I don't understand a presumably easy argument in my textbook.
Let $L: \mathbb R^n \to \mathbb R^m$ be a linear map, $n \geq m$, $A \subset \mathbb R^n$ $\lambda^n$-measurable. We assume that $\dim L(\mathbb R^n) < m$. The author claims that $A \cap L^{-1}\{y\} = \emptyset$ for $\lambda^m$-a.e $y \in \mathbb R^m$ and consequently $\mathcal H^{n-m}(A \cap L^{-1}\{y\}) = 0$ $\lambda^m$-a.e.
Why does this hold? Can someone enlighten me?
P.S.: $\mathcal H^{n-m}$ stands for the (n-m)-dimensional Hausdorff measure.
Since $L$ is linear, $L(\mathbb R^n)$ is a subspace of $\mathbb R^m.$ Since $\dim L(\mathbb R^n) < m,$ $L(\mathbb R^n)$ is a proper subspace of $\mathbb R^m.$ It's "easy" to see that if $X\subseteq \mathbb R^m$ is a proper linear subspace, then $\lambda^m(X) = 0$ (we can assume that $X$ is spanned by some coordinate vectors of $\mathbb R^m$ by applying an orthogonal linear map, then we can cover $X$ with a sequence of "blocks" (i.e. cartesian products of compact intervals) of arbitrarily small total Lebesgue measure). So we conclude that $\lambda^m(L(\mathbb R^n)) = 0$ and that in fact we have $$ L^{-1}(y) = \emptyset\qquad for\ \lambda^m-a.e.\ y\in\mathbb R^m, $$ namely for all those $y\in\mathbb R^m$ not contained in $L(\mathbb R^n).$