I would appreciate it if someone could refer me to a proof (or simply give one here) for the statement in the title. That is:
If $k<n$, then every $k-$dimensional subspace of $\mathbb{R}^{n}$ has $n-$dimensional Lebesgue measure zero.
I've seen some proofs that use Sard's lemma but I'm not really familiar with that subject and I've never seen a proof of said lemma so I'd appreciate a proof that doesn't use it if possible.
Thanks in advance!
A $k-$dimensional subspace $S$ of $\mathbb R^n$, with $k<n$, is the range of linear transformation of rank $k$. That means that there exists a matrix $A\in\mathbb{R}^{n\times n}$, with $\mathrm{rank}(A)=k<n$, such that $\mathrm{Ran}(A)=S$. But then $\det A=0$. The theorem of change of variables $$ m(S)=\int_{\mathbb R^n}\lvert\det A\rvert\,dx=0. $$
If you are looking for something more elementary, try to show this for a line in $\mathbb R^2$, where it is not hard to show that every bounded piece of it can be covered by ${\mathcal O}(n)$ squares with sides of length $1/n$, and hence of area $1/n^2$. That means that its 2-dimensional Lebesgue measure is bounded by $kn\cdot \dfrac{1}{n^2}$, for every $n$, and thus it has to vanish.
Note, that similarly $m(S)=0$, for every $k-$dimensional hyperplane in $\mathbb R^n$, for $k<n$.
The "standard" argument is somewhat like this: You can use the fact that a countable , disjoint union of sets of measure zero has measure zero (you can see this as a corollary of the fact that the tail (sum) of a convergent series goes to $0$). You can start inductively, by partitioning the Real line into disjoint subintervals $(a,a+1]$ , where a goes from $-\infty$ to $+\infty$ (as Integers), and then do something similar for higher dimensions.
( For a bit more rigor, you know that , in the (completion) of the product measure for $\mathbb R^2$, the subset $\mathbb R\times {0} $ (same with $\mathbb R^n \times 0^{n-1}$ ) is a product of measurable sets. And then, by monotonicity of the measure, any $(n-k)$-dimensional measurable subset $S$ of $\mathbb R^n$ will also have Lebesgue measure $0$. )
To show each has (product )measure zero, you can enclose each $(a,a+1]$ in a box of height $2\epsilon$ , i.e., enclose them in the box $[+\epsilon, -\epsilon] \times [a,a+1]$ , and then let $\epsilon \rightarrow 0$. Note that for different types of dimension; specifically Hausdorff dimension, you may have subsets of Hausdorff dimension less than $n$ (in an ambient space $\mathbb R^n$) , which have non-negative Lebesgue measure; examples of these are Fat Cantor sets ( where 'Fat' applies to the set, not to Cantor ;))
The OP seemed to ask for a proof using Sard's Theorem, and it seemed that OP is not quite familiar with abstract measure theory (otherwise he/she should know the change of variable formula).
So here is a straightforward (but a little bit constructive) proof using Sard's Theorem.
Firstly, different textbooks use different version of Sard's Theorem, but the most straightforward one which we will use is this:
[Sard's Theorem]: Let $f:M\longrightarrow N$ be a smooth function. Then the set of regular values of $f$ is an everywhere dense open set. Equivalently, the set of critical values of $f$ has measure $0$.
Denote the Jacobian matrix of $f$ to be $(Df)$. It can also be called the push forward of $f$, denoted by $f_{*}$, but let us keep everything simple.
Then, recall the definition of critical points and values as follows:
[Definition]: Let $f:M\longrightarrow N$ be a smooth map between two smooth manifolds. A point $p\in M$ is said to be a Critical Point of $f$ if $(Df):T_{p}M\longrightarrow T_{q}N$ is not surjective. This is equivalent to saying that $\text{rank}(Df)<\text{dim}(N)$.
[Definition]: A point $q\in N$ is said to be a Critical Value of $f$ if the set $f^{-1}(q)$ contains a critical point of $f$.
Now let us prove!
Proof:
Let $L$ be subspace you mentioned. It is actual the sub-manifold given by a vector subspace of $\mathbb{R}^{n}$.
Also, we require $\dim(L)=k< n-1$.
We want to show $\mu(L)=0$, where $\mu$ is the regular Lebesgue measure on $\mathbb{R}^{n}$.
Let $\{e_{i}\}_{i=1}^{k}$ be a basis of $L$. Then, consider the map defined by $$f:\mathbb{R}^{k}\longrightarrow\mathbb{R}^{n}\ \ f(x_{1},\cdots, x_{k}):=(x_{1},\cdots, x_{k})\cdot(e_{1},\cdots, e_{k})^{\intercal}.$$
This map is clearly $C^{\infty}$.
Further, we see that $f(\mathbb{R}^{k})=L$, since the RHS is just a linear combination of scalars from $\mathbb{R}^{k}$ with basis elements of $L$.
Now, if we write for each $e_{i}$ that $e_{i}:=(a_{i,1},\cdots, a_{i,k}),$, then $f(x_{1},\cdots, x_{k})=(f_{1},\cdots, f_{k})$ where $$f_{1}=x_{1}a_{1,1}+x_{2}a_{1,2}+\cdots+x_{k}a_{1,k}$$ $$f_{2}=x_{1}a_{2,1}+x_{2}a_{2,2}+\cdots+x_{k}a_{2,k},$$ $$\cdots$$ $$f_{k}=x_{1}a_{k,1}+x_{2}a_{k,2}+\cdots+x_{k}a_{k,k},$$ so that the Jacobian is $$(Df)= \begin{pmatrix} a_{1,1}&a_{1,2}&\cdots&a_{1,k}\\ a_{2,1}&a_{2,2}&\cdots&a_{2,k}\\ \cdots&\cdots&\cdots&\cdots\\ a_{k,1}&a_{k,2}&\cdots&a_{k,k} \end{pmatrix} = \begin{pmatrix} e_{1}\\ e_{2}\\ \cdots\\ e_{k} \end{pmatrix}.$$
Since $\{e_{i}\}_{i=1}^{k}$ is the basis, we know that the rows of $(Df)$ are linearly independent and thus this $k\times k$ matrix has full rank. But then $$\text{rank}(Df)=k<\dim(\mathbb{T}_{f(p)}\mathbb{R}^{n})=n,\ \text{by hypothesis}.$$
Thus, $(Df)$ is not surjective, which implies every point in the domain of $f$ is a critical point. In other words, the set of critical points is $\mathbb{R}^{k}$.
But we have seen that $f(\mathbb{R}^{k})=L$, so the set of critical values if $L$.
It then follows immediately from Sard's Theorem that $\mu(L)=0$, as desired.
[Remark 1:] The idea of this exercise is actually really intuitive. If have you constructed the lebesgue measure using high dimension cubes, then it is really easy to cover some lower dimensional things using these cubes with only a little bit volumes so that the $\sup$ of the sum of the volumes of these cube coverings is $<\epsilon$.
Think about if you construct lebesgue measure in $\mathbb{R}^{3}$ using cubes, then to cover some line in $\mathbb{R}^{3}$, these cubes can be really really really small, since it is just a line.
[Remark 2:] If you are not familiar with the $\mathbb{R}^{n}-$construction of Lebesgue measure using cubes, read Stein's real analysis. Unlike other introductory measure theory book which construct the Lebesgue measure on the real line only, he directly constructed in $\mathbb{R}^{n}$ using cubes.
I asked myself the same question and came up with this proof, which doesn't use Sard's Theorem or a change of variables. Here are the steps (I skimped on some details to save space):
1.) Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation. Then there is a constant $c$ such that for any closed cube $Q$ in $\mathbb{R}^n$ (a rectangle with sides all of the same length), $T(Q)$ is contained in a cube $R$ with volume $c \, \text{vol} \, (Q)$.
Pf. Put $Q = [a_1 - \lambda, a_1 + \lambda] \times \cdots \times [a_n - \lambda, a_n + \lambda] $ and let $(b_{ij})$ be the matrix representation of $T$. Let $|T| = \text{max} \{|b_{ij}| : 1 \leq i, j \leq n \} $ and let $|x| = \text{max} \, \{|x_i|, \ldots, |x_n| \}$ for any $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$ (sup metric in $\mathbb{R}^n$). For any $x \in Q$, $|Tx| \leq n |T||x| $, hence $x \in Q$ implies
$ |Tx - Ta| = |T(x-a)| \leq n|T| |x-a| \leq n|T| \lambda $.
It follows that $T(Q)$ is contained in a cube $R$ centered at $Ta$ with sides of length $2d\lambda$, where $d= n|T|$. $\text{vol} \, R = c \, \text{vol} \, (Q) $, where $c = d^n$.
2.) Let $Y$ be a subspace of $\mathbb{R}^n$ of dimension $n-1$. Then $Y$ has "measure zero" (not necessarily Lebesgue measure zero) in the sense that, for any $\epsilon > 0$, it can be covered by a countable set of rectangles with total volume less than $\epsilon$.
Pf. Let $\{w_1, \ldots, w_{n-1} \}$ be a basis for $Y$. Define $ T: \mathbb{R}^n \rightarrow \mathbb{R}^n$, $ Tx = \sum_{i=1}^{n-1} x_iw_i . $
$T$ is linear and $T(\mathbb{R}^n)=Y=T(\mathbb{R}^{n-1} \times 0)$. The subspace $\mathbb{R}^{n-1} \times 0$ is a "measure zero" set in $\mathbb{R}^n$ in the sense defined above. To see this, note that this set can be covered by a countable collection of rectangles of the form $[-N, N] \times \ldots [-N, N] \times [-\delta, \delta]$, where $\delta > 0$ is arbitrarily small. It follows that for arbitrary $\epsilon$, we can cover $\mathbb{R}^{n-1} \times 0$ by closed cubes $\{Q_i\}_{i \in \mathbb{N}}$ such that $\sum \text{vol} \, (Q_i) < \epsilon $. Each $T(Q_i)$ is contained within a cube $R_i$ with $\text{vol} \, R_i = c \, \text{vol} \, Q_i $. Hence
$ Y \subset T \left( \bigcup Q_i \right) = \bigcup T(Q_i) \subset \bigcup R_i $ and $\sum \text{vol}\, (R_i) = \sum c \, \text{vol} \, (Q_i) < c \epsilon $.
So $Y$ is a set of measure zero in $\mathbb{R}^n$ as well.
3.) Let $Y$ be as in 2.) and let $\epsilon$ be arbitrary. $Y$ is a measure zero set in the sense defined in 2.), so it can be covered by rectangles $\{R_i\}$ such that $\sum \text{vol}\, (R_i) < \epsilon$. If $m$ is Lebesgue measure on $\mathbb{R}^n$, it follows that
$ m(Y) \leq m \left( \bigcup R_i \right) \leq \sum m(R_i) < \epsilon . $
So $Y$ has Lebesgue measure zero, i.e. $m(Y) = 0.$ (Note we assumed here $Y$ is Lebesgue measurable. This is indeed the case but it wasn't obvious to me at first.)
I am not sure if the below approach is correct. Rudin Real and Complex Analysis book states the result asked in the question within the proof of Theorem 2.20 regarding the Lebesgue measure on ${\boldsymbol{R^{k}}}$. Hence trying to use only results upto this portion of the book.
Part (a) of Theorem 2.20 states that $m\left(W\right)=\text{vol}\left(W\right)$ for every $k$-cell $W$. Using this, we can wrap or cover every compact (closed and bounded) set, $E$, in the lower subspace, $Y$, with dimension, $m<k$, with a suitable $k$-cell from the higher dimensional space. The additional parameters for this $k$-cell in the higher dimension can be $-\frac{\epsilon}{2},\frac{\epsilon}{2}$ with $\epsilon>0$. This gives $m\left(E\right)=\text{vol}\left(E\right)\leq\epsilon^{k-m}\text{volm}$. Here, volm is the volume of the wrapper of $E$, in the $m$-dimensional sub space, $Y$.
Since $\epsilon$ was arbitraty, $m\left(E\right)=\text{vol}\left(E\right)=0$. The entire space (and hence the subspace) is $\sigma-$compact so it can be written as the countable union of compact sets (which are closed and bounded and hence covered like the set $E$ above) all of which have measure zero, so the measure of the entire subspace is zero.
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Let $V$ be a $k$-dimensional subspace of $\mathbb{R}^n$, where $k<n$.
$V$ is contained in an $(n-1)$-dimensional subspace of $\mathbb{R}^n$.
So it is sufficient for us to show any $(n-1)$-dimensional subspace of $\mathbb{R}^n$ has measure zero.
Let $V$ be an $(n-1)$-dimensional subspace of $\mathbb{R}^n$.
Let $a_1,\dots, a_{n-1}$ be a basis of $V$.
Let $a_i=(a_{1,i},\dots, a_{n,i})^T$.
Let $A:=(a_1,\dots, a_{n-1})$ be an $n$ by $n-1$ matrix.
$\operatorname{rank} A=n-1$.
Without loss of generality, we can assume that the first $n-1$ rows of $A$ are linearly independent.
$V=\{x\in\mathbb{R}^n:x=At, t\in\mathbb{R}^{n-1}\}$.
Let $A':=\begin{pmatrix}a_{1,1} & \cdots &a_{1,n-1}\\
\vdots & \cdots & \vdots\\
a_{n-1,1} &\cdots &a_{n-1,n-1}\end{pmatrix}$.
Then, $A'$ is non-singular.
Let $(x_1,\dots,x_n)^T\in V$.
Then, there is $t\in\mathbb{R}^{n-1}$ such that $(x_1,\dots,x_n)^T=At$.
Then, $t=(A')^{-1}(x_1,\dots,x_{n-1})^T$.
Then, $x_n=(a_{n,1},\dots,a_{n,n-1})(A')^{-1}(x_1,\dots,x_{n-1})^T$.
Let $(a_{n,1},\dots,a_{n,n-1})(A')^{-1}(x_1,\dots,x_{n-1})^T=b_1x_1+\cdots+b_{n-1}x_{n-1}$.
Then, $x\in\{(x_1,\dots,x_n)^T\in\mathbb{R}^n:x_n=b_1x_1+\cdots+b_{n-1}x_{n-1}\}$.
So, $V\subset\{(x_1,\dots,x_n)^T\in\mathbb{R}^n:x_n=b_1x_1+\cdots+b_{n-1}x_{n-1}\}$.
And it is obvious $V=\{(x_1,\dots,x_n)^T\in\mathbb{R}^n:x_n=b_1x_1+\cdots+b_{n-1}x_{n-1}\}$.
Obviously $\mathbb{R}^{n-1}=\bigcup_{m_1\in\mathbb{Z},\dots,m_{n-1}\in\mathbb{Z}}[m_1,m_1+1]\times\cdots\times [m_{n-1},m_{n-1}+1]$ holds.
Let $m_1\in\mathbb{Z},\dots,m_{n-1}\in\mathbb{Z}$.
It is sufficient to show $$V':=\{(x_1,\dots,x_n)^T\in\mathbb{R}^n:x_n=b_1x_1+\cdots+b_{n-1}x_{n-1},(x_1,\dots,x_{n-1})^T\in [m_1,m_1+1]\times\cdots\times [m_{n-1},m_{n-1}+1]\}$$ has measure zero.
Divide $[m_1,m_1+1]\times\cdots\times [m_{n-1},m_{n-1}+1]$ into $k^{n-1}$ cubes whose width is $\frac{1}{k}$.
Then, it is easy to show that $V'$ is covered by rectangles whose total volume is $\frac{(|b_1|+\cdots+|b_{n-1}|)}{k}$.
So, $V'$ has measure zero.
