this post for instance
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:44




$begingroup$
@Tsemo see this post for instance
$endgroup$
– Omnomnomnom
Dec 9 '18 at 2:44










2 Answers
2






active

oldest

votes


















3












$begingroup$

Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You mean the pre-image of a lebesque by a absolutely continuous?
    $endgroup$
    – Non
    Dec 17 '18 at 2:13



















0












$begingroup$

Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13
















    3












    $begingroup$

    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13














    3












    3








    3





    $begingroup$

    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.






    share|cite|improve this answer











    $endgroup$



    Yes, $ f circ sin$ is measurable. The composition of measurable functions is in general not measurable, so there is something to do. You allude to using Lipschitzness, but I think the relevant property is absolute continuity (which sine satisfies). We just need to show that absolutely continuous functions take Lebesgue measurable sets to Lebesgue measurable sets. This is answered here: Absolutely continuous maps measurable sets to measurable sets. The basic point is that absolutely continuous functions take null sets to null sets.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 13:23









    jgon

    15.6k32143




    15.6k32143










    answered Dec 9 '18 at 4:07









    zoidbergzoidberg

    1,080113




    1,080113












    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13


















    • $begingroup$
      You mean the pre-image of a lebesque by a absolutely continuous?
      $endgroup$
      – Non
      Dec 17 '18 at 2:13
















    $begingroup$
    You mean the pre-image of a lebesque by a absolutely continuous?
    $endgroup$
    – Non
    Dec 17 '18 at 2:13




    $begingroup$
    You mean the pre-image of a lebesque by a absolutely continuous?
    $endgroup$
    – Non
    Dec 17 '18 at 2:13











    0












    $begingroup$

    Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.






        share|cite|improve this answer









        $endgroup$



        Here is the bit of math that was missing to complete the demonstration that $f circ sin$ is measurable: Image of Lebesgue measurable set by $C^1$ function is measurable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 4:43









        NonNon

        538




        538






























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