using measure theory in a proof of invariance of domain https://terrytao.wordpress.com/2011/06/13
Terence Tao, in the blog cited above, derives invariance of domain
from the fixed point theorem
without reliance on algebraic-topology. He does, however, appeal to measure theory
for a result that I am unable to replicate. If I understand him he asserts:
If $P$ is a polynomial in $
\mathbb{R}^n$ then there exists an $y\in \mathbb{R}^n$ and a $\delta > 0$
such that $P(x+\epsilon y)\neq 0$ for $x\in S^{n-1}$ and $0 < \epsilon < \delta$.
He says this follows from the fact that $S^{n-1}$ and $P(S^{n-1})$ both
have measure $0$ in $\mathbb{R}^n$.
Do I interpret him correctly? His assertions sound plausible but I need a hint or two to
make a convincing argument of my own. Is it true that an infinitesimal displacement is
sufficient to make 2 sets of measure 0 disjoint?
 A: Here is my understanding of his argument. First, let me recall some of the context of the proof. At this point, Tao has constructed a polynomial $P:\mathbb{R}^n \to \mathbb{R}^n$ that is non-zero on $\Sigma_1$, which is some compact set. We have another set $\Sigma_2$, which is of measure $0$, and we would like our polynomial $P$ to be non-zero on $\Sigma_2$ as well. (The precise definitions of $\Sigma_1$ and $\Sigma_2$ are of no importance for this argument.) Tao claims that we can ensure this by perturbing $P$ a little, that is by considering a shifted polynomial
$$
P_{\lambda}(y) = P(y) - \lambda
$$
for some arbitrarily small constant $\lambda \in \mathbb{R}^n$.
Actually, $P_{\lambda}$ will be non-zero on $\Sigma_2$ for almost every $\lambda$ (or, as Tao puts it, for generic $\lambda$). This means that the set $E$ of $\lambda$'s such that $0 \in P_{\lambda}(\Sigma_2)$ has measure $0$. Indeed, since $\Sigma_2$ has measure $0$ and $P$ is smooth, the image $P(\Sigma_2)$ has measure $0$. If $\lambda \in E$, then $0 \in P_{\lambda}(\Sigma_2) =P(\Sigma_2)- \lambda$. So $\lambda \in P(\Sigma_2)$. This implies that $E \subset P(\Sigma_2)$. Thus $E$ has measure $0$.
In particular, we can choose $\lambda$ as small as we like. We must choose it small enough so that $P_{\lambda}$ is non-zero on both $\Sigma_1$ and $\Sigma_2$. We know that $P(\Sigma_1)$ does not contain $0$. Since $\Sigma_1$ is compact, its image $P(\Sigma_1)$ is also compact. Hence there is some $\delta$ such that $||P(y)|| > \delta >0$ on $\Sigma_1$. Choosing $\lambda$ such that $|| \lambda || < \delta$, we are sure that $P_{\lambda}$ is also non-zero on $\Sigma_1$.
