Question about sets of Jordan measure zero












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If $A ⊆ ℝ^n$ , we say that A is a set of zero content if for every $ε > 0$ there are compact rectangles $R_1,…,R_m$ such that $A ⊆ R_1∪…∪R_m$ and $vol(R_1)+…+vol(R_m) < ε$ , with $vol([a_1,b_1]×…×[a_n,b_n]) = (b_1-a_1)⋅…⋅(b_n-a_n)$.



Is it true that if $A$ is not of zero content, then there is a compact rectangle $R ⊆ ℝ^n$ such that $R ⊆ A$ and $vol(R) > 0$?










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    $begingroup$


    If $A ⊆ ℝ^n$ , we say that A is a set of zero content if for every $ε > 0$ there are compact rectangles $R_1,…,R_m$ such that $A ⊆ R_1∪…∪R_m$ and $vol(R_1)+…+vol(R_m) < ε$ , with $vol([a_1,b_1]×…×[a_n,b_n]) = (b_1-a_1)⋅…⋅(b_n-a_n)$.



    Is it true that if $A$ is not of zero content, then there is a compact rectangle $R ⊆ ℝ^n$ such that $R ⊆ A$ and $vol(R) > 0$?










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      1





      $begingroup$


      If $A ⊆ ℝ^n$ , we say that A is a set of zero content if for every $ε > 0$ there are compact rectangles $R_1,…,R_m$ such that $A ⊆ R_1∪…∪R_m$ and $vol(R_1)+…+vol(R_m) < ε$ , with $vol([a_1,b_1]×…×[a_n,b_n]) = (b_1-a_1)⋅…⋅(b_n-a_n)$.



      Is it true that if $A$ is not of zero content, then there is a compact rectangle $R ⊆ ℝ^n$ such that $R ⊆ A$ and $vol(R) > 0$?










      share|cite|improve this question









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      If $A ⊆ ℝ^n$ , we say that A is a set of zero content if for every $ε > 0$ there are compact rectangles $R_1,…,R_m$ such that $A ⊆ R_1∪…∪R_m$ and $vol(R_1)+…+vol(R_m) < ε$ , with $vol([a_1,b_1]×…×[a_n,b_n]) = (b_1-a_1)⋅…⋅(b_n-a_n)$.



      Is it true that if $A$ is not of zero content, then there is a compact rectangle $R ⊆ ℝ^n$ such that $R ⊆ A$ and $vol(R) > 0$?







      real-analysis measure-theory






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      asked Jan 19 at 5:54









      sawesawe

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          $begingroup$

          As a counterexample, take $n=1$ and $A = [0,1]setminusmathbb{Q}$ -- the irrationals in the unit interval. Since $A$ has nonzero measure it has nonzero content since a set of zero content must be of zero measure.



          However, $A$ contains no non-degenerate compact interval, since the rationals are dense.



          Note though that $A$ is not "Jordan measurable" in the sense of zero-measure boundary. Every point in $[0,1]$ is a boundary point of $A$.






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            1 Answer
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            $begingroup$

            As a counterexample, take $n=1$ and $A = [0,1]setminusmathbb{Q}$ -- the irrationals in the unit interval. Since $A$ has nonzero measure it has nonzero content since a set of zero content must be of zero measure.



            However, $A$ contains no non-degenerate compact interval, since the rationals are dense.



            Note though that $A$ is not "Jordan measurable" in the sense of zero-measure boundary. Every point in $[0,1]$ is a boundary point of $A$.






            share|cite|improve this answer











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              0












              $begingroup$

              As a counterexample, take $n=1$ and $A = [0,1]setminusmathbb{Q}$ -- the irrationals in the unit interval. Since $A$ has nonzero measure it has nonzero content since a set of zero content must be of zero measure.



              However, $A$ contains no non-degenerate compact interval, since the rationals are dense.



              Note though that $A$ is not "Jordan measurable" in the sense of zero-measure boundary. Every point in $[0,1]$ is a boundary point of $A$.






              share|cite|improve this answer











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                0





                $begingroup$

                As a counterexample, take $n=1$ and $A = [0,1]setminusmathbb{Q}$ -- the irrationals in the unit interval. Since $A$ has nonzero measure it has nonzero content since a set of zero content must be of zero measure.



                However, $A$ contains no non-degenerate compact interval, since the rationals are dense.



                Note though that $A$ is not "Jordan measurable" in the sense of zero-measure boundary. Every point in $[0,1]$ is a boundary point of $A$.






                share|cite|improve this answer











                $endgroup$



                As a counterexample, take $n=1$ and $A = [0,1]setminusmathbb{Q}$ -- the irrationals in the unit interval. Since $A$ has nonzero measure it has nonzero content since a set of zero content must be of zero measure.



                However, $A$ contains no non-degenerate compact interval, since the rationals are dense.



                Note though that $A$ is not "Jordan measurable" in the sense of zero-measure boundary. Every point in $[0,1]$ is a boundary point of $A$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 19 at 9:46

























                answered Jan 19 at 9:07









                RRLRRL

                53.9k52675




                53.9k52675






























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