Is a simply connected subset of $mathbb{R}^n$ necessarily measurable?












7












$begingroup$


Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.



Is $S$ necessarily Lebesgue measurable?



Is it necessarily Borel measurable?



What's the proof?



(Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)






general-topology measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$

















    7












    $begingroup$


    Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.



    Is $S$ necessarily Lebesgue measurable?



    Is it necessarily Borel measurable?



    What's the proof?



    (Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)






    general-topology measure-theory lebesgue-measure







    share|cite|improve this question









    $endgroup$















      7












      7








      7


      3



      $begingroup$


      Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.



      Is $S$ necessarily Lebesgue measurable?



      Is it necessarily Borel measurable?



      What's the proof?



      (Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)






      general-topology measure-theory lebesgue-measure







      share|cite|improve this question









      $endgroup$




      Let $Ssubsetmathbb{R}^n$ be a subset that is simply connected as a topological space.



      Is $S$ necessarily Lebesgue measurable?



      Is it necessarily Borel measurable?



      What's the proof?



      (Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)






      general-topology measure-theory lebesgue-measure




      general-topology measure-theory lebesgue-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 16 at 8:09









      Ben Blum-SmithBen Blum-Smith

      10.1k23086




      10.1k23086






















          1 Answer
          1






          active

          oldest

          votes


















          11












          $begingroup$

          Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.



          To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.



          To see that this set is simply connected is best to just draw it.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
            $endgroup$
            – stressed out
            Jan 16 at 8:24






          • 3




            $begingroup$
            the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
            $endgroup$
            – postmortes
            Jan 16 at 9:33






          • 2




            $begingroup$
            My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
            $endgroup$
            – DanielWainfleet
            Jan 16 at 23:11






          • 1




            $begingroup$
            @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:02










          • $begingroup$
            +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:07













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          11












          $begingroup$

          Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.



          To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.



          To see that this set is simply connected is best to just draw it.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
            $endgroup$
            – stressed out
            Jan 16 at 8:24






          • 3




            $begingroup$
            the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
            $endgroup$
            – postmortes
            Jan 16 at 9:33






          • 2




            $begingroup$
            My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
            $endgroup$
            – DanielWainfleet
            Jan 16 at 23:11






          • 1




            $begingroup$
            @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:02










          • $begingroup$
            +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:07


















          11












          $begingroup$

          Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.



          To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.



          To see that this set is simply connected is best to just draw it.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
            $endgroup$
            – stressed out
            Jan 16 at 8:24






          • 3




            $begingroup$
            the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
            $endgroup$
            – postmortes
            Jan 16 at 9:33






          • 2




            $begingroup$
            My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
            $endgroup$
            – DanielWainfleet
            Jan 16 at 23:11






          • 1




            $begingroup$
            @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:02










          • $begingroup$
            +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:07
















          11












          11








          11





          $begingroup$

          Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.



          To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.



          To see that this set is simply connected is best to just draw it.






          share|cite|improve this answer











          $endgroup$



          Let $Vsubseteq [0,1]$ be a nonmeasurable subset of $Bbb R$, such as the Vitali set, and consider the "comb" $(Vtimes[0,1])cup ([0,1]times{0})$, which is a simply connected nonmeasurable subset of $Bbb R^2$.



          To see that this set is nonmeasurable remember that if $(X,mathcal A,mu)$ and $(Y,mathcal B,nu)$ are two measure spaces then for a measurable $Esubseteq Xtimes Y$ we have that the slices $E_y={xin Xmid (x,y)in E}$ are $mu$-measurable for $nu$-almost every $y$ and that the slices $E_x={yin Ymid (x,y)in E}$ are $nu$-measurable for $mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.



          To see that this set is simply connected is best to just draw it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 23:10









          DanielWainfleet

          35k31648




          35k31648










          answered Jan 16 at 8:18









          Alessandro CodenottiAlessandro Codenotti

          3,81311539




          3,81311539








          • 2




            $begingroup$
            Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
            $endgroup$
            – stressed out
            Jan 16 at 8:24






          • 3




            $begingroup$
            the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
            $endgroup$
            – postmortes
            Jan 16 at 9:33






          • 2




            $begingroup$
            My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
            $endgroup$
            – DanielWainfleet
            Jan 16 at 23:11






          • 1




            $begingroup$
            @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:02










          • $begingroup$
            +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:07
















          • 2




            $begingroup$
            Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
            $endgroup$
            – stressed out
            Jan 16 at 8:24






          • 3




            $begingroup$
            the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
            $endgroup$
            – postmortes
            Jan 16 at 9:33






          • 2




            $begingroup$
            My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
            $endgroup$
            – DanielWainfleet
            Jan 16 at 23:11






          • 1




            $begingroup$
            @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:02










          • $begingroup$
            +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
            $endgroup$
            – Ben Blum-Smith
            Jan 17 at 5:07










          2




          2




          $begingroup$
          Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
          $endgroup$
          – stressed out
          Jan 16 at 8:24




          $begingroup$
          Could you elaborate on your answer please? Why is $V times [0,1] cup [0,1] times {0}$ simply connected and non-measurable? Moreover, what does the last ${0}$ do exactly? Why is it there?
          $endgroup$
          – stressed out
          Jan 16 at 8:24




          3




          3




          $begingroup$
          the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
          $endgroup$
          – postmortes
          Jan 16 at 9:33




          $begingroup$
          the last ${0}$ is part of a product, $[0,1]times {0}$ and it's there to get the backbone of the comb in place (which makes it all simply connected). If you draw the set, as suggested in the answer, you'll see that each point of the Vitali set provides a 'tooth' of the comb along the interval $[0,1]$ and you can walk along the backbone and up and down each tooth
          $endgroup$
          – postmortes
          Jan 16 at 9:33




          2




          2




          $begingroup$
          My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
          $endgroup$
          – DanielWainfleet
          Jan 16 at 23:11




          $begingroup$
          My edit was merely to add some brackets to the def'n of the "comb" to make it easier to read it.
          $endgroup$
          – DanielWainfleet
          Jan 16 at 23:11




          1




          1




          $begingroup$
          @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
          $endgroup$
          – Ben Blum-Smith
          Jan 17 at 5:02




          $begingroup$
          @stressedout - in case further elaboration is useful: Allesandro Codenotti's answer is set in $mathbb{R}^2$. Thus to specify points you have to specify both the $x$ and $y$ coordinates. The $[0,1]times {0}$ is the set of points with $x$ coordinate in $[0,1]$ and whose $y$ coordinate is $0$. Without the $times{0}$, it's not a subset of $mathbb{R}^2$.
          $endgroup$
          – Ben Blum-Smith
          Jan 17 at 5:02












          $begingroup$
          +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
          $endgroup$
          – Ben Blum-Smith
          Jan 17 at 5:07






          $begingroup$
          +1 Beautiful! (And, accepted.) Also, in order so that this question/answer pair contains a formal proof of simple connectedness: the set given is homotopy equivalent to an interval, since it deformation retracts to $[0,1]times{0}$ via the family of maps $(x,y)mapsto (x,(1-t)y)$ for $tin [0,1]$.
          $endgroup$
          – Ben Blum-Smith
          Jan 17 at 5:07




















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