Integrability of log of distance functionRelationship between Erdos and Falconer distance problemsAn infimum of integrals of a positive function.Compact Riemannian manifold with maximum average distanceMultivariate monotonic functionExotic Lebesgue Measurable FunctionTranscendental distance setsIntegral form of maximal function estimate on variable exponent spacesDerivative of distance function to a closed, rectifiable set
Integrability of log of distance function
Relationship between Erdos and Falconer distance problemsAn infimum of integrals of a positive function.Compact Riemannian manifold with maximum average distanceMultivariate monotonic functionExotic Lebesgue Measurable FunctionTranscendental distance setsIntegral form of maximal function estimate on variable exponent spacesDerivative of distance function to a closed, rectifiable set
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Let $Esubset B_1(0)subset mathbbR^n$ be a compact set s.t. $lambda(E)=0$, where $lambda$ is the Lebesgue measure, and $B_1(0)$ is the Euclidean unit ball centered at the origin. Is the following integral finite:
$$int_B_1(0)-log d(x,E)dlambda(x)<infty?$$
Although this question seems trivial, I have failed to find a reference to it or to variations of it in previous discussions. I was not able to come up with a counter-example nor a proof. I also asked in mathstackexchange a variation of it, but didn’t get a sufficient answer.
Thanks ahead
geometric-measure-theory lebesgue-measure
asked 11 hours ago
BOSBOS
1817 bronze badges
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Let $Esubset B_1(0)subset mathbbR^n$ be a compact set s.t. $lambda(E)=0$, where $lambda$ is the Lebesgue measure, and $B_1(0)$ is the Euclidean unit ball centered at the origin. Is the following integral finite:
$$int_B_1(0)-log d(x,E)dlambda(x)<infty?$$
Although this question seems trivial, I have failed to find a reference to it or to variations of it in previous discussions. I was not able to come up with a counter-example nor a proof. I also asked in mathstackexchange a variation of it, but didn’t get a sufficient answer.
Thanks ahead
geometric-measure-theory lebesgue-measure
asked 11 hours ago
BOSBOS
1817 bronze badges
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$begingroup$
Correct, thank you.
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– BOS
10 hours ago
add a comment
|
$begingroup$
Let $Esubset B_1(0)subset mathbbR^n$ be a compact set s.t. $lambda(E)=0$, where $lambda$ is the Lebesgue measure, and $B_1(0)$ is the Euclidean unit ball centered at the origin. Is the following integral finite:
$$int_B_1(0)-log d(x,E)dlambda(x)<infty?$$
Although this question seems trivial, I have failed to find a reference to it or to variations of it in previous discussions. I was not able to come up with a counter-example nor a proof. I also asked in mathstackexchange a variation of it, but didn’t get a sufficient answer.
Thanks ahead
geometric-measure-theory lebesgue-measure
asked 11 hours ago
BOSBOS
1817 bronze badges
$endgroup$
Let $Esubset B_1(0)subset mathbbR^n$ be a compact set s.t. $lambda(E)=0$, where $lambda$ is the Lebesgue measure, and $B_1(0)$ is the Euclidean unit ball centered at the origin. Is the following integral finite:
$$int_B_1(0)-log d(x,E)dlambda(x)<infty?$$
Although this question seems trivial, I have failed to find a reference to it or to variations of it in previous discussions. I was not able to come up with a counter-example nor a proof. I also asked in mathstackexchange a variation of it, but didn’t get a sufficient answer.
Thanks ahead
geometric-measure-theory lebesgue-measure
geometric-measure-theory lebesgue-measure
asked 11 hours ago
BOSBOS
1817 bronze badges
asked 11 hours ago
BOSBOS
1817 bronze badges
edited 10 hours ago
BOS
asked 11 hours ago
BOSBOS
1817 bronze badges
asked 11 hours ago
BOSBOS
1817 bronze badges
asked 11 hours ago
BOSBOS
1817 bronze badges
1817 bronze badges
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Correct, thank you.
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– BOS
10 hours ago
add a comment
|
$begingroup$
Correct, thank you.
$endgroup$
– BOS
10 hours ago
$begingroup$
Correct, thank you.
$endgroup$
– BOS
10 hours ago
$begingroup$
Correct, thank you.
$endgroup$
– BOS
10 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
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The integral in question is finite for most sets of measure zero, but can diverge to $infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^k-1$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^-k/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^-k/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
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Something like this is what I thought of a couple minutes ago. :-)
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– Iosif Pinelis
9 hours ago
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@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
add a comment
|
$begingroup$
If $Eneemptyset$, then $d(x,E)le2$ for all $xin B_1(0)$. So, your integral is $lelambda(B_1(0))ln2<infty$.
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
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1
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
1
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integral in question is finite for most sets of measure zero, but can diverge to $infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^k-1$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^-k/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^-k/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
$endgroup$
$begingroup$
Something like this is what I thought of a couple minutes ago. :-)
$endgroup$
– Iosif Pinelis
9 hours ago
$begingroup$
@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
add a comment
|
$begingroup$
The integral in question is finite for most sets of measure zero, but can diverge to $infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^k-1$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^-k/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^-k/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
$endgroup$
$begingroup$
Something like this is what I thought of a couple minutes ago. :-)
$endgroup$
– Iosif Pinelis
9 hours ago
$begingroup$
@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
add a comment
|
$begingroup$
The integral in question is finite for most sets of measure zero, but can diverge to $infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^k-1$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^-k/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^-k/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
$endgroup$
The integral in question is finite for most sets of measure zero, but can diverge to $infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^k-1$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^-k/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^-k/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
answered 9 hours ago
Yuval PeresYuval Peres
2,90212 silver badges14 bronze badges
2,90212 silver badges14 bronze badges
$begingroup$
Something like this is what I thought of a couple minutes ago. :-)
$endgroup$
– Iosif Pinelis
9 hours ago
$begingroup$
@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
add a comment
|
$begingroup$
Something like this is what I thought of a couple minutes ago. :-)
$endgroup$
– Iosif Pinelis
9 hours ago
$begingroup$
@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
$begingroup$
Something like this is what I thought of a couple minutes ago. :-)
$endgroup$
– Iosif Pinelis
9 hours ago
$begingroup$
Something like this is what I thought of a couple minutes ago. :-)
$endgroup$
– Iosif Pinelis
9 hours ago
$begingroup$
@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
$begingroup$
@YuvalPeres thank you for this very nice construction.
$endgroup$
– BOS
6 hours ago
add a comment
|
$begingroup$
If $Eneemptyset$, then $d(x,E)le2$ for all $xin B_1(0)$. So, your integral is $lelambda(B_1(0))ln2<infty$.
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
$endgroup$
1
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
1
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
add a comment
|
$begingroup$
If $Eneemptyset$, then $d(x,E)le2$ for all $xin B_1(0)$. So, your integral is $lelambda(B_1(0))ln2<infty$.
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
$endgroup$
1
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
1
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
add a comment
|
$begingroup$
If $Eneemptyset$, then $d(x,E)le2$ for all $xin B_1(0)$. So, your integral is $lelambda(B_1(0))ln2<infty$.
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
$endgroup$
If $Eneemptyset$, then $d(x,E)le2$ for all $xin B_1(0)$. So, your integral is $lelambda(B_1(0))ln2<infty$.
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
answered 10 hours ago
Iosif PinelisIosif Pinelis
25.6k3 gold badges30 silver badges69 bronze badges
25.6k3 gold badges30 silver badges69 bronze badges
1
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
1
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
add a comment
|
1
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
1
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
1
1
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
You answered the question somehow, but I think the problem occurs where $d(x,E)$ is small, so integrability is still an issue (i.e. the problem is if the integral exists since it may be $-infty$ )…
$endgroup$
– Dirk
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
@Dirk : This is what I am thinking about now. :-)
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
$begingroup$
The integral is $>-infty$ if $E$ is the Cantor set. Are there compact sets of zero Lebesgue measure that are much bigger than the Cantor set?
$endgroup$
– Iosif Pinelis
10 hours ago
1
1
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
$begingroup$
@Iosif Yes, there are.
$endgroup$
– Yuval Peres
9 hours ago
add a comment
|
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$begingroup$
Correct, thank you.
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– BOS
10 hours ago