Equality of integrals of differential forms

Question

I have two $(n-1)$-forms $\omega_{1}$ and $\omega_{2}$ on $\mathbb{R}^n$ and a smooth function $g(x) \colon \mathbb{R}^n \to \mathbb{R}$ ($dg$ doesn't vanish anywhere) such that $dg \wedge \omega_1 = dg \wedge \omega_2$ holds. Let $M = \{x \in \mathbb{R}^n \mid g(x) = 0 \}$. Is it true that $$ \int\limits_{M} \omega_1 = \int\limits_{M} \omega_2 $$

Answer 1

Considering $\omega = \omega_1 - \omega_2$ we have to show that if $dg \wedge \omega = 0$ then $\int_{M} \omega = 0$.

1. $dg(X_p) = 0$ iff $X \in T_{p}M$ because $dg(X_p) = \langle \nabla g(p), X \rangle$.
2. Let $X^1,...,X^n$ be such that $X^1,...,X^{n-1}$ are from $T_pM$ and $X^n$ is transversal to $T_pM$. Then $$ 0 = dg \wedge \omega (X^1_p,...,X^n_p) = \pm dg(X^n_p) \omega(X^1_p,...,X^{n-1}_p) $$ but $dg(X^n_p) \neq 0$ then $\omega$ vanishes on tangent spaces to $M$, hence $\int_{M} \omega = 0$.