Lebesgue measure of a subspace of lower dimension is 0

Question

I'm currently reading Rudin's book Real and Complex analysis. In page 52 he says

To prove (e) let $T:R^k\to R^k$ be linear. If the range of T is a subspace Y of lower dimension then $m(Y)=0$.

I don't quite get that part. I'm guessing it could be deduced from the determinant formula given in page 54 but that seems like a circular reasoning. Is there a direct argument without using determinants and the decomposition of a linear map as a finite product of the type of linear maps given in page 54?

Answer 1

One could decompose the subspace into countably many unit hypercubes of the same dimension as the subspace. Then note that each hypercube is contained in an open set of arbitrarily small measure and hence has measure 0. Then use countable additivity.

Answer 2

Let $k < n$, where $k,n \in \Bbb N$ and $n > 1$. Let $Q \subset \Bbb R^n$ be a $k$-dimensional subspace of $\Bbb R^n$. Let $\{\mathrm e_i| 1 \le i \le n\}$ denote the unit vectors along the axes.

Consider the k-dimensional hyperplane $K$, spanned by the unit vectors $\mathrm e_i, 1 \le i \le k$. Define $\forall m \in \Bbb Z, \; I_m := [m,m+1)$.

We have

$$K = \bigcup_{m_i \in \Bbb Z} (\Pi_{i=1}^k I_{m_i} \times \Pi_{i=k+1}^n \{0\})$$

$K$ is covered by a countable collections of hypersquares with hyperarea $1$. Suppose $S_r$ is an enumeration of these hypersquares. Consider the hypercubes $C_r :=S_r \times \Pi_{k+1}^{n-1} (-\frac{1}{2},\frac{1}{2}) \times (-\frac{\varepsilon}{2^r},\frac{\varepsilon}{2^r})$, with hypervolume $\frac{\varepsilon}{2^r}$. We have:

$$K \subset \bigcup_{r = 1}^\infty C_r \\ \mu(K) \le \sum_{r=1}^\infty = \varepsilon $$

As $\varepsilon$ was chosen arbitrarily, $\mu(K) = 0$.

Let Q be an arbitrary k-dimensional subspace of $\Bbb R^n$ and $\mathrm e'_i, 1 \le i \le k$ be an orthonormal basis of $Q$, obtained through Gram-Schmidt process. Extend this basis to $\{\mathrm e'_i| 1 \le i \le k\} \cup \{\mathrm w_j, k+1 \le j \le n\}$ with $\{\mathrm w_j| k+1 \le j \le n\}$ being orthogonal to $Q$. Denote the space spanned by $\{\mathrm w_j, k+1 \le j \le n\}$, by $Q^{\bot}$. Applying Gram-Shmidt process to $\{\mathrm w_j | k+1 \le j \le n\}$, we obtain obtain the orthonormal set $\{\mathrm e'_j | k+1 \le j \le n\}$ spanning $Q^{\bot}$.

Let $x_i$ and $x'_i, 1 \le i \le n$ denote the coordinates of $\mathrm x \in \Bbb R^n$ with respect to $\{\mathrm e_i| 1 \le i \le n\}$ and $\{\mathrm e'_i| 1 \le i \le n\}$ respectively, i.e. $\mathrm x = \sum_1^n x'_i \mathrm e'_i = \sum_1^n x_i \mathrm e_i$.

For $x, y \in \Bbb R^n$, $\sum_1^n (x_i -y_i)^2 = \sum_1^n (x'_i -y'_i)^2$. Therefore we obtain the same metric space by defining distance function to be $d(x,y) = \sum_1^n (x'_i -y'_i)^2$ with the lebesgue measure defined on this metric coincinding with the original one and the same reasoning as in case of co-ordinate hyperplane K, gives $\mu(Q) = 0$.

Answer 3

First prove that if $B$ is a box in $\mathbf{R}^k$, then the set $T(B)=\{Tx:x\in B\}$ is lebesgue measureable, and $$m(T(B))=\det(T)|B|.$$ then uses the definition of outer measure, it easy to see that for any set $E$ $$m^*(T(E))\leq \det(T)m^*(E)$$ Note that if $m(E)$ is infinite, we can write $E=\cup_{n=1}^\infty E\cap B(0,n)$.

Answer 4

Here's how I think Rudin wants you to think about it.

First, it suffices to show that the k-dimensional measure of a lower dimensional hypersquare is always zero, because then the lower dimensional subspace can be written as a countable union of these squares.

You then consider partitions of the whole k-dimensional space into 2^-n boxes, and keep only those finitely many boxes that intersect your hypersquare. Each time you increment n, the union of the finite collection either stays the same or shrinks. All you need to show is that it shrinks to arbitrarily small measure as n goes to infinity. This is pretty obvious. For example, take any n, and note that there will come a point when at least half of the mass in that union will be discarded.

Answer 5

You can prove this by first proving the lemma that if for all i belonging to some set A:

1. Ei belong to M (where M is a sigma algebra)
2. mu(Ei) > 0 (where mu is a positive sigma-finite measure on M)
3. Ei and Ej are disjoint if i!=j and j belongs to A

then: A is at-most countable A.

This can be proved via Lema 6.9 (page 121 of Rudin's RCA) and then using the logic regarding at-most countability at the end of section 4.15 (page 84 of Rudin's RCA).

Once we have proved the above:

Let Y be a subspace of lower dimension in Rk (the Euclidian space of dimension k). There exists P belonging to Rk that does not belong to Y.

For all t belonging to R1 (the set of real numbers), let Qt = Y + P*t

if m(Y) > 0, then Qt satisfy the conditions of the lemma at the beginning. Therefore, m(Qt) > 0 on at-most countable t, which is a contradiction, because m(Qt) = m(Y+t) = m(Y) for all t belonging to R1 (which is uncountable).