For a given positive real number $x$, construct a smooth function $f:\mathbb{R}^n\to \mathbb{R}^m$ such that the Lebesgue measure of the critical points is $x$.
I was trying to do by using the bump function idea. First thing, enough to construct a function $f:\mathbb{R}^n\to \mathbb{R}$. If we construct a bump function which is taking value $1$ on a ball whose measure is $x$ then the critical values are the ball and the points where $f$ is zero. 
But how to get the critical points whose measure is $x$?
Take the same kind of function which is not $0$ at infinity. If $r$ is such that $B_0(r)$ has measure $x$, the function looks like this:
For an explicit formula in the case $f:\Bbb R^n\to \Bbb R$, take $h:\Bbb R\to \Bbb R$ such that $h(x)=1$ if $x<0$ and $h(x)=e^{-1/x^2}$ if $x\geq 0$. The function $h$ is smooth with set of critical points $\Bbb R_-$. Then take $$f(x)=h(\Vert x\Vert^2-r^2).$$ Because $$\nabla f(x)=2h^\prime(\Vert x\Vert^2-r^2)\cdot x,$$ the set of critical points of $f$ is $B_0(r)$.