Lebesgue measure of a subspace of lower dimension is 0

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?

Solution 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.