Given $Y$ is a proper subspace of $X$ ($\dim Y < k$), then $m(Y) = 0$. It's intuitively clear, but it's not trivial!!
It is true for $R^k$, because the Lebesgue measure is outer-regular (ie: for every $A$ measurable, $m(A) = \inf\{m(E): A\subseteq E,\, E\text{ open}\}$.
$Y$ is a proper subspace of $X$, by linear transformation, you can move it to another subspace $E$ with the last coordinate is $0$. $E = \{(x_1,x_2,\ldots ,x_k): x_k =0\}$, and it is sufficient to show $m(E) = 0$. (" the sufficient part here " is actually not trivial: there is theorem in Rudin said that, if you have a linear transformation $T$, for any set $E$, $m(T(E)) = \lambda m(E)$, $\lambda$ is constant, and I need this theorem to use this linear transformation.)
The set of $S = \{(x_1,x_2,\ldots ,x_{k-1},0)\}$, with every $x_i$ rational, is countable, and can be listed as $y_1,y_2,y_3,\ldots$
For each point $y_i = (x_1,x_2,...x_{k-1},0)$, we put it inside a box $W_i$ $W_i = (x_1-1/2,x_1+1/2) \times (x_2-1/2,x_2+1/2)\times \ldots\times (x_{k-1}-1/2,x_{k-1}+1/2) \times(-\epsilon/2^{i+1},\epsilon/2^{i+1})$
The volume/measure of $W_i$ is $\epsilon/2^i$.
$E = \bigcup W_i \implies m(E) \le \sum\limits_{i=1}^\infty m(W_i) = \epsilon \implies m(E) =0.$