Is the function measurable with respect to Lebesgue measure












3












$begingroup$


Let $h: [a,b]rightarrow mathbb{R}$ be a continuous function. For every $yin mathbb{R}$, denote
$$
S(y)=begin{cases} 0 & text{, if } h^{-1}(y) = emptyset \
# h^{-1}(y) &text{, if } h^{-1}(y) text{ is finite}\
infty, &text{, if } h^{-1}(y) text{ is infinite}
end{cases}
$$
Is the function measurable (as a function from $mathbb{R} mapsto mathbb{R}$) with respect to the Lebesgue measure in $mathbb{R}$?






real-analysis measure-theory lebesgue-measure







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3












$begingroup$


Let $h: [a,b]rightarrow mathbb{R}$ be a continuous function. For every $yin mathbb{R}$, denote
$$
S(y)=begin{cases} 0 & text{, if } h^{-1}(y) = emptyset \
# h^{-1}(y) &text{, if } h^{-1}(y) text{ is finite}\
infty, &text{, if } h^{-1}(y) text{ is infinite}
end{cases}
$$
Is the function measurable (as a function from $mathbb{R} mapsto mathbb{R}$) with respect to the Lebesgue measure in $mathbb{R}$?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
    $endgroup$
    – André 3000
    Dec 9 '18 at 3:29














3












3








3


3



$begingroup$


Let $h: [a,b]rightarrow mathbb{R}$ be a continuous function. For every $yin mathbb{R}$, denote
$$
S(y)=begin{cases} 0 & text{, if } h^{-1}(y) = emptyset \
# h^{-1}(y) &text{, if } h^{-1}(y) text{ is finite}\
infty, &text{, if } h^{-1}(y) text{ is infinite}
end{cases}
$$
Is the function measurable (as a function from $mathbb{R} mapsto mathbb{R}$) with respect to the Lebesgue measure in $mathbb{R}$?






real-analysis measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




Let $h: [a,b]rightarrow mathbb{R}$ be a continuous function. For every $yin mathbb{R}$, denote
$$
S(y)=begin{cases} 0 & text{, if } h^{-1}(y) = emptyset \
# h^{-1}(y) &text{, if } h^{-1}(y) text{ is finite}\
infty, &text{, if } h^{-1}(y) text{ is infinite}
end{cases}
$$
Is the function measurable (as a function from $mathbb{R} mapsto mathbb{R}$) with respect to the Lebesgue measure in $mathbb{R}$?






real-analysis measure-theory lebesgue-measure




real-analysis measure-theory lebesgue-measure






share|cite|improve this question















share|cite|improve this question













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edited Dec 9 '18 at 4:39









dantopa

6,46942243




6,46942243










asked Dec 8 '18 at 21:43









Enn YadavEnn Yadav

212




212












  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
    $endgroup$
    – André 3000
    Dec 9 '18 at 3:29


















  • $begingroup$
    Welcome to math.se. Here are some tips on how to ask a good question.
    $endgroup$
    – André 3000
    Dec 9 '18 at 3:29
















$begingroup$
Welcome to math.se. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Dec 9 '18 at 3:29




$begingroup$
Welcome to math.se. Here are some tips on how to ask a good question.
$endgroup$
– André 3000
Dec 9 '18 at 3:29










1 Answer
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Here's a neat proof that $f$ is Borel-measurable:



It suffices to show that $S^{-1}({n})$ is Borel for all $n geq 0$, as the image of $S$ is contained in ${0,1,2,ldots}cup{infty}$. For $n = 0$, $S^{-1}({0}) = mathbb{R} setminus f([a,b])$, which is open. For $n geq 1$, let $g = ftimes cdots times f: [a,b]^n to mathbb{R}^n$. For $Delta subseteq [a,b]^n$ the set of points where at least two coordinates are equal (the "big diagonal") and $D: mathbb{R} to mathbb{R}^n$ the map $D: t mapsto (t,ldots, t)$ (the diagonal map), I claim
$$ S^{-1}({n}) = D^{-1}(g([a,b]^n setminus Delta)).$$
As $g$ is continuous and $[a,b]^n setminus Delta$ is a countable union of compact sets, $g([a,b]^n setminus Delta)$ is Borel, and hence $S^{-1}({n})$ is.






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    Here's a neat proof that $f$ is Borel-measurable:



    It suffices to show that $S^{-1}({n})$ is Borel for all $n geq 0$, as the image of $S$ is contained in ${0,1,2,ldots}cup{infty}$. For $n = 0$, $S^{-1}({0}) = mathbb{R} setminus f([a,b])$, which is open. For $n geq 1$, let $g = ftimes cdots times f: [a,b]^n to mathbb{R}^n$. For $Delta subseteq [a,b]^n$ the set of points where at least two coordinates are equal (the "big diagonal") and $D: mathbb{R} to mathbb{R}^n$ the map $D: t mapsto (t,ldots, t)$ (the diagonal map), I claim
    $$ S^{-1}({n}) = D^{-1}(g([a,b]^n setminus Delta)).$$
    As $g$ is continuous and $[a,b]^n setminus Delta$ is a countable union of compact sets, $g([a,b]^n setminus Delta)$ is Borel, and hence $S^{-1}({n})$ is.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Here's a neat proof that $f$ is Borel-measurable:



      It suffices to show that $S^{-1}({n})$ is Borel for all $n geq 0$, as the image of $S$ is contained in ${0,1,2,ldots}cup{infty}$. For $n = 0$, $S^{-1}({0}) = mathbb{R} setminus f([a,b])$, which is open. For $n geq 1$, let $g = ftimes cdots times f: [a,b]^n to mathbb{R}^n$. For $Delta subseteq [a,b]^n$ the set of points where at least two coordinates are equal (the "big diagonal") and $D: mathbb{R} to mathbb{R}^n$ the map $D: t mapsto (t,ldots, t)$ (the diagonal map), I claim
      $$ S^{-1}({n}) = D^{-1}(g([a,b]^n setminus Delta)).$$
      As $g$ is continuous and $[a,b]^n setminus Delta$ is a countable union of compact sets, $g([a,b]^n setminus Delta)$ is Borel, and hence $S^{-1}({n})$ is.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Here's a neat proof that $f$ is Borel-measurable:



        It suffices to show that $S^{-1}({n})$ is Borel for all $n geq 0$, as the image of $S$ is contained in ${0,1,2,ldots}cup{infty}$. For $n = 0$, $S^{-1}({0}) = mathbb{R} setminus f([a,b])$, which is open. For $n geq 1$, let $g = ftimes cdots times f: [a,b]^n to mathbb{R}^n$. For $Delta subseteq [a,b]^n$ the set of points where at least two coordinates are equal (the "big diagonal") and $D: mathbb{R} to mathbb{R}^n$ the map $D: t mapsto (t,ldots, t)$ (the diagonal map), I claim
        $$ S^{-1}({n}) = D^{-1}(g([a,b]^n setminus Delta)).$$
        As $g$ is continuous and $[a,b]^n setminus Delta$ is a countable union of compact sets, $g([a,b]^n setminus Delta)$ is Borel, and hence $S^{-1}({n})$ is.






        share|cite|improve this answer









        $endgroup$



        Here's a neat proof that $f$ is Borel-measurable:



        It suffices to show that $S^{-1}({n})$ is Borel for all $n geq 0$, as the image of $S$ is contained in ${0,1,2,ldots}cup{infty}$. For $n = 0$, $S^{-1}({0}) = mathbb{R} setminus f([a,b])$, which is open. For $n geq 1$, let $g = ftimes cdots times f: [a,b]^n to mathbb{R}^n$. For $Delta subseteq [a,b]^n$ the set of points where at least two coordinates are equal (the "big diagonal") and $D: mathbb{R} to mathbb{R}^n$ the map $D: t mapsto (t,ldots, t)$ (the diagonal map), I claim
        $$ S^{-1}({n}) = D^{-1}(g([a,b]^n setminus Delta)).$$
        As $g$ is continuous and $[a,b]^n setminus Delta$ is a countable union of compact sets, $g([a,b]^n setminus Delta)$ is Borel, and hence $S^{-1}({n})$ is.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 21:53









        Joshua MundingerJoshua Mundinger

        2,5141026




        2,5141026






























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