Measure of solution set

Question

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 $1/2$.

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• 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..

Answer 1

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 $[a_1, b_1]\subset\mathbb{R}$, which is $b_1 - a_1$,
• area of a rectangle $[a_1, b_1]\times [a_2, b_2]\subset\mathbb{R}^2$, 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.

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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 $[0, \sqrt{2}]\times [0,0]$, rotated by $45^\circ$ around the point $(1, 0)$). Of course, we can compute its length ($ =\sqrt{2} $), but for $\mathbb{R}^2$, this number is irrelevant.

Answer 2

For $n\geq 2,$ if $T\subset \Bbb R^n $ is a translate of an $(n-1)$-dimensional linear subspace $S\subset\Bbb R^n$, 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

$A\subset \cup_{k\in \Bbb Z^+} S(k,r)$

where $S(k,r)=[-k,k]^{(n-1)}\times [-f(k,r),f(k,r)]$

where $f(k,r)=r\cdot 2^{-k-2}\cdot (2k)^{-(n-1)}.$

So the $n$-dimensional Lebesgue measure $m(A)$ is not more than $\sum_{k\in \Bbb Z^+}m(S(k,r))=r.$ Since $r>0$ is arbitrary, therefore $m(A)=0.$

Answer 3

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, \mathbb{R}^d) $ with $x_0 \in \mathbb{R}^d$ 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 $\mathbb{R}^n$ and of (say) Hausdorff dimension $(n-d)$.