Mixing
Measure of solution set
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
If I want to solve
$$x = 1$$
for $x$, the solution set has exactly one solution, $ x = 1$, and (Lebesgue) measure 0.
Say I have $x_1, x_2$ and they satisfy
$$sum_i=1^2 x_i = 1 \
x_i in [0, 1] ,, forall i$$
The solutions are given by $x_1, x_2 : x_1 in [0, 1], x_2 = 1-x_1$. The first variable is free on $[0, 1]$, and the second variable is determined given the first. Measure: $1$.
Generalize and let $N$ be the number of variables, and solve
$$sum_i=1^N x_i = 1 \
x_i in [0, 1] ,, forall i = 1 dots N$$
The continuation of the previous series is $N=3$, and here we have
$$x_1 in [0, 1]\
x_2 in [0, 1 - x_1] \
x_3 = 1 - x_1 - x_2 $$
If I draw $x_2 = x_2(x_1)$, it's clear that the size of the solution set is half of the square with length $1$, and the size of that is $1/2$.
- How would I generally find the size of the solution set as $N$ increases?
- I find it slightly counter-intuitive that the measure of the solution set is larger when $N=2$, compared to $N=3$ ($1 > 1/2$) -- did I miss something here?
- How could one think about the size of the solution set as $N to infty$? How would I approach that problem? I'm not familiar with Hilbert spaces but I suppose I need to read into that topic for this..
measure-theory lebesgue-measure
asked Jul 17 at 12:09
FooBar
526312
add a comment |Â
up vote
2
down vote
favorite
If I want to solve
$$x = 1$$
for $x$, the solution set has exactly one solution, $ x = 1$, and (Lebesgue) measure 0.
Say I have $x_1, x_2$ and they satisfy
$$sum_i=1^2 x_i = 1 \
x_i in [0, 1] ,, forall i$$
The solutions are given by $x_1, x_2 : x_1 in [0, 1], x_2 = 1-x_1$. The first variable is free on $[0, 1]$, and the second variable is determined given the first. Measure: $1$.
Generalize and let $N$ be the number of variables, and solve
$$sum_i=1^N x_i = 1 \
x_i in [0, 1] ,, forall i = 1 dots N$$
The continuation of the previous series is $N=3$, and here we have
$$x_1 in [0, 1]\
x_2 in [0, 1 - x_1] \
x_3 = 1 - x_1 - x_2 $$
If I draw $x_2 = x_2(x_1)$, it's clear that the size of the solution set is half of the square with length $1$, and the size of that is $1/2$.
- How would I generally find the size of the solution set as $N$ increases?
- I find it slightly counter-intuitive that the measure of the solution set is larger when $N=2$, compared to $N=3$ ($1 > 1/2$) -- did I miss something here?
- How could one think about the size of the solution set as $N to infty$? How would I approach that problem? I'm not familiar with Hilbert spaces but I suppose I need to read into that topic for this..
measure-theory lebesgue-measure
asked Jul 17 at 12:09
FooBar
526312
2
For $N=2$, are you asking about $lambdabig((x_1, x_2): x_1 + x_2 = 1, x_1,x_2 in [0,1] big)$, where $lambda$ is Lebesgue measure on $mathbbR^2$? If so, that set has measure $0$: the points in that set form a line segment in $mathbbR^2$.
– aduh
Jul 17 at 12:36
@aduh That cleared up a lot. I suppose the answer to the second bullet point then is: "All of them have measure $0$ on the relevant $mathbb R^N$.
– FooBar
Jul 17 at 12:57
@aduh Ultimatively, I'm interested in characterizing the solution set as $N$ increases. If the Lebesgue measure is $0$ for all of these, perhaps there is a more meaningful way of doing that..
– FooBar
Jul 17 at 12:58
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If I want to solve
$$x = 1$$
for $x$, the solution set has exactly one solution, $ x = 1$, and (Lebesgue) measure 0.
Say I have $x_1, x_2$ and they satisfy
$$sum_i=1^2 x_i = 1 \
x_i in [0, 1] ,, forall i$$
The solutions are given by $x_1, x_2 : x_1 in [0, 1], x_2 = 1-x_1$. The first variable is free on $[0, 1]$, and the second variable is determined given the first. Measure: $1$.
Generalize and let $N$ be the number of variables, and solve
$$sum_i=1^N x_i = 1 \
x_i in [0, 1] ,, forall i = 1 dots N$$
The continuation of the previous series is $N=3$, and here we have
$$x_1 in [0, 1]\
x_2 in [0, 1 - x_1] \
x_3 = 1 - x_1 - x_2 $$
If I draw $x_2 = x_2(x_1)$, it's clear that the size of the solution set is half of the square with length $1$, and the size of that is $1/2$.
- How would I generally find the size of the solution set as $N$ increases?
- I find it slightly counter-intuitive that the measure of the solution set is larger when $N=2$, compared to $N=3$ ($1 > 1/2$) -- did I miss something here?
- How could one think about the size of the solution set as $N to infty$? How would I approach that problem? I'm not familiar with Hilbert spaces but I suppose I need to read into that topic for this..
measure-theory lebesgue-measure
asked Jul 17 at 12:09
FooBar
526312
If I want to solve
$$x = 1$$
for $x$, the solution set has exactly one solution, $ x = 1$, and (Lebesgue) measure 0.
Say I have $x_1, x_2$ and they satisfy
$$sum_i=1^2 x_i = 1 \
x_i in [0, 1] ,, forall i$$
The solutions are given by $x_1, x_2 : x_1 in [0, 1], x_2 = 1-x_1$. The first variable is free on $[0, 1]$, and the second variable is determined given the first. Measure: $1$.
Generalize and let $N$ be the number of variables, and solve
$$sum_i=1^N x_i = 1 \
x_i in [0, 1] ,, forall i = 1 dots N$$
The continuation of the previous series is $N=3$, and here we have
$$x_1 in [0, 1]\
x_2 in [0, 1 - x_1] \
x_3 = 1 - x_1 - x_2 $$
If I draw $x_2 = x_2(x_1)$, it's clear that the size of the solution set is half of the square with length $1$, and the size of that is $1/2$.
- How would I generally find the size of the solution set as $N$ increases?
- I find it slightly counter-intuitive that the measure of the solution set is larger when $N=2$, compared to $N=3$ ($1 > 1/2$) -- did I miss something here?
- How could one think about the size of the solution set as $N to infty$? How would I approach that problem? I'm not familiar with Hilbert spaces but I suppose I need to read into that topic for this..
measure-theory lebesgue-measure
asked Jul 17 at 12:09
FooBar
526312
asked Jul 17 at 12:09
FooBar
526312
asked Jul 17 at 12:09
FooBar
526312
asked Jul 17 at 12:09
FooBar
526312
526312
2
For $N=2$, are you asking about $lambdabig((x_1, x_2): x_1 + x_2 = 1, x_1,x_2 in [0,1] big)$, where $lambda$ is Lebesgue measure on $mathbbR^2$? If so, that set has measure $0$: the points in that set form a line segment in $mathbbR^2$.
– aduh
Jul 17 at 12:36
@aduh That cleared up a lot. I suppose the answer to the second bullet point then is: "All of them have measure $0$ on the relevant $mathbb R^N$.
– FooBar
Jul 17 at 12:57
@aduh Ultimatively, I'm interested in characterizing the solution set as $N$ increases. If the Lebesgue measure is $0$ for all of these, perhaps there is a more meaningful way of doing that..
– FooBar
Jul 17 at 12:58
add a comment |Â
2
For $N=2$, are you asking about $lambdabig((x_1, x_2): x_1 + x_2 = 1, x_1,x_2 in [0,1] big)$, where $lambda$ is Lebesgue measure on $mathbbR^2$? If so, that set has measure $0$: the points in that set form a line segment in $mathbbR^2$.
– aduh
Jul 17 at 12:36
@aduh That cleared up a lot. I suppose the answer to the second bullet point then is: "All of them have measure $0$ on the relevant $mathbb R^N$.
– FooBar
Jul 17 at 12:57
@aduh Ultimatively, I'm interested in characterizing the solution set as $N$ increases. If the Lebesgue measure is $0$ for all of these, perhaps there is a more meaningful way of doing that..
– FooBar
Jul 17 at 12:58
2
2
For $N=2$, are you asking about $lambdabig((x_1, x_2): x_1 + x_2 = 1, x_1,x_2 in [0,1] big)$, where $lambda$ is Lebesgue measure on $mathbbR^2$? If so, that set has measure $0$: the points in that set form a line segment in $mathbbR^2$.
– aduh
Jul 17 at 12:36
For $N=2$, are you asking about $lambdabig((x_1, x_2): x_1 + x_2 = 1, x_1,x_2 in [0,1] big)$, where $lambda$ is Lebesgue measure on $mathbbR^2$? If so, that set has measure $0$: the points in that set form a line segment in $mathbbR^2$.
– aduh
Jul 17 at 12:36
@aduh That cleared up a lot. I suppose the answer to the second bullet point then is: "All of them have measure $0$ on the relevant $mathbb R^N$.
– FooBar
Jul 17 at 12:57
@aduh That cleared up a lot. I suppose the answer to the second bullet point then is: "All of them have measure $0$ on the relevant $mathbb R^N$.
– FooBar
Jul 17 at 12:57
@aduh Ultimatively, I'm interested in characterizing the solution set as $N$ increases. If the Lebesgue measure is $0$ for all of these, perhaps there is a more meaningful way of doing that..
– FooBar
Jul 17 at 12:58
@aduh Ultimatively, I'm interested in characterizing the solution set as $N$ increases. If the Lebesgue measure is $0$ for all of these, perhaps there is a more meaningful way of doing that..
– FooBar
Jul 17 at 12:58
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
For $ngeq 2,$ if $Tsubset Bbb R^n $ is a translate of an $(n-1)$-dimensional linear subspace $SsubsetBbb R^n$, then $S$ is isometrically isomorphic to $A=Bbb R^(n-1)times 0. $ So $T,S$ and $A$ each have the same $n$-dimensional Lebesgue measure. For any $r>0$ we have
$Asubset cup_kin Bbb Z^+ S(k,r)$
where $S(k,r)=[-k,k]^(n-1)times [-f(k,r),f(k,r)]$
where $f(k,r)=rcdot 2^-k-2cdot (2k)^-(n-1).$
So the $n$-dimensional Lebesgue measure $m(A)$ is not more than $sum_kin Bbb Z^+m(S(k,r))=r.$ Since $r>0$ is arbitrary, therefore $m(A)=0.$
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
add a comment |Â
up vote
2
down vote
By definition, the Lebesgue measure (volume) of the hyper-rectangle $[a_1, b_1]times cdots times [a_N, b_N]$ is $(b_1 - a_1)cdots (b_N - a_N)$.
Special cases of this definition are
- length of a line segment $[a_1, b_1]subsetmathbbR$, which is $b_1 - a_1$,
- area of a rectangle $[a_1, b_1]times [a_2, b_2]subsetmathbbR^2$, which is $(b_2 - a_2)(b_1 - a_1)$.
For all $N> 1$, the measure of $[a_1, b_1]times cdots times [a_N -1, b_N-1]times[a_N, a_N]$ is $0$.
Your solution set is a subset of such a hyper-rectangle, rotated so that it is a subset of a hyperplane with normal $(1, 1, dots, 1)$. Rotation of an object does not change its volume.
Some comments for the case $N = 2$. In this case, the area of your solution set (which is a line segment) is $0$ (it can be covered by the interval $[0, sqrt2]times [0,0]$, rotated by $45^circ$ around the point $(1, 0)$). Of course, we can compute its length ($ =sqrt2 $), but for $mathbbR^2$, this number is irrelevant.
answered Jul 17 at 14:51
Antoine
2,485925
add a comment |Â
up vote
1
down vote
Notice that your function ($(x_1,x_2,x_3) mapsto x_1+x_2+x_3$) is continuously differentiable and the derivative is surjective in the sense of a linear map. Hence the https://en.wikipedia.org/wiki/Implicit_function_theorem applies and it can be used to show the following theorem (Inverse of regular value is a submanifold, Milnor's proof) :
Let $f in C^1(mathbb R^n, mathbbR^d) $ with $x_0 in mathbbR^d$ and $n>d$. Let furthermore for any $x in f^-1(f(x_0))$, $D_x f$ be surjective.
Then: $f^-1(x_0)$ is a manifold of dimension $(n-d)$. Hence it is of Lebesgue measure $0$ in $mathbbR^n$ and of (say) Hausdorff dimension $(n-d)$.
answered Jul 17 at 14:26
Marko Karbevski
1,035821
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
For $ngeq 2,$ if $Tsubset Bbb R^n $ is a translate of an $(n-1)$-dimensional linear subspace $SsubsetBbb R^n$, then $S$ is isometrically isomorphic to $A=Bbb R^(n-1)times 0. $ So $T,S$ and $A$ each have the same $n$-dimensional Lebesgue measure. For any $r>0$ we have
$Asubset cup_kin Bbb Z^+ S(k,r)$
where $S(k,r)=[-k,k]^(n-1)times [-f(k,r),f(k,r)]$
where $f(k,r)=rcdot 2^-k-2cdot (2k)^-(n-1).$
So the $n$-dimensional Lebesgue measure $m(A)$ is not more than $sum_kin Bbb Z^+m(S(k,r))=r.$ Since $r>0$ is arbitrary, therefore $m(A)=0.$
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
add a comment |Â
up vote
2
down vote
For $ngeq 2,$ if $Tsubset Bbb R^n $ is a translate of an $(n-1)$-dimensional linear subspace $SsubsetBbb R^n$, then $S$ is isometrically isomorphic to $A=Bbb R^(n-1)times 0. $ So $T,S$ and $A$ each have the same $n$-dimensional Lebesgue measure. For any $r>0$ we have
$Asubset cup_kin Bbb Z^+ S(k,r)$
where $S(k,r)=[-k,k]^(n-1)times [-f(k,r),f(k,r)]$
where $f(k,r)=rcdot 2^-k-2cdot (2k)^-(n-1).$
So the $n$-dimensional Lebesgue measure $m(A)$ is not more than $sum_kin Bbb Z^+m(S(k,r))=r.$ Since $r>0$ is arbitrary, therefore $m(A)=0.$
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For $ngeq 2,$ if $Tsubset Bbb R^n $ is a translate of an $(n-1)$-dimensional linear subspace $SsubsetBbb R^n$, then $S$ is isometrically isomorphic to $A=Bbb R^(n-1)times 0. $ So $T,S$ and $A$ each have the same $n$-dimensional Lebesgue measure. For any $r>0$ we have
$Asubset cup_kin Bbb Z^+ S(k,r)$
where $S(k,r)=[-k,k]^(n-1)times [-f(k,r),f(k,r)]$
where $f(k,r)=rcdot 2^-k-2cdot (2k)^-(n-1).$
So the $n$-dimensional Lebesgue measure $m(A)$ is not more than $sum_kin Bbb Z^+m(S(k,r))=r.$ Since $r>0$ is arbitrary, therefore $m(A)=0.$
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
For $ngeq 2,$ if $Tsubset Bbb R^n $ is a translate of an $(n-1)$-dimensional linear subspace $SsubsetBbb R^n$, then $S$ is isometrically isomorphic to $A=Bbb R^(n-1)times 0. $ So $T,S$ and $A$ each have the same $n$-dimensional Lebesgue measure. For any $r>0$ we have
$Asubset cup_kin Bbb Z^+ S(k,r)$
where $S(k,r)=[-k,k]^(n-1)times [-f(k,r),f(k,r)]$
where $f(k,r)=rcdot 2^-k-2cdot (2k)^-(n-1).$
So the $n$-dimensional Lebesgue measure $m(A)$ is not more than $sum_kin Bbb Z^+m(S(k,r))=r.$ Since $r>0$ is arbitrary, therefore $m(A)=0.$
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
edited Jul 17 at 17:08
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
answered Jul 17 at 17:02
DanielWainfleet
31.7k31644
31.7k31644
add a comment |Â
add a comment |Â
up vote
2
down vote
By definition, the Lebesgue measure (volume) of the hyper-rectangle $[a_1, b_1]times cdots times [a_N, b_N]$ is $(b_1 - a_1)cdots (b_N - a_N)$.
Special cases of this definition are
- length of a line segment $[a_1, b_1]subsetmathbbR$, which is $b_1 - a_1$,
- area of a rectangle $[a_1, b_1]times [a_2, b_2]subsetmathbbR^2$, which is $(b_2 - a_2)(b_1 - a_1)$.
For all $N> 1$, the measure of $[a_1, b_1]times cdots times [a_N -1, b_N-1]times[a_N, a_N]$ is $0$.
Your solution set is a subset of such a hyper-rectangle, rotated so that it is a subset of a hyperplane with normal $(1, 1, dots, 1)$. Rotation of an object does not change its volume.
Some comments for the case $N = 2$. In this case, the area of your solution set (which is a line segment) is $0$ (it can be covered by the interval $[0, sqrt2]times [0,0]$, rotated by $45^circ$ around the point $(1, 0)$). Of course, we can compute its length ($ =sqrt2 $), but for $mathbbR^2$, this number is irrelevant.
answered Jul 17 at 14:51
Antoine
2,485925
add a comment |Â
up vote
2
down vote
By definition, the Lebesgue measure (volume) of the hyper-rectangle $[a_1, b_1]times cdots times [a_N, b_N]$ is $(b_1 - a_1)cdots (b_N - a_N)$.
Special cases of this definition are
- length of a line segment $[a_1, b_1]subsetmathbbR$, which is $b_1 - a_1$,
- area of a rectangle $[a_1, b_1]times [a_2, b_2]subsetmathbbR^2$, which is $(b_2 - a_2)(b_1 - a_1)$.
For all $N> 1$, the measure of $[a_1, b_1]times cdots times [a_N -1, b_N-1]times[a_N, a_N]$ is $0$.
Your solution set is a subset of such a hyper-rectangle, rotated so that it is a subset of a hyperplane with normal $(1, 1, dots, 1)$. Rotation of an object does not change its volume.
Some comments for the case $N = 2$. In this case, the area of your solution set (which is a line segment) is $0$ (it can be covered by the interval $[0, sqrt2]times [0,0]$, rotated by $45^circ$ around the point $(1, 0)$). Of course, we can compute its length ($ =sqrt2 $), but for $mathbbR^2$, this number is irrelevant.
answered Jul 17 at 14:51
Antoine
2,485925
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By definition, the Lebesgue measure (volume) of the hyper-rectangle $[a_1, b_1]times cdots times [a_N, b_N]$ is $(b_1 - a_1)cdots (b_N - a_N)$.
Special cases of this definition are
- length of a line segment $[a_1, b_1]subsetmathbbR$, which is $b_1 - a_1$,
- area of a rectangle $[a_1, b_1]times [a_2, b_2]subsetmathbbR^2$, which is $(b_2 - a_2)(b_1 - a_1)$.
For all $N> 1$, the measure of $[a_1, b_1]times cdots times [a_N -1, b_N-1]times[a_N, a_N]$ is $0$.
Your solution set is a subset of such a hyper-rectangle, rotated so that it is a subset of a hyperplane with normal $(1, 1, dots, 1)$. Rotation of an object does not change its volume.
Some comments for the case $N = 2$. In this case, the area of your solution set (which is a line segment) is $0$ (it can be covered by the interval $[0, sqrt2]times [0,0]$, rotated by $45^circ$ around the point $(1, 0)$). Of course, we can compute its length ($ =sqrt2 $), but for $mathbbR^2$, this number is irrelevant.
answered Jul 17 at 14:51
Antoine
2,485925
By definition, the Lebesgue measure (volume) of the hyper-rectangle $[a_1, b_1]times cdots times [a_N, b_N]$ is $(b_1 - a_1)cdots (b_N - a_N)$.
Special cases of this definition are
- length of a line segment $[a_1, b_1]subsetmathbbR$, which is $b_1 - a_1$,
- area of a rectangle $[a_1, b_1]times [a_2, b_2]subsetmathbbR^2$, which is $(b_2 - a_2)(b_1 - a_1)$.
For all $N> 1$, the measure of $[a_1, b_1]times cdots times [a_N -1, b_N-1]times[a_N, a_N]$ is $0$.
Your solution set is a subset of such a hyper-rectangle, rotated so that it is a subset of a hyperplane with normal $(1, 1, dots, 1)$. Rotation of an object does not change its volume.
Some comments for the case $N = 2$. In this case, the area of your solution set (which is a line segment) is $0$ (it can be covered by the interval $[0, sqrt2]times [0,0]$, rotated by $45^circ$ around the point $(1, 0)$). Of course, we can compute its length ($ =sqrt2 $), but for $mathbbR^2$, this number is irrelevant.
answered Jul 17 at 14:51
Antoine
2,485925
edited Jul 23 at 17:04
answered Jul 17 at 14:51
Antoine
2,485925
answered Jul 17 at 14:51
Antoine
2,485925
answered Jul 17 at 14:51
Antoine
2,485925
2,485925
add a comment |Â
add a comment |Â
up vote
1
down vote
Notice that your function ($(x_1,x_2,x_3) mapsto x_1+x_2+x_3$) is continuously differentiable and the derivative is surjective in the sense of a linear map. Hence the https://en.wikipedia.org/wiki/Implicit_function_theorem applies and it can be used to show the following theorem (Inverse of regular value is a submanifold, Milnor's proof) :
Let $f in C^1(mathbb R^n, mathbbR^d) $ with $x_0 in mathbbR^d$ and $n>d$. Let furthermore for any $x in f^-1(f(x_0))$, $D_x f$ be surjective.
Then: $f^-1(x_0)$ is a manifold of dimension $(n-d)$. Hence it is of Lebesgue measure $0$ in $mathbbR^n$ and of (say) Hausdorff dimension $(n-d)$.
answered Jul 17 at 14:26
Marko Karbevski
1,035821
add a comment |Â
up vote
1
down vote
Notice that your function ($(x_1,x_2,x_3) mapsto x_1+x_2+x_3$) is continuously differentiable and the derivative is surjective in the sense of a linear map. Hence the https://en.wikipedia.org/wiki/Implicit_function_theorem applies and it can be used to show the following theorem (Inverse of regular value is a submanifold, Milnor's proof) :
Let $f in C^1(mathbb R^n, mathbbR^d) $ with $x_0 in mathbbR^d$ and $n>d$. Let furthermore for any $x in f^-1(f(x_0))$, $D_x f$ be surjective.
Then: $f^-1(x_0)$ is a manifold of dimension $(n-d)$. Hence it is of Lebesgue measure $0$ in $mathbbR^n$ and of (say) Hausdorff dimension $(n-d)$.
answered Jul 17 at 14:26
Marko Karbevski
1,035821
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that your function ($(x_1,x_2,x_3) mapsto x_1+x_2+x_3$) is continuously differentiable and the derivative is surjective in the sense of a linear map. Hence the https://en.wikipedia.org/wiki/Implicit_function_theorem applies and it can be used to show the following theorem (Inverse of regular value is a submanifold, Milnor's proof) :
Let $f in C^1(mathbb R^n, mathbbR^d) $ with $x_0 in mathbbR^d$ and $n>d$. Let furthermore for any $x in f^-1(f(x_0))$, $D_x f$ be surjective.
Then: $f^-1(x_0)$ is a manifold of dimension $(n-d)$. Hence it is of Lebesgue measure $0$ in $mathbbR^n$ and of (say) Hausdorff dimension $(n-d)$.
answered Jul 17 at 14:26
Marko Karbevski
1,035821
Notice that your function ($(x_1,x_2,x_3) mapsto x_1+x_2+x_3$) is continuously differentiable and the derivative is surjective in the sense of a linear map. Hence the https://en.wikipedia.org/wiki/Implicit_function_theorem applies and it can be used to show the following theorem (Inverse of regular value is a submanifold, Milnor's proof) :
Let $f in C^1(mathbb R^n, mathbbR^d) $ with $x_0 in mathbbR^d$ and $n>d$. Let furthermore for any $x in f^-1(f(x_0))$, $D_x f$ be surjective.
Then: $f^-1(x_0)$ is a manifold of dimension $(n-d)$. Hence it is of Lebesgue measure $0$ in $mathbbR^n$ and of (say) Hausdorff dimension $(n-d)$.
answered Jul 17 at 14:26
Marko Karbevski
1,035821
answered Jul 17 at 14:26
Marko Karbevski
1,035821
answered Jul 17 at 14:26
Marko Karbevski
1,035821
answered Jul 17 at 14:26
Marko Karbevski
1,035821
1,035821
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854436%2fmeasure-of-solution-set%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
For $N=2$, are you asking about $lambdabig((x_1, x_2): x_1 + x_2 = 1, x_1,x_2 in [0,1] big)$, where $lambda$ is Lebesgue measure on $mathbbR^2$? If so, that set has measure $0$: the points in that set form a line segment in $mathbbR^2$.
– aduh
Jul 17 at 12:36
@aduh That cleared up a lot. I suppose the answer to the second bullet point then is: "All of them have measure $0$ on the relevant $mathbb R^N$.
– FooBar
Jul 17 at 12:57
@aduh Ultimatively, I'm interested in characterizing the solution set as $N$ increases. If the Lebesgue measure is $0$ for all of these, perhaps there is a more meaningful way of doing that..
– FooBar
Jul 17 at 12:58