Lebesgue measure of sets in $\mathbb{R}^N$

Question

Let $\Omega\subseteq \mathbb{R}^N$ be an open, bounded and connected set (it can be assumed with smooth boundary if necessary).

Consider $\phi:\Omega\to\mathbb{R}$, $\phi\in C^1(\overline{\Omega})$ (the Banach space of continuous functions on the closure of $\Omega$ having continuous partial derivatives on the closure of $\Omega$) satisfying:

$\bullet\ \Vert \phi\Vert_{C^1(\overline{\Omega})}\leq 1$

$\bullet\ \phi^{-1}(0)=\{x\in \Omega\ |\ \phi(x)=0\}\neq \emptyset$

$\bullet\ \nabla\phi(x)\neq 0$ for all $x\in\phi^{-1}(0)$.

Is it true that for each $\varepsilon>0$ there is a constant $\delta=\delta(\varepsilon)>0$ (depending just on $\varepsilon$) such that:

$$\lambda (\phi^{-1}(-\delta,\delta))<\varepsilon $$

?

I denote with $\lambda$ the Lebesgue measure on $\mathbb{R}^N$.

P.S. Our assumption guarantee that any level set $\phi=c$ has a null Lebesgue measure.

Answer 1

The answer is no even for $N = 1$ and $\Omega = (-1,1)$, thus $\bar \Omega = [-1,1]$. Let $\phi_1 \colon [-1,1] \to [-1,1]$ be any increasing function (which may even be in $C^\infty$) with $\phi_1(0) = 0$ and $\phi_1'(x) > 0$ for all $x \in \bar \Omega$. Let $\phi_t := t \cdot \phi_1$. Then given $\epsilon > 0$ and $C > 0$ any fixed constant we get $\lambda(\phi_t^{-1}((-\epsilon,\epsilon)) = 2$ for $t < \epsilon$. Note that the assumptions are fulfilled for any $t \in (0,1)$.