Question about Jordan Measure
$begingroup$
I'm using the term "content" for Jordan measure and the term "measure" for Lebesgue measure.
The definition of content that I was given is:
A bounded set $D⊆ℝ^n$ has content if $x ↦ 1$ is Riemann integrable on $D$ and in that case we say that $vol(D) = ∫_D1$ is the content of $D$.
I know (without proof) that $D$ has content if and only if $D$ is bounded and $∂D$ has measure zero.
Now if $∂D$ has content zero, then $D$ has content (because a zero content set is also a zero measure set). However it should be false the if $D$ has content then $∂D$ has content zero. So the following proof of this statement must contain a mistake. Can someone help me find it?
If $D$ has content then there is an interval $I⊇D$ such that $vol(D) = ∫_Dχ_D$ exists (and we can assume $ClD ⊆ IntI$). This means that for each $ε > 0$ there is a partition $P$ of $I$ such that $U(χ_D,P) - L(χ_D,P) < ε$. I'm denoting with $J$ the generic subinterval of $I$ determined by $P$. Let $C$ be the set of those $J$s which contain both points of $D$ and points of $I-D$, and C' the set of those $J$s which aren't in C. If $x ∈ ∂D$ then $x ∈ IntI$, so we can find a $J$ containing both $x$ and a point of $D$ and a $J$ containing both $x$ and a point of $I-D$. It follows that $∂D ⊆ ∪C$. But
$$ ε > U(χ_D,P) - L(χ_D,P) = ∑_J osc(χ_D,J)vol(J) = ∑_{J∈C} osc(χ_D,J)vol(J) + ∑_{J∈C'} osc(χ_D,J)vol(J) = ∑_{J∈C} vol(J).$$
real-analysis measure-theory
asked Jan 21 at 8:08
sawesawe
345
$endgroup$
add a comment |
$begingroup$
I'm using the term "content" for Jordan measure and the term "measure" for Lebesgue measure.
The definition of content that I was given is:
A bounded set $D⊆ℝ^n$ has content if $x ↦ 1$ is Riemann integrable on $D$ and in that case we say that $vol(D) = ∫_D1$ is the content of $D$.
I know (without proof) that $D$ has content if and only if $D$ is bounded and $∂D$ has measure zero.
Now if $∂D$ has content zero, then $D$ has content (because a zero content set is also a zero measure set). However it should be false the if $D$ has content then $∂D$ has content zero. So the following proof of this statement must contain a mistake. Can someone help me find it?
If $D$ has content then there is an interval $I⊇D$ such that $vol(D) = ∫_Dχ_D$ exists (and we can assume $ClD ⊆ IntI$). This means that for each $ε > 0$ there is a partition $P$ of $I$ such that $U(χ_D,P) - L(χ_D,P) < ε$. I'm denoting with $J$ the generic subinterval of $I$ determined by $P$. Let $C$ be the set of those $J$s which contain both points of $D$ and points of $I-D$, and C' the set of those $J$s which aren't in C. If $x ∈ ∂D$ then $x ∈ IntI$, so we can find a $J$ containing both $x$ and a point of $D$ and a $J$ containing both $x$ and a point of $I-D$. It follows that $∂D ⊆ ∪C$. But
$$ ε > U(χ_D,P) - L(χ_D,P) = ∑_J osc(χ_D,J)vol(J) = ∑_{J∈C} osc(χ_D,J)vol(J) + ∑_{J∈C'} osc(χ_D,J)vol(J) = ∑_{J∈C} vol(J).$$
real-analysis measure-theory
asked Jan 21 at 8:08
sawesawe
345
$endgroup$
$begingroup$
I believe it is correct since I show below in a different way that the boundary must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:25
1
$begingroup$
In general a measure zero set is not necessarily of content zero. However, a compact measure zero set must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:31
add a comment |
$begingroup$
I'm using the term "content" for Jordan measure and the term "measure" for Lebesgue measure.
The definition of content that I was given is:
A bounded set $D⊆ℝ^n$ has content if $x ↦ 1$ is Riemann integrable on $D$ and in that case we say that $vol(D) = ∫_D1$ is the content of $D$.
I know (without proof) that $D$ has content if and only if $D$ is bounded and $∂D$ has measure zero.
Now if $∂D$ has content zero, then $D$ has content (because a zero content set is also a zero measure set). However it should be false the if $D$ has content then $∂D$ has content zero. So the following proof of this statement must contain a mistake. Can someone help me find it?
If $D$ has content then there is an interval $I⊇D$ such that $vol(D) = ∫_Dχ_D$ exists (and we can assume $ClD ⊆ IntI$). This means that for each $ε > 0$ there is a partition $P$ of $I$ such that $U(χ_D,P) - L(χ_D,P) < ε$. I'm denoting with $J$ the generic subinterval of $I$ determined by $P$. Let $C$ be the set of those $J$s which contain both points of $D$ and points of $I-D$, and C' the set of those $J$s which aren't in C. If $x ∈ ∂D$ then $x ∈ IntI$, so we can find a $J$ containing both $x$ and a point of $D$ and a $J$ containing both $x$ and a point of $I-D$. It follows that $∂D ⊆ ∪C$. But
$$ ε > U(χ_D,P) - L(χ_D,P) = ∑_J osc(χ_D,J)vol(J) = ∑_{J∈C} osc(χ_D,J)vol(J) + ∑_{J∈C'} osc(χ_D,J)vol(J) = ∑_{J∈C} vol(J).$$
real-analysis measure-theory
asked Jan 21 at 8:08
sawesawe
345
$endgroup$
I'm using the term "content" for Jordan measure and the term "measure" for Lebesgue measure.
The definition of content that I was given is:
A bounded set $D⊆ℝ^n$ has content if $x ↦ 1$ is Riemann integrable on $D$ and in that case we say that $vol(D) = ∫_D1$ is the content of $D$.
I know (without proof) that $D$ has content if and only if $D$ is bounded and $∂D$ has measure zero.
Now if $∂D$ has content zero, then $D$ has content (because a zero content set is also a zero measure set). However it should be false the if $D$ has content then $∂D$ has content zero. So the following proof of this statement must contain a mistake. Can someone help me find it?
If $D$ has content then there is an interval $I⊇D$ such that $vol(D) = ∫_Dχ_D$ exists (and we can assume $ClD ⊆ IntI$). This means that for each $ε > 0$ there is a partition $P$ of $I$ such that $U(χ_D,P) - L(χ_D,P) < ε$. I'm denoting with $J$ the generic subinterval of $I$ determined by $P$. Let $C$ be the set of those $J$s which contain both points of $D$ and points of $I-D$, and C' the set of those $J$s which aren't in C. If $x ∈ ∂D$ then $x ∈ IntI$, so we can find a $J$ containing both $x$ and a point of $D$ and a $J$ containing both $x$ and a point of $I-D$. It follows that $∂D ⊆ ∪C$. But
$$ ε > U(χ_D,P) - L(χ_D,P) = ∑_J osc(χ_D,J)vol(J) = ∑_{J∈C} osc(χ_D,J)vol(J) + ∑_{J∈C'} osc(χ_D,J)vol(J) = ∑_{J∈C} vol(J).$$
real-analysis measure-theory
real-analysis measure-theory
asked Jan 21 at 8:08
sawesawe
345
asked Jan 21 at 8:08
sawesawe
345
asked Jan 21 at 8:08
sawesawe
345
asked Jan 21 at 8:08
sawesawe
345
asked Jan 21 at 8:08
sawesawe
345
345
$begingroup$
I believe it is correct since I show below in a different way that the boundary must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:25
1
$begingroup$
In general a measure zero set is not necessarily of content zero. However, a compact measure zero set must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:31
add a comment |
$begingroup$
I believe it is correct since I show below in a different way that the boundary must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:25
1
$begingroup$
In general a measure zero set is not necessarily of content zero. However, a compact measure zero set must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:31
$begingroup$
I believe it is correct since I show below in a different way that the boundary must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:25
$begingroup$
I believe it is correct since I show below in a different way that the boundary must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:25
1
1
$begingroup$
In general a measure zero set is not necessarily of content zero. However, a compact measure zero set must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:31
$begingroup$
In general a measure zero set is not necessarily of content zero. However, a compact measure zero set must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:31
add a comment |
1 Answer
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Actually if $D subset I$ is bounded and $vol(D) = int_Ichi_D$ exists then the boundary $partial D$ has content zero.
The closure $bar{D} = D cup partial D$ is closed and bounded and, hence, compact. Let $mathcal{O} = {O_n}$ be an open cover of $partial D.$ Then $mathcal{O'}=mathcal{O} cup {int(D)}$ is an open cover of $bar{D}$. Since $bar{D}$ is compact there exists a finite subcollection $mathcal{G} ={G_1,G_2,ldots, G_n} subset mathcal{O'}$ which covers $bar{D}$ as well as $partial D subset bar{D}$. If $int(D) notin mathcal{G}$ then we have found an open subcover $mathcal{G} subset mathcal{O}$ of $partial D$. If $int(D) in mathcal{G}$ then one of the sets $G_j = int(D)$ and ${G_k : k neq j} subset mathcal{O}$ is an open subcover of $partial D$. Therefore, $partial D$ is compact.
Since $chi_D$ is integrable over $I$ and discontinuous on $partial D$, we must have $partial D$ of measure zero. Since $partial D$ is compact it must also be of content zero.
answered Jan 21 at 9:24
RRLRRL
51.7k42573
$endgroup$
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
add a comment |
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$begingroup$
Actually if $D subset I$ is bounded and $vol(D) = int_Ichi_D$ exists then the boundary $partial D$ has content zero.
The closure $bar{D} = D cup partial D$ is closed and bounded and, hence, compact. Let $mathcal{O} = {O_n}$ be an open cover of $partial D.$ Then $mathcal{O'}=mathcal{O} cup {int(D)}$ is an open cover of $bar{D}$. Since $bar{D}$ is compact there exists a finite subcollection $mathcal{G} ={G_1,G_2,ldots, G_n} subset mathcal{O'}$ which covers $bar{D}$ as well as $partial D subset bar{D}$. If $int(D) notin mathcal{G}$ then we have found an open subcover $mathcal{G} subset mathcal{O}$ of $partial D$. If $int(D) in mathcal{G}$ then one of the sets $G_j = int(D)$ and ${G_k : k neq j} subset mathcal{O}$ is an open subcover of $partial D$. Therefore, $partial D$ is compact.
Since $chi_D$ is integrable over $I$ and discontinuous on $partial D$, we must have $partial D$ of measure zero. Since $partial D$ is compact it must also be of content zero.
answered Jan 21 at 9:24
RRLRRL
51.7k42573
$endgroup$
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
add a comment |
$begingroup$
Actually if $D subset I$ is bounded and $vol(D) = int_Ichi_D$ exists then the boundary $partial D$ has content zero.
The closure $bar{D} = D cup partial D$ is closed and bounded and, hence, compact. Let $mathcal{O} = {O_n}$ be an open cover of $partial D.$ Then $mathcal{O'}=mathcal{O} cup {int(D)}$ is an open cover of $bar{D}$. Since $bar{D}$ is compact there exists a finite subcollection $mathcal{G} ={G_1,G_2,ldots, G_n} subset mathcal{O'}$ which covers $bar{D}$ as well as $partial D subset bar{D}$. If $int(D) notin mathcal{G}$ then we have found an open subcover $mathcal{G} subset mathcal{O}$ of $partial D$. If $int(D) in mathcal{G}$ then one of the sets $G_j = int(D)$ and ${G_k : k neq j} subset mathcal{O}$ is an open subcover of $partial D$. Therefore, $partial D$ is compact.
Since $chi_D$ is integrable over $I$ and discontinuous on $partial D$, we must have $partial D$ of measure zero. Since $partial D$ is compact it must also be of content zero.
answered Jan 21 at 9:24
RRLRRL
51.7k42573
$endgroup$
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
add a comment |
$begingroup$
Actually if $D subset I$ is bounded and $vol(D) = int_Ichi_D$ exists then the boundary $partial D$ has content zero.
The closure $bar{D} = D cup partial D$ is closed and bounded and, hence, compact. Let $mathcal{O} = {O_n}$ be an open cover of $partial D.$ Then $mathcal{O'}=mathcal{O} cup {int(D)}$ is an open cover of $bar{D}$. Since $bar{D}$ is compact there exists a finite subcollection $mathcal{G} ={G_1,G_2,ldots, G_n} subset mathcal{O'}$ which covers $bar{D}$ as well as $partial D subset bar{D}$. If $int(D) notin mathcal{G}$ then we have found an open subcover $mathcal{G} subset mathcal{O}$ of $partial D$. If $int(D) in mathcal{G}$ then one of the sets $G_j = int(D)$ and ${G_k : k neq j} subset mathcal{O}$ is an open subcover of $partial D$. Therefore, $partial D$ is compact.
Since $chi_D$ is integrable over $I$ and discontinuous on $partial D$, we must have $partial D$ of measure zero. Since $partial D$ is compact it must also be of content zero.
answered Jan 21 at 9:24
RRLRRL
51.7k42573
$endgroup$
Actually if $D subset I$ is bounded and $vol(D) = int_Ichi_D$ exists then the boundary $partial D$ has content zero.
The closure $bar{D} = D cup partial D$ is closed and bounded and, hence, compact. Let $mathcal{O} = {O_n}$ be an open cover of $partial D.$ Then $mathcal{O'}=mathcal{O} cup {int(D)}$ is an open cover of $bar{D}$. Since $bar{D}$ is compact there exists a finite subcollection $mathcal{G} ={G_1,G_2,ldots, G_n} subset mathcal{O'}$ which covers $bar{D}$ as well as $partial D subset bar{D}$. If $int(D) notin mathcal{G}$ then we have found an open subcover $mathcal{G} subset mathcal{O}$ of $partial D$. If $int(D) in mathcal{G}$ then one of the sets $G_j = int(D)$ and ${G_k : k neq j} subset mathcal{O}$ is an open subcover of $partial D$. Therefore, $partial D$ is compact.
Since $chi_D$ is integrable over $I$ and discontinuous on $partial D$, we must have $partial D$ of measure zero. Since $partial D$ is compact it must also be of content zero.
answered Jan 21 at 9:24
RRLRRL
51.7k42573
answered Jan 21 at 9:24
RRLRRL
51.7k42573
answered Jan 21 at 9:24
RRLRRL
51.7k42573
answered Jan 21 at 9:24
RRLRRL
51.7k42573
51.7k42573
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
add a comment |
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
$begingroup$
Nice and concise proof. +1
$endgroup$
– Paramanand Singh
Jan 21 at 11:39
add a comment |
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$begingroup$
I believe it is correct since I show below in a different way that the boundary must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:25
1
$begingroup$
In general a measure zero set is not necessarily of content zero. However, a compact measure zero set must be of content zero.
$endgroup$
– RRL
Jan 21 at 9:31