Let $f:R^m\to R^n$ be a $C^k$ function, where $k>\max \{0,,m-n\}$. Then the lebesgue measure of the set of singular values is $0$.
I've been trying to prove this. I came up with the following argument: let $n=1$, and $m>1$. Then the set of singular points are those where $\frac{\partial f}{\partial x_i}=0$, where $1\leq i\leq m$. These describe hyperplanes in $R^m$, and generally they will be transverse to each other. Hence, the intersection of all these hyperplanes would of measure $0$, and any $C^1$ funtion maps a set of measure $0$ to a set of measure $0$. Hence, the set of singular values has measure $0$.
Is there any way to generalize this argument for $n\neq 1$? Is this where the function being $C^k$ is used (as opposed to just using the $C^1$ property of the function)?