If $T:\mathbb R^k\to \mathbb R^k$ is linear and $\dim\operatorname{range}T < k$ then $m(\operatorname{range}T) = 0$ 
If $T:\mathbb R^k\to \mathbb R^k$ is linear and the range of $T$ is a subspace of lower dimension, i.e. $\dim\operatorname{range}T < k$ then prove that $m(\operatorname{range}T) = 0$, where $m$ is the Lebesgue measure on $\mathbb R^k$.

Rudin uses this in his proof of Theorem 2.20(e) of Real and Complex Analysis.

To prove this, I have two ideas in mind, but I haven't been able to complete any one of them. Following is my work, which I request you to help me complete (or suggest alternatives):
Attempt $1$ (Induction): I am trying to do induction on $k$. The base case $k=1$ is trivial since the only subspace of $\mathbb R$ of lower dimension is $\{0\}$ and $m(\{0\}) = 0$. Assume the statement holds for all $k < n$. To complete the induction step, we must prove it for $k = n$. Assume $\dim\operatorname{range}T < n$ for some linear $T:\mathbb R^n\to\mathbb R^n$. What do I do next?
Attempt $2$ (Direct Proof): I feel we can take arbitrarily thin sets (in the sense that their measure is small) which enclose the subspace $\operatorname{range}T$. By monotonicity of the Lebesgue measure, we should be able to arrange $m(\operatorname{range}T) < \epsilon$ for every $\epsilon > 0$, implying $m(\operatorname{range}T) = 0$. Some vague thoughts are about using cosets - which are essentially parallely-shifted copies of the subspace in consideration.
Attempt $3$ (Inner regularity of $m$): Let $Y = \operatorname{range} T$. It suffices to show that $m(K) = 0$ for every compact $K\subset Y$. Translation invariance of the Lebesgue measure has already been established (but not rotation yet, don't use that). Suppose $K\subset Y$ is compact. I want to cover $K$ by $\epsilon$-balls of full dimension, and use a monotonicity argument to deduce $m(K) = 0$. This seems hard to do without information on the exact form of $K$ (or $Y$).
Thank you!
 A: I'd go with that "parallel-shifted copies" idea:
With $U:=T(\Bbb R^k)$ pick $v\notin U$ and $w\in\Bbb Z^k$. Let $S_w=U\cap (w+[0,1]^k)$. Then the infnitely many translated copies $S_w+\lambda v$, $|\lambda|<\frac1{|v|}$ are pairwise disjoint, but all contained in $w+[-1,2]^k$ (which has finite measuer $3^k$). It follows that $m(S_w)=0$. As $U=\bigcup_{w\in\Bbb Z^k}S_w$ is the union of countably many zero sets, the desired result follows.
A: Let $l<k$ be the dimension of $Y=T(\mathbb R^k).$ Then there exists a bijective linear map $S:\mathbb R^l\to Y.$ If $K\subset Y$ is compact, then $S^{-1}(K)$ is a compact subset of $\mathbb R^l.$ It thus suffices to show $m_k(S(E))=0$ for every compact subset $E$ of $\mathbb R^l.$
Let such an $E$ be given. Choose $a>0$ such that $E\subset [-a,a]^l.$ Let $N$ be a large positive integer. Subdivide $[-a,a]$ into $2N$ subintervals $I_j$, each of length $a/N.$ Then
$$\tag 1 [-a,a]^l = \bigcup I_{j_1}\times I_{j_2} \cdots \times I_{j_l},$$
where $j_1, \dots ,j_{l}$ each run independently through $1,\dots, 2N.$ There are exactly $(2N)^l$ cubes on the right of $(1).$
Now every linear map on $\mathbb R^l$ is Lipschitz, so $S$ is Lipschitz. Thus there exists $M>0$ such that $|S(y)-S(x)|\le M|y-x|$ for all $x,y \in \mathbb R^l.$
Let $C$ be any of the cubes in the union in $(1).$ Then $\text {diam }C=\sqrt la/N.$ The Lipschitz condition then implies $\text {diam }S(C) \le M\sqrt l a/N.$ Take any point $p\in S(C).$ Then $S(C)$ is contained in a regular $k$-cube centered at $p,$ of side length $2M\sqrt l a/N.$  The volume of this last cube equals $(2M\sqrt l a/N)^k.$
We're in good shape: $S(E)$ is contained in the union of $(2N)^l$ $k$-cubes whose $k$-measures are each $(2M\sqrt l a/N)^k.$ The volume of this union is thus bounded above by
$$(2N)^l \frac{(2M\sqrt l a)^k}{N^k} = (2M\sqrt l a)^kN^{l-k}.$$
This is true for every $N.$ Because $l<k,$ the right side $\to 0$ as $N\to\infty.$ This shows the $k$-volume of $S(E)$ is $0,$ and we're done.
A: Indeed, pick $K$ compact in the image of $T$. Let $D$ be unit disk in $\mathbb{R}^{k-1}\subset \mathbb{R}^k$. Then:
Lemma: There exists a linear map $L:\mathbb{R}^k\to \mathbb{R}^k$ such that $K \subset L(D)$.
From the lemma, the result is deduced as follows. $L$ has finite operator norm i.e. maps any ball of radius $r$ inside a ball of radius $Mr$ for some fixed $M$. Given $\epsilon>0$, cover $D$ inside $\mathbb{R}^{k-1}$ by balls of total volume $<\epsilon/M^k$ ($D$ is inside the cube in $\mathbb{R}^{k-1}$ of side length 2; cut the cube into $N^{k-1}$ subcubes of side $2/N$, each covered by a ball of radius $\sqrt{k} 1/N$, so of volume $C/N^k$; as $N$ grows the total volume of all balls $C/N$ becomes arbitrarily small). Then the image of $D$ is covered by images of the balls, contained in balls of total volume $<\epsilon$. Thus we are done.
Proof of lemma is linear algebra. There exists a linear bijection $A$ between $Y$ and some $\mathbb{R}^l$ with $l\leq k-1$. The preimage of $K$ under this bijection is compact, so lies in some ball, and thus, after rescaling the map $A$, in the unit ball. Extending $A$ to a linear map from $\mathbb{R}^k$ by (setting the extension to be zero on the standard basis vectors $e_{l+1}, \ldots, e_{k}$) we get the map $L$ as wanted in the lemma.
