Lebesque measure of a compact set and a line












2












$begingroup$


I am trying to work out the problem below. My only idea at the moment is that the compactness of $K$ gives me convergent subsequences. Any help would be appreciated.



Suppose that $K$ is a compact subset of the unit square $[0, 1]^2$
in $R^2$ and the two-dimensional Lebesgue measure of $K$ is greater than 1/2. Show that there are infinitely
many lines $L$ in $R^2$
such that $m_L(K ∩ L) > 1/2$, where $m_L$ is the one-dimensional
Lebesgue measure on the line $L$.






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you know Fubini?
    $endgroup$
    – zhw.
    Dec 20 '18 at 23:59
















2












$begingroup$


I am trying to work out the problem below. My only idea at the moment is that the compactness of $K$ gives me convergent subsequences. Any help would be appreciated.



Suppose that $K$ is a compact subset of the unit square $[0, 1]^2$
in $R^2$ and the two-dimensional Lebesgue measure of $K$ is greater than 1/2. Show that there are infinitely
many lines $L$ in $R^2$
such that $m_L(K ∩ L) > 1/2$, where $m_L$ is the one-dimensional
Lebesgue measure on the line $L$.






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you know Fubini?
    $endgroup$
    – zhw.
    Dec 20 '18 at 23:59














2












2








2





$begingroup$


I am trying to work out the problem below. My only idea at the moment is that the compactness of $K$ gives me convergent subsequences. Any help would be appreciated.



Suppose that $K$ is a compact subset of the unit square $[0, 1]^2$
in $R^2$ and the two-dimensional Lebesgue measure of $K$ is greater than 1/2. Show that there are infinitely
many lines $L$ in $R^2$
such that $m_L(K ∩ L) > 1/2$, where $m_L$ is the one-dimensional
Lebesgue measure on the line $L$.






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




I am trying to work out the problem below. My only idea at the moment is that the compactness of $K$ gives me convergent subsequences. Any help would be appreciated.



Suppose that $K$ is a compact subset of the unit square $[0, 1]^2$
in $R^2$ and the two-dimensional Lebesgue measure of $K$ is greater than 1/2. Show that there are infinitely
many lines $L$ in $R^2$
such that $m_L(K ∩ L) > 1/2$, where $m_L$ is the one-dimensional
Lebesgue measure on the line $L$.






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 20 '18 at 23:58







user628226

















asked Dec 20 '18 at 23:48









user628226user628226

575




575












  • $begingroup$
    do you know Fubini?
    $endgroup$
    – zhw.
    Dec 20 '18 at 23:59


















  • $begingroup$
    do you know Fubini?
    $endgroup$
    – zhw.
    Dec 20 '18 at 23:59
















$begingroup$
do you know Fubini?
$endgroup$
– zhw.
Dec 20 '18 at 23:59




$begingroup$
do you know Fubini?
$endgroup$
– zhw.
Dec 20 '18 at 23:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

Denote by $mu$ the Lebesgue measure over $[0,1]$. The 2-dimensional Lebesgue measure over $[0,1]^2$ is given by the product measure $lambda=mutimes mu$ and using Fubini we have:



$$lambda(K)=int 1_K(x,y) dlambda(x,y)=int_{0}^1left (int_0^1 1_K(x,y) dmu(x)right ) dmu(y)$$



Now if there are only finitely many lines $L$ such that $Lcap K$ has length bigger than $frac{1}{2}$ we would have $$int_0^1 1_K(x,y) dmu(x)leq frac{1}{2}$$



from where $$lambda(K)leqint_{0}^1frac{1}{2} dmu=frac{1}{2}$$
and this contradicts the hypothesis.






share|cite|improve this answer









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    $begingroup$

    Denote by $mu$ the Lebesgue measure over $[0,1]$. The 2-dimensional Lebesgue measure over $[0,1]^2$ is given by the product measure $lambda=mutimes mu$ and using Fubini we have:



    $$lambda(K)=int 1_K(x,y) dlambda(x,y)=int_{0}^1left (int_0^1 1_K(x,y) dmu(x)right ) dmu(y)$$



    Now if there are only finitely many lines $L$ such that $Lcap K$ has length bigger than $frac{1}{2}$ we would have $$int_0^1 1_K(x,y) dmu(x)leq frac{1}{2}$$



    from where $$lambda(K)leqint_{0}^1frac{1}{2} dmu=frac{1}{2}$$
    and this contradicts the hypothesis.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Denote by $mu$ the Lebesgue measure over $[0,1]$. The 2-dimensional Lebesgue measure over $[0,1]^2$ is given by the product measure $lambda=mutimes mu$ and using Fubini we have:



      $$lambda(K)=int 1_K(x,y) dlambda(x,y)=int_{0}^1left (int_0^1 1_K(x,y) dmu(x)right ) dmu(y)$$



      Now if there are only finitely many lines $L$ such that $Lcap K$ has length bigger than $frac{1}{2}$ we would have $$int_0^1 1_K(x,y) dmu(x)leq frac{1}{2}$$



      from where $$lambda(K)leqint_{0}^1frac{1}{2} dmu=frac{1}{2}$$
      and this contradicts the hypothesis.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Denote by $mu$ the Lebesgue measure over $[0,1]$. The 2-dimensional Lebesgue measure over $[0,1]^2$ is given by the product measure $lambda=mutimes mu$ and using Fubini we have:



        $$lambda(K)=int 1_K(x,y) dlambda(x,y)=int_{0}^1left (int_0^1 1_K(x,y) dmu(x)right ) dmu(y)$$



        Now if there are only finitely many lines $L$ such that $Lcap K$ has length bigger than $frac{1}{2}$ we would have $$int_0^1 1_K(x,y) dmu(x)leq frac{1}{2}$$



        from where $$lambda(K)leqint_{0}^1frac{1}{2} dmu=frac{1}{2}$$
        and this contradicts the hypothesis.






        share|cite|improve this answer









        $endgroup$



        Denote by $mu$ the Lebesgue measure over $[0,1]$. The 2-dimensional Lebesgue measure over $[0,1]^2$ is given by the product measure $lambda=mutimes mu$ and using Fubini we have:



        $$lambda(K)=int 1_K(x,y) dlambda(x,y)=int_{0}^1left (int_0^1 1_K(x,y) dmu(x)right ) dmu(y)$$



        Now if there are only finitely many lines $L$ such that $Lcap K$ has length bigger than $frac{1}{2}$ we would have $$int_0^1 1_K(x,y) dmu(x)leq frac{1}{2}$$



        from where $$lambda(K)leqint_{0}^1frac{1}{2} dmu=frac{1}{2}$$
        and this contradicts the hypothesis.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 0:04









        yamete kudasaiyamete kudasai

        1,155818




        1,155818






























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