Let me motivate my question with this example.

The volume integral of a ball $\int_{B(0,R)} dx$ can be written as an integral over the surface of balls, i.e.

$$\int_{B(0,R)} dx = \int_0^R \int_{\partial B(0,r)}dS dr.$$

This shows, that the derivative w.r.t. $R$ is just the surface-integral

$$\frac{d}{dR} \int_{B(0,R)} dx = \int_{\partial B(0,R)}dS$$

Now, what happens if we generalize this: Let $F: \mathbb{R}^n \rightarrow \mathbb{R}$ and suppose that $\lambda( F^{-1}(-\infty, R]) < \infty$ for all $R$ then I see two ways to generalize my example:

(1) $\mu((-\infty,R]):=\int_{F^{-1}(-\infty, R]} dx$ defines a measure. The measure $\mu$ is a.c. with respect to the Lebesgue measure, i.e. $\mu(A)= \int_{A} fdx$ for some measurable $f$. In my example, the function $f(r)= \int_{\partial B(0,r)} dS.$

(2) But the most obvious generalization would be probably: $$\int_{F^{-1}(-\infty, R]} dx = \int_{-\infty}^{R} \int_{F^{-1}(\{r\})} \frac{1}{||\nabla F(x)||}dS(x) dr?$$

Apparently, (1) is more general than (2). I suspect (although do not know a proof of this) that (2) holds for submersions. Now my question is: How big is the difference between (1) and (2), i.e. does (1) only hold, if the representation in (2) holds almost everywhere? Under what precise conditions does (2) hold?