Every subset of a subspace of $mathbb{R}^n$ of dim $<n$ has measure 0
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I have read in my teacher’s book the following statement.
- Every subset A of a subspace C of $mathbb{R}^n$ with $dim(C)<n$ has measure 0
And I’m having issues proving it and mainly understanding why. Is it because measure space $(mathbb{R}^n , M( mathbb{R}^n), lambda_n)$ where $M( mathbb{R}^n)$ is the $sigma$-algebra of the measurable sets of $mathbb{R}^n$, and $lambda_n$ is Lebesgue Measure is a complete measure space?
measure-theory lebesgue-measure
asked Nov 16 at 19:36
M. Navarro
516
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up vote
0
down vote
favorite
I have read in my teacher’s book the following statement.
- Every subset A of a subspace C of $mathbb{R}^n$ with $dim(C)<n$ has measure 0
And I’m having issues proving it and mainly understanding why. Is it because measure space $(mathbb{R}^n , M( mathbb{R}^n), lambda_n)$ where $M( mathbb{R}^n)$ is the $sigma$-algebra of the measurable sets of $mathbb{R}^n$, and $lambda_n$ is Lebesgue Measure is a complete measure space?
measure-theory lebesgue-measure
asked Nov 16 at 19:36
M. Navarro
516
I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question.
– Michał Miśkiewicz
Nov 16 at 19:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have read in my teacher’s book the following statement.
- Every subset A of a subspace C of $mathbb{R}^n$ with $dim(C)<n$ has measure 0
And I’m having issues proving it and mainly understanding why. Is it because measure space $(mathbb{R}^n , M( mathbb{R}^n), lambda_n)$ where $M( mathbb{R}^n)$ is the $sigma$-algebra of the measurable sets of $mathbb{R}^n$, and $lambda_n$ is Lebesgue Measure is a complete measure space?
measure-theory lebesgue-measure
asked Nov 16 at 19:36
M. Navarro
516
I have read in my teacher’s book the following statement.
- Every subset A of a subspace C of $mathbb{R}^n$ with $dim(C)<n$ has measure 0
And I’m having issues proving it and mainly understanding why. Is it because measure space $(mathbb{R}^n , M( mathbb{R}^n), lambda_n)$ where $M( mathbb{R}^n)$ is the $sigma$-algebra of the measurable sets of $mathbb{R}^n$, and $lambda_n$ is Lebesgue Measure is a complete measure space?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Nov 16 at 19:36
M. Navarro
516
asked Nov 16 at 19:36
M. Navarro
516
asked Nov 16 at 19:36
M. Navarro
516
asked Nov 16 at 19:36
M. Navarro
516
asked Nov 16 at 19:36
M. Navarro
516
516
I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question.
– Michał Miśkiewicz
Nov 16 at 19:40
add a comment |
I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question.
– Michał Miśkiewicz
Nov 16 at 19:40
I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question.
– Michał Miśkiewicz
Nov 16 at 19:40
I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question.
– Michał Miśkiewicz
Nov 16 at 19:40
add a comment |
1 Answer
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If you recall from calculus, if we integrated over a region, say $displaystyleint_{[a,b]}dx$, we would get the area of a rectangle with base length (b-a) if we were in $mathbb{R}^2$ or $displaystyleint_{[a,b] times [c,d]}dx$, the volume of a parallelepiped whose base has length (b-a) and width (d-c).
In $mathbb{R}^3$, the measure of A is the volume of the set A. If dim C $<$n, then one of the dimensions of the higher dimensional rectangles is 0.
So if the set A is a square, it has volume 0. Thus the square would have measure 0. Now generalize this idea to higher dimensions.
answered Nov 16 at 20:41
Joel Pereira
3967
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If you recall from calculus, if we integrated over a region, say $displaystyleint_{[a,b]}dx$, we would get the area of a rectangle with base length (b-a) if we were in $mathbb{R}^2$ or $displaystyleint_{[a,b] times [c,d]}dx$, the volume of a parallelepiped whose base has length (b-a) and width (d-c).
In $mathbb{R}^3$, the measure of A is the volume of the set A. If dim C $<$n, then one of the dimensions of the higher dimensional rectangles is 0.
So if the set A is a square, it has volume 0. Thus the square would have measure 0. Now generalize this idea to higher dimensions.
answered Nov 16 at 20:41
Joel Pereira
3967
add a comment |
up vote
0
down vote
accepted
If you recall from calculus, if we integrated over a region, say $displaystyleint_{[a,b]}dx$, we would get the area of a rectangle with base length (b-a) if we were in $mathbb{R}^2$ or $displaystyleint_{[a,b] times [c,d]}dx$, the volume of a parallelepiped whose base has length (b-a) and width (d-c).
In $mathbb{R}^3$, the measure of A is the volume of the set A. If dim C $<$n, then one of the dimensions of the higher dimensional rectangles is 0.
So if the set A is a square, it has volume 0. Thus the square would have measure 0. Now generalize this idea to higher dimensions.
answered Nov 16 at 20:41
Joel Pereira
3967
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If you recall from calculus, if we integrated over a region, say $displaystyleint_{[a,b]}dx$, we would get the area of a rectangle with base length (b-a) if we were in $mathbb{R}^2$ or $displaystyleint_{[a,b] times [c,d]}dx$, the volume of a parallelepiped whose base has length (b-a) and width (d-c).
In $mathbb{R}^3$, the measure of A is the volume of the set A. If dim C $<$n, then one of the dimensions of the higher dimensional rectangles is 0.
So if the set A is a square, it has volume 0. Thus the square would have measure 0. Now generalize this idea to higher dimensions.
answered Nov 16 at 20:41
Joel Pereira
3967
If you recall from calculus, if we integrated over a region, say $displaystyleint_{[a,b]}dx$, we would get the area of a rectangle with base length (b-a) if we were in $mathbb{R}^2$ or $displaystyleint_{[a,b] times [c,d]}dx$, the volume of a parallelepiped whose base has length (b-a) and width (d-c).
In $mathbb{R}^3$, the measure of A is the volume of the set A. If dim C $<$n, then one of the dimensions of the higher dimensional rectangles is 0.
So if the set A is a square, it has volume 0. Thus the square would have measure 0. Now generalize this idea to higher dimensions.
answered Nov 16 at 20:41
Joel Pereira
3967
answered Nov 16 at 20:41
Joel Pereira
3967
answered Nov 16 at 20:41
Joel Pereira
3967
answered Nov 16 at 20:41
Joel Pereira
3967
3967
add a comment |
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I assume "measure $0$" means "Lebesgue measure $0$". But what kind of dimension do you use? If you look at the definition, the question should answer itself. Anyway, you should include it in the question.
– Michał Miśkiewicz
Nov 16 at 19:40