Regularity of Lebesgue measure












3












$begingroup$


Let $Asubseteq mathbb{R}$ be a Lebesgue measurable set of measure $m(A)=p>0$. Then for all $0<q<p$, show that there is a subset $Bsubseteq A$ with $m(B)=q$.



Which Theorem do i have to use here, regularity or density of Lebesgue measure or something else ?






measure-theory lebesgue-measure







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$endgroup$








  • 1




    $begingroup$
    Look at this
    $endgroup$
    – leo
    Nov 13 '13 at 16:40
















3












$begingroup$


Let $Asubseteq mathbb{R}$ be a Lebesgue measurable set of measure $m(A)=p>0$. Then for all $0<q<p$, show that there is a subset $Bsubseteq A$ with $m(B)=q$.



Which Theorem do i have to use here, regularity or density of Lebesgue measure or something else ?






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Look at this
    $endgroup$
    – leo
    Nov 13 '13 at 16:40














3












3








3





$begingroup$


Let $Asubseteq mathbb{R}$ be a Lebesgue measurable set of measure $m(A)=p>0$. Then for all $0<q<p$, show that there is a subset $Bsubseteq A$ with $m(B)=q$.



Which Theorem do i have to use here, regularity or density of Lebesgue measure or something else ?






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




Let $Asubseteq mathbb{R}$ be a Lebesgue measurable set of measure $m(A)=p>0$. Then for all $0<q<p$, show that there is a subset $Bsubseteq A$ with $m(B)=q$.



Which Theorem do i have to use here, regularity or density of Lebesgue measure or something else ?






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 Nov 13 '13 at 18:08







derivative

















asked Nov 13 '13 at 15:44









derivativederivative

1,140924




1,140924








  • 1




    $begingroup$
    Look at this
    $endgroup$
    – leo
    Nov 13 '13 at 16:40














  • 1




    $begingroup$
    Look at this
    $endgroup$
    – leo
    Nov 13 '13 at 16:40








1




1




$begingroup$
Look at this
$endgroup$
– leo
Nov 13 '13 at 16:40




$begingroup$
Look at this
$endgroup$
– leo
Nov 13 '13 at 16:40










1 Answer
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$begingroup$

Consider $f(x) = int_A 1_{(-infty, x]}$. Then $lim_{x to -infty} f(x) = 0$, $lim_{x to +infty} f(x) = p$ and $f$ is continuous.



Then, use the intermediate value theorem.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
    $endgroup$
    – derivative
    Nov 15 '13 at 14:01








  • 2




    $begingroup$
    No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
    $endgroup$
    – copper.hat
    Nov 15 '13 at 16:32











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6












$begingroup$

Consider $f(x) = int_A 1_{(-infty, x]}$. Then $lim_{x to -infty} f(x) = 0$, $lim_{x to +infty} f(x) = p$ and $f$ is continuous.



Then, use the intermediate value theorem.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
    $endgroup$
    – derivative
    Nov 15 '13 at 14:01








  • 2




    $begingroup$
    No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
    $endgroup$
    – copper.hat
    Nov 15 '13 at 16:32
















6












$begingroup$

Consider $f(x) = int_A 1_{(-infty, x]}$. Then $lim_{x to -infty} f(x) = 0$, $lim_{x to +infty} f(x) = p$ and $f$ is continuous.



Then, use the intermediate value theorem.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
    $endgroup$
    – derivative
    Nov 15 '13 at 14:01








  • 2




    $begingroup$
    No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
    $endgroup$
    – copper.hat
    Nov 15 '13 at 16:32














6












6








6





$begingroup$

Consider $f(x) = int_A 1_{(-infty, x]}$. Then $lim_{x to -infty} f(x) = 0$, $lim_{x to +infty} f(x) = p$ and $f$ is continuous.



Then, use the intermediate value theorem.






share|cite|improve this answer











$endgroup$



Consider $f(x) = int_A 1_{(-infty, x]}$. Then $lim_{x to -infty} f(x) = 0$, $lim_{x to +infty} f(x) = p$ and $f$ is continuous.



Then, use the intermediate value theorem.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 16:07









Viktor Glombik

8211527




8211527










answered Nov 13 '13 at 15:54









copper.hatcopper.hat

127k559160




127k559160












  • $begingroup$
    Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
    $endgroup$
    – derivative
    Nov 15 '13 at 14:01








  • 2




    $begingroup$
    No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
    $endgroup$
    – copper.hat
    Nov 15 '13 at 16:32


















  • $begingroup$
    Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
    $endgroup$
    – derivative
    Nov 15 '13 at 14:01








  • 2




    $begingroup$
    No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
    $endgroup$
    – copper.hat
    Nov 15 '13 at 16:32
















$begingroup$
Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
$endgroup$
– derivative
Nov 15 '13 at 14:01






$begingroup$
Why is f continuous ? I think we can't prove it with $epsilon-delta$ criterion ( $m(A)<deltaRightarrowint_{A} mathbb{1}_{(-infty,x]}<epsilon $) because this is precisely the exercise, to show that such a $delta$ exists.
$endgroup$
– derivative
Nov 15 '13 at 14:01






2




2




$begingroup$
No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
$endgroup$
– copper.hat
Nov 15 '13 at 16:32




$begingroup$
No, this is not precisely the exercise. Suppose $x>y$, then $|f(x)-f(y)| = int_A 1_{(y,x]} le int 1_{(y,x]} = |x-y|$. Similarly for $x>y$. Hence $f$ is Lipschitz continuous with rank 1.
$endgroup$
– copper.hat
Nov 15 '13 at 16:32


















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