Does the coarea formula hold for delta-function?

Question

Let $\Omega \subset \mathbb R^n$ be an open bounded domain, $u \colon \Omega \to \mathbb R$ be a Lipshitz function and suppose that $\nabla u (x) \neq 0$ for $x \in \Omega$. The coarea formula tells us that for any measurable bounded $f \colon \Omega \to \mathbb R$ we have $$ \int_\Omega f(x) \, dx = \int_{\mathbb R} \int_{u^{-1}(t)} f(x) \, \frac{dH^{n-1}(x)}{|\nabla u(x)|}\,dt, $$ where $dH^{n-1}$ is the Hausdorff measure. My question is whether this formula remains valid for $u \in C^\infty(\Omega)$ (with bounded derivatives) but with $f(x) = \delta(u(x))$? Is it true, in particular, that for $g \in C^\infty_c(\Omega)$ the following sequence of equalities holds? $$ \int_{u^{-1}(0)} g(x) \frac{dH^{n-1}(x)}{|\nabla u(x)|}=\int_{\mathbb R} \delta(t) \int_{u^{-1}(t)}g(x) \frac{dH^{n-1}}{|\nabla u(x)|} dt \\ = \frac{1}{2\pi}\int_{\mathbb R}\int_{\mathbb R} e^{its}\int_{u^{-1}(t)}g(x) \frac{dH^{n-1}(x)}{|\nabla u(x)|} ds dt \\ = \frac{1}{2\pi} \int_{\mathbb R} \int_\Omega e^{isu(x)} g(x) \, dx ds \quad ? $$