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 $vol(R) > 0$?
real-analysis measure-theory
asked Jan 19 at 5:54
sawesawe
345
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add a comment |
$begingroup$
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 $vol(R) > 0$?
real-analysis measure-theory
asked Jan 19 at 5:54
sawesawe
345
$endgroup$
add a comment |
$begingroup$
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 $vol(R) > 0$?
real-analysis measure-theory
asked Jan 19 at 5:54
sawesawe
345
$endgroup$
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
real-analysis measure-theory
asked Jan 19 at 5:54
sawesawe
345
asked Jan 19 at 5:54
sawesawe
345
asked Jan 19 at 5:54
sawesawe
345
asked Jan 19 at 5:54
sawesawe
345
asked Jan 19 at 5:54
sawesawe
345
345
add a comment |
add a comment |
1 Answer
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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 $A$.
answered Jan 19 at 9:07
RRLRRL
51.3k42573
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1 Answer
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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 $A$.
answered Jan 19 at 9:07
RRLRRL
51.3k42573
$endgroup$
add a comment |
$begingroup$
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 $A$.
answered Jan 19 at 9:07
RRLRRL
51.3k42573
$endgroup$
add a comment |
$begingroup$
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 $A$.
answered Jan 19 at 9:07
RRLRRL
51.3k42573
$endgroup$
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 $A$.
answered Jan 19 at 9:07
RRLRRL
51.3k42573
edited Jan 19 at 9:46
answered Jan 19 at 9:07
RRLRRL
51.3k42573
answered Jan 19 at 9:07
RRLRRL
51.3k42573
answered Jan 19 at 9:07
RRLRRL
51.3k42573
51.3k42573
add a comment |
add a comment |
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