Prove a composite function involving cantor set is measurable












1














Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by



$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$



and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.



I know that $f$ is a measurable function. Than I proceed by



$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.



I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question
























  • Is $f=chi_K$ ?
    – Chinnapparaj R
    Nov 27 '18 at 3:36












  • yes $f$ is characterstic function of $K$
    – Ricky Ng
    Nov 27 '18 at 3:41












  • What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
    – Xander Henderson
    Nov 27 '18 at 3:55










  • I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
    – Ricky Ng
    Nov 27 '18 at 3:58


















1














Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by



$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$



and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.



I know that $f$ is a measurable function. Than I proceed by



$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.



I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question
























  • Is $f=chi_K$ ?
    – Chinnapparaj R
    Nov 27 '18 at 3:36












  • yes $f$ is characterstic function of $K$
    – Ricky Ng
    Nov 27 '18 at 3:41












  • What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
    – Xander Henderson
    Nov 27 '18 at 3:55










  • I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
    – Ricky Ng
    Nov 27 '18 at 3:58
















1












1








1







Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by



$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$



and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.



I know that $f$ is a measurable function. Than I proceed by



$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.



I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question















Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by



$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$



and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.



I know that $f$ is a measurable function. Than I proceed by



$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.



I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?






real-analysis measure-theory lebesgue-measure




real-analysis 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 Nov 27 '18 at 4:01

























asked Nov 27 '18 at 3:31









Ricky Ng

438




438












  • Is $f=chi_K$ ?
    – Chinnapparaj R
    Nov 27 '18 at 3:36












  • yes $f$ is characterstic function of $K$
    – Ricky Ng
    Nov 27 '18 at 3:41












  • What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
    – Xander Henderson
    Nov 27 '18 at 3:55










  • I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
    – Ricky Ng
    Nov 27 '18 at 3:58




















  • Is $f=chi_K$ ?
    – Chinnapparaj R
    Nov 27 '18 at 3:36












  • yes $f$ is characterstic function of $K$
    – Ricky Ng
    Nov 27 '18 at 3:41












  • What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
    – Xander Henderson
    Nov 27 '18 at 3:55










  • I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
    – Ricky Ng
    Nov 27 '18 at 3:58


















Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36






Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36














yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41






yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41














What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55




What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55












I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58






I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58












2 Answers
2






active

oldest

votes


















1














For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.






share|cite|improve this answer





















  • How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
    – Ricky Ng
    Nov 27 '18 at 6:13












  • @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
    – Guacho Perez
    Nov 27 '18 at 6:28



















1














If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$



$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    1














    For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
    Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.






    share|cite|improve this answer





















    • How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
      – Ricky Ng
      Nov 27 '18 at 6:13












    • @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
      – Guacho Perez
      Nov 27 '18 at 6:28
















    1














    For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
    Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.






    share|cite|improve this answer





















    • How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
      – Ricky Ng
      Nov 27 '18 at 6:13












    • @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
      – Guacho Perez
      Nov 27 '18 at 6:28














    1












    1








    1






    For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
    Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.






    share|cite|improve this answer












    For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
    Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 4:04









    Guacho Perez

    3,88911131




    3,88911131












    • How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
      – Ricky Ng
      Nov 27 '18 at 6:13












    • @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
      – Guacho Perez
      Nov 27 '18 at 6:28


















    • How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
      – Ricky Ng
      Nov 27 '18 at 6:13












    • @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
      – Guacho Perez
      Nov 27 '18 at 6:28
















    How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
    – Ricky Ng
    Nov 27 '18 at 6:13






    How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
    – Ricky Ng
    Nov 27 '18 at 6:13














    @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
    – Guacho Perez
    Nov 27 '18 at 6:28




    @RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
    – Guacho Perez
    Nov 27 '18 at 6:28











    1














    If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$



    $g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set






    share|cite|improve this answer


























      1














      If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$



      $g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set






      share|cite|improve this answer
























        1












        1








        1






        If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$



        $g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set






        share|cite|improve this answer












        If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$



        $g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 3:59









        ZAF

        4307




        4307






























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