Mixing
Regularity of Lebesgue measure
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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
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
$endgroup$
add a comment |
$begingroup$
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
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
$endgroup$
1
$begingroup$
Look at this
$endgroup$
– leo
Nov 13 '13 at 16:40
add a comment |
$begingroup$
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
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
$endgroup$
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
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
edited Nov 13 '13 at 18:08
derivative
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
asked Nov 13 '13 at 15:44
derivativederivative
1,140924
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
add a comment |
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
add a comment |
1 Answer
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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.
edited Jan 10 at 16:07
Viktor Glombik
8211527
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
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
add a comment |
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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.
edited Jan 10 at 16:07
Viktor Glombik
8211527
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
edited Jan 10 at 16:07
Viktor Glombik
8211527
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
edited Jan 10 at 16:07
Viktor Glombik
8211527
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
$endgroup$
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.
edited Jan 10 at 16:07
Viktor Glombik
8211527
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
edited Jan 10 at 16:07
Viktor Glombik
8211527
edited Jan 10 at 16:07
Viktor Glombik
8211527
edited Jan 10 at 16:07
Viktor Glombik
8211527
8211527
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
answered Nov 13 '13 at 15:54
copper.hatcopper.hat
127k559160
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
add a comment |
$begingroup$
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
add a comment |
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Look at this
$endgroup$
– leo
Nov 13 '13 at 16:40