Jordan measure and singular matrix My question is,
Define $V$ as the set of all singular $n\times n$ matrices, we can think V as a subset of Euclidean space $\mathbf{R}^{n^2}$. I need to prove that V can be written as,
$$V=\bigcup^{\infty}_{i=1}A_i$$
where Jordan measure of each $A_i$ is zero in Euclidean space $\mathbf{R}^{n^2}$.
The key point is that the union is countable. I have tried to use the rational numbers to overcome this obstacle, but didn't prove it. 
Thanks in advance!
 A: Note that Jordan measure is only defined for bounded sets. So we are asked to prove that the Jordan measure of $V$ is "$\>\sigma$-finite".
It is instructive to consider first the case $n=2$. Then
$$V=\bigl\{(x_1,x_2,x_3,x_4)\in{\mathbb R}^4\>|\>x_1x_4-x_2x_3=0\bigr\}\ .$$
The function
$$f(x):=x_1x_4-x_2x_3$$
has gradient $\nabla f(x)=(x_4,-x_3,-x_2,x_1)$ which is $\ne0$ at all points of $\dot V:=V\setminus\{0\}$. It follows that $\dot V$ is a smooth hypersurface in ${\mathbb R}^4$.
Tile ${\mathbb R}^4$ with closed unit cubes $C_i$, $i\in{\mathbb N}_{\geq1}$. Then $A_i:=C_i\cap\dot V$ is a (maybe empty) set of zero Jordan measure, and so  is $A_0:=\{0\}$. It follows that one has $V=\bigcup_{i\geq0} A_i$ as required.
The case $n>2$ can be dealt with similarly. One has to stratify $V$ according to the rank of the constituent matrices: $V=\bigcup_{r=0}^{n-1} V_r$. (In the case $n=2$ we have $V_0=\{0\}$ and $V_1=\dot V$.)  Each $V_r$ is then a submanifold of ${\mathbb R}^{n\times n}$ of a certain dimension $<n^2$, and therefore possesses a representation $V_r=\bigcup_{i\geq1} A_{r,i}$ of the required kind.
