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The $m$-dimensional Hausdorff measure agrees with the Lebesgue measure on $mathbbR^n$ when $m=n$?
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Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $A subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $A$ is defined as:
$$
H^m(A) = lim_delta to 0 inf bigg A subset bigcup_j S_j, textdiam(S_j) < delta bigg quad quad (*)
$$
I'm trying to understand this definition and how the various components of it are motivated.
Suppose $n=m=3$ and that $A$ is the the closed unit ball in $mathbbR^3$. From what I've read if $m=n$ the Hausdorff measure agrees with the Lebesgue measure so $(*)$ should evaluate to simply $alpha_m = alpha_n = alpha_3 = frac43pi$. How can this evaluation be performed starting from the definition of the Hausdorff measure in $(*)$?
measure-theory lebesgue-measure hausdorff-measure
asked Jul 31 at 15:04
csss
1,22811221
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show 10 more comments
up vote
1
down vote
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Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $A subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $A$ is defined as:
$$
H^m(A) = lim_delta to 0 inf bigg A subset bigcup_j S_j, textdiam(S_j) < delta bigg quad quad (*)
$$
I'm trying to understand this definition and how the various components of it are motivated.
Suppose $n=m=3$ and that $A$ is the the closed unit ball in $mathbbR^3$. From what I've read if $m=n$ the Hausdorff measure agrees with the Lebesgue measure so $(*)$ should evaluate to simply $alpha_m = alpha_n = alpha_3 = frac43pi$. How can this evaluation be performed starting from the definition of the Hausdorff measure in $(*)$?
measure-theory lebesgue-measure hausdorff-measure
asked Jul 31 at 15:04
csss
1,22811221
1
In the case $m=n$, once $delta$ is large enough you can simply attain the infimum by setting $S_1=A,S_j=emptyset$ for $j>1$, at which point the sum simply reduces to $alpha_m (mathrmdiam(A)/2)^m$ which is of course the actual Lebesgue measure of the ball. This is not very insightful of course.
– Ian
Jul 31 at 15:10
1
There is an error in the definition. $delta$ should be taken to zero, not infinity. Note that this does have an effect on the validity of @Ian's approach.
– Xander Henderson
Jul 31 at 15:11
1
With @XanderHenderson's correction (thanks), the situation for $m=n$ is now the same as the usual construction of the Lebesgue measure: cover the ball by tiny balls and sum up their volumes, and then as the balls becomes smaller, reduce the overlap and reduce the amount of volume located outside the original ball. Again, this isn't very insightful to what this construction means for purposes of dealing with $m neq n$.
– Ian
Jul 31 at 15:14
1
As for the case $m neq n$, the idea is to cover $A$ by sets $S_j$ and treat the $m$-dimensional volume of the $S_j$ as being approximated by the volume of a $m$-dimensional ball with the same diameter as the $S_j$. So for example, you might cover a square in the $(x,y)$ plane ($m=2$) with rectangular prisms ($n=3$) which are very short in the $z$ direction and are equal in size in the $x$ and $y$ directions.
– Ian
Jul 31 at 15:18
1
(Cont.) Now $alpha_2 (mathrmdiam(S_j)/2)^2$ is the area of a disk whose diameter is the same as the diameter of the projection of these prisms into the plane. For small disks this area is close to the true area of the projection (which is to say the area of the corresponding square), so you are nearly summing up the area of the squares, which is exactly what you want to do.
– Ian
Jul 31 at 15:18
 |Â
show 10 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $A subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $A$ is defined as:
$$
H^m(A) = lim_delta to 0 inf bigg A subset bigcup_j S_j, textdiam(S_j) < delta bigg quad quad (*)
$$
I'm trying to understand this definition and how the various components of it are motivated.
Suppose $n=m=3$ and that $A$ is the the closed unit ball in $mathbbR^3$. From what I've read if $m=n$ the Hausdorff measure agrees with the Lebesgue measure so $(*)$ should evaluate to simply $alpha_m = alpha_n = alpha_3 = frac43pi$. How can this evaluation be performed starting from the definition of the Hausdorff measure in $(*)$?
measure-theory lebesgue-measure hausdorff-measure
asked Jul 31 at 15:04
csss
1,22811221
Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $A subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $A$ is defined as:
$$
H^m(A) = lim_delta to 0 inf bigg A subset bigcup_j S_j, textdiam(S_j) < delta bigg quad quad (*)
$$
I'm trying to understand this definition and how the various components of it are motivated.
Suppose $n=m=3$ and that $A$ is the the closed unit ball in $mathbbR^3$. From what I've read if $m=n$ the Hausdorff measure agrees with the Lebesgue measure so $(*)$ should evaluate to simply $alpha_m = alpha_n = alpha_3 = frac43pi$. How can this evaluation be performed starting from the definition of the Hausdorff measure in $(*)$?
measure-theory lebesgue-measure hausdorff-measure
asked Jul 31 at 15:04
csss
1,22811221
edited Jul 31 at 15:11
asked Jul 31 at 15:04
csss
1,22811221
asked Jul 31 at 15:04
csss
1,22811221
asked Jul 31 at 15:04
csss
1,22811221
1,22811221
1
In the case $m=n$, once $delta$ is large enough you can simply attain the infimum by setting $S_1=A,S_j=emptyset$ for $j>1$, at which point the sum simply reduces to $alpha_m (mathrmdiam(A)/2)^m$ which is of course the actual Lebesgue measure of the ball. This is not very insightful of course.
– Ian
Jul 31 at 15:10
1
There is an error in the definition. $delta$ should be taken to zero, not infinity. Note that this does have an effect on the validity of @Ian's approach.
– Xander Henderson
Jul 31 at 15:11
1
With @XanderHenderson's correction (thanks), the situation for $m=n$ is now the same as the usual construction of the Lebesgue measure: cover the ball by tiny balls and sum up their volumes, and then as the balls becomes smaller, reduce the overlap and reduce the amount of volume located outside the original ball. Again, this isn't very insightful to what this construction means for purposes of dealing with $m neq n$.
– Ian
Jul 31 at 15:14
1
As for the case $m neq n$, the idea is to cover $A$ by sets $S_j$ and treat the $m$-dimensional volume of the $S_j$ as being approximated by the volume of a $m$-dimensional ball with the same diameter as the $S_j$. So for example, you might cover a square in the $(x,y)$ plane ($m=2$) with rectangular prisms ($n=3$) which are very short in the $z$ direction and are equal in size in the $x$ and $y$ directions.
– Ian
Jul 31 at 15:18
1
(Cont.) Now $alpha_2 (mathrmdiam(S_j)/2)^2$ is the area of a disk whose diameter is the same as the diameter of the projection of these prisms into the plane. For small disks this area is close to the true area of the projection (which is to say the area of the corresponding square), so you are nearly summing up the area of the squares, which is exactly what you want to do.
– Ian
Jul 31 at 15:18
 |Â
show 10 more comments
1
In the case $m=n$, once $delta$ is large enough you can simply attain the infimum by setting $S_1=A,S_j=emptyset$ for $j>1$, at which point the sum simply reduces to $alpha_m (mathrmdiam(A)/2)^m$ which is of course the actual Lebesgue measure of the ball. This is not very insightful of course.
– Ian
Jul 31 at 15:10
1
There is an error in the definition. $delta$ should be taken to zero, not infinity. Note that this does have an effect on the validity of @Ian's approach.
– Xander Henderson
Jul 31 at 15:11
1
With @XanderHenderson's correction (thanks), the situation for $m=n$ is now the same as the usual construction of the Lebesgue measure: cover the ball by tiny balls and sum up their volumes, and then as the balls becomes smaller, reduce the overlap and reduce the amount of volume located outside the original ball. Again, this isn't very insightful to what this construction means for purposes of dealing with $m neq n$.
– Ian
Jul 31 at 15:14
1
As for the case $m neq n$, the idea is to cover $A$ by sets $S_j$ and treat the $m$-dimensional volume of the $S_j$ as being approximated by the volume of a $m$-dimensional ball with the same diameter as the $S_j$. So for example, you might cover a square in the $(x,y)$ plane ($m=2$) with rectangular prisms ($n=3$) which are very short in the $z$ direction and are equal in size in the $x$ and $y$ directions.
– Ian
Jul 31 at 15:18
1
(Cont.) Now $alpha_2 (mathrmdiam(S_j)/2)^2$ is the area of a disk whose diameter is the same as the diameter of the projection of these prisms into the plane. For small disks this area is close to the true area of the projection (which is to say the area of the corresponding square), so you are nearly summing up the area of the squares, which is exactly what you want to do.
– Ian
Jul 31 at 15:18
1
1
In the case $m=n$, once $delta$ is large enough you can simply attain the infimum by setting $S_1=A,S_j=emptyset$ for $j>1$, at which point the sum simply reduces to $alpha_m (mathrmdiam(A)/2)^m$ which is of course the actual Lebesgue measure of the ball. This is not very insightful of course.
– Ian
Jul 31 at 15:10
In the case $m=n$, once $delta$ is large enough you can simply attain the infimum by setting $S_1=A,S_j=emptyset$ for $j>1$, at which point the sum simply reduces to $alpha_m (mathrmdiam(A)/2)^m$ which is of course the actual Lebesgue measure of the ball. This is not very insightful of course.
– Ian
Jul 31 at 15:10
1
1
There is an error in the definition. $delta$ should be taken to zero, not infinity. Note that this does have an effect on the validity of @Ian's approach.
– Xander Henderson
Jul 31 at 15:11
There is an error in the definition. $delta$ should be taken to zero, not infinity. Note that this does have an effect on the validity of @Ian's approach.
– Xander Henderson
Jul 31 at 15:11
1
1
With @XanderHenderson's correction (thanks), the situation for $m=n$ is now the same as the usual construction of the Lebesgue measure: cover the ball by tiny balls and sum up their volumes, and then as the balls becomes smaller, reduce the overlap and reduce the amount of volume located outside the original ball. Again, this isn't very insightful to what this construction means for purposes of dealing with $m neq n$.
– Ian
Jul 31 at 15:14
With @XanderHenderson's correction (thanks), the situation for $m=n$ is now the same as the usual construction of the Lebesgue measure: cover the ball by tiny balls and sum up their volumes, and then as the balls becomes smaller, reduce the overlap and reduce the amount of volume located outside the original ball. Again, this isn't very insightful to what this construction means for purposes of dealing with $m neq n$.
– Ian
Jul 31 at 15:14
1
1
As for the case $m neq n$, the idea is to cover $A$ by sets $S_j$ and treat the $m$-dimensional volume of the $S_j$ as being approximated by the volume of a $m$-dimensional ball with the same diameter as the $S_j$. So for example, you might cover a square in the $(x,y)$ plane ($m=2$) with rectangular prisms ($n=3$) which are very short in the $z$ direction and are equal in size in the $x$ and $y$ directions.
– Ian
Jul 31 at 15:18
As for the case $m neq n$, the idea is to cover $A$ by sets $S_j$ and treat the $m$-dimensional volume of the $S_j$ as being approximated by the volume of a $m$-dimensional ball with the same diameter as the $S_j$. So for example, you might cover a square in the $(x,y)$ plane ($m=2$) with rectangular prisms ($n=3$) which are very short in the $z$ direction and are equal in size in the $x$ and $y$ directions.
– Ian
Jul 31 at 15:18
1
1
(Cont.) Now $alpha_2 (mathrmdiam(S_j)/2)^2$ is the area of a disk whose diameter is the same as the diameter of the projection of these prisms into the plane. For small disks this area is close to the true area of the projection (which is to say the area of the corresponding square), so you are nearly summing up the area of the squares, which is exactly what you want to do.
– Ian
Jul 31 at 15:18
(Cont.) Now $alpha_2 (mathrmdiam(S_j)/2)^2$ is the area of a disk whose diameter is the same as the diameter of the projection of these prisms into the plane. For small disks this area is close to the true area of the projection (which is to say the area of the corresponding square), so you are nearly summing up the area of the squares, which is exactly what you want to do.
– Ian
Jul 31 at 15:18
 |Â
show 10 more comments
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1
In the case $m=n$, once $delta$ is large enough you can simply attain the infimum by setting $S_1=A,S_j=emptyset$ for $j>1$, at which point the sum simply reduces to $alpha_m (mathrmdiam(A)/2)^m$ which is of course the actual Lebesgue measure of the ball. This is not very insightful of course.
– Ian
Jul 31 at 15:10
1
There is an error in the definition. $delta$ should be taken to zero, not infinity. Note that this does have an effect on the validity of @Ian's approach.
– Xander Henderson
Jul 31 at 15:11
1
With @XanderHenderson's correction (thanks), the situation for $m=n$ is now the same as the usual construction of the Lebesgue measure: cover the ball by tiny balls and sum up their volumes, and then as the balls becomes smaller, reduce the overlap and reduce the amount of volume located outside the original ball. Again, this isn't very insightful to what this construction means for purposes of dealing with $m neq n$.
– Ian
Jul 31 at 15:14
1
As for the case $m neq n$, the idea is to cover $A$ by sets $S_j$ and treat the $m$-dimensional volume of the $S_j$ as being approximated by the volume of a $m$-dimensional ball with the same diameter as the $S_j$. So for example, you might cover a square in the $(x,y)$ plane ($m=2$) with rectangular prisms ($n=3$) which are very short in the $z$ direction and are equal in size in the $x$ and $y$ directions.
– Ian
Jul 31 at 15:18
1
(Cont.) Now $alpha_2 (mathrmdiam(S_j)/2)^2$ is the area of a disk whose diameter is the same as the diameter of the projection of these prisms into the plane. For small disks this area is close to the true area of the projection (which is to say the area of the corresponding square), so you are nearly summing up the area of the squares, which is exactly what you want to do.
– Ian
Jul 31 at 15:18