Sub-dimensional linear subspaces of $\mathbb{R}^{n}$ have measure zero.

Question

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!

Answer 1

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 $k<n$.

Answer 2

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 ;))

Answer 3

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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