Bogachev change of variables theorem.












2














I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:



Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
$$
F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
$$
where $|r(x,y)|leq epsilon |x-y|$.



Here is where I am stucked:



Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
$$
lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
$$
where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.



Can someone give me some help?



Comment: I think the argument pass over to give some inferior bound for
$text{diam}( F(U))$ by using in some way the inequality
$|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question





























    2














    I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:



    Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
    $$
    F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
    $$
    where $|r(x,y)|leq epsilon |x-y|$.



    Here is where I am stucked:



    Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
    $$
    lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
    $$
    where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.



    Can someone give me some help?



    Comment: I think the argument pass over to give some inferior bound for
    $text{diam}( F(U))$ by using in some way the inequality
    $|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.






    measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question



























      2












      2








      2







      I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:



      Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
      $$
      F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
      $$
      where $|r(x,y)|leq epsilon |x-y|$.



      Here is where I am stucked:



      Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
      $$
      lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
      $$
      where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.



      Can someone give me some help?



      Comment: I think the argument pass over to give some inferior bound for
      $text{diam}( F(U))$ by using in some way the inequality
      $|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.






      measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question















      I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:



      Let $F:Usubset mathbb{R}^nto mathbb{R}^n$ an injective $C¹$ map. Given $epsilon>$ there exist $delta>0$ s.t $forall x,y in U,~|x-y|<delta$ implies
      $$
      F(x)-F(y)=F^{'}(x)(y-x)+r(x,y)
      $$
      where $|r(x,y)|leq epsilon |x-y|$.



      Here is where I am stucked:



      Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $delta.$ If $|det F'(x_0)|leq sqrt{epsilon}$ then we have that
      $$
      lambda (L(Q))-lambda(F(Q))leq sqrt{epsilon}lambda (Q)
      $$
      where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $lambda$ is the $n$-dimensional lebesgue measure.



      Can someone give me some help?



      Comment: I think the argument pass over to give some inferior bound for
      $text{diam}( F(U))$ by using in some way the inequality
      $|F(x)-F(y)|geq |F^{'}(x_0)(x-y)|-|r(x,y|$.






      measure-theory lebesgue-integral lebesgue-measure




      measure-theory lebesgue-integral lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 18 at 16:22

























      asked Nov 18 at 15:35









      Eduardo

      551112




      551112






















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          I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.






          share|cite|improve this answer





















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            I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.






            share|cite|improve this answer


























              1














              I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.






              share|cite|improve this answer
























                1












                1








                1






                I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.






                share|cite|improve this answer












                I don't know enough measure theory but since $lambda(L(Q))=det L cdot lambda(Q)$, you could try to prove that $det L le frac{lambda(F(Q))}{lambda(Q)} + |det F'(x_0)|$.







                share|cite|improve this answer












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                answered Nov 18 at 17:06









                lzralbu

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