Lebesgue measure of a subspace of lower dimension I'm aware that this question has been asked before by Zero.
However, there is a step in the answers provided by others that I've yet to understand.
Let $T:\mathbb{R^n} \to \mathbb{R^n}$ be linear. If the range of $T$ is a subspace $Y$ of lower dimension, then $m(Y) = 0$.
I understand that the subspace $\mathbb{R}^k \subset \mathbb{R}^n$ of dimension $k < n$ spanned by standard bases ${e_1,...,e_k}$ has measure 0. But how does this extend to a subspace is spanned by other vectors ${b_1,...,b_k} = {U(e_1),...,U(e_k)}$ for some linear map $U$?
i.e. Prove measure of $T(X) = Y \subseteq U(\operatorname{span}({e_1,...,e_k}))$ is 0.
Below Zero's question, I believe somitra spoke of the Gram-Shmidt process to transform $b_i$ into an orthonormal basis, which seems relevant, but I have a rather limited background in linear algebra, so is there some elementary alternative?
 A: 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.
Related Questions:
Sub-dimensional linear subspaces of $\mathbb{R}^{n}$ have measure zero.
Lebesgue measure of a subspace of lower dimension is 0
Any linear subspace has measure zero
Every subset of a subspace of $\mathbb{R}^n$ of dim $<n$ has measure 0
proof that the lebesgue measure of a subspace of lower dimension is 0.
