Equality between volume and surface

Question

Let $u:\mathbb{R}^n \rightarrow [0,1]$. Does there exist some relation between the volume $|\{u > t\}|$ ($|\cdot|$ denotes the Lebesgue measure) and the surface $H^{n-1}(u = s)$, possibly in integral form?

EDIT: I had this question after learning the isoperimetric inequality since it seems that there should be some sort of equality between the volume and surface and not just an inequality.

Answer 1

The isoperimetric inequality: for $V\subset\Bbb R^n$, the measures of the body and the hypersurface verify $$m_{n-1}(\partial V)\ge n\,m_n(V)^{(n-1)/n}m_n(B_1)^{1/n}$$ with $B_1$ the unit ball.

Interesting integrals-related fact:

The $n$-dimensional isoperimetric inequality is equivalent (for sufficiently smooth domains) to the Sobolev inequality on $\Bbb R^n$ with optimal constant.

See Isoperimetric inequality

EDIT: the coarea formula gives a relation between the measure of a body composed of level sets and the measures of the level sets. $$ \operatorname{Vol}(M) = \int_M d\operatorname{Vol}_M = \int_{-\infty}^\infty\frac1{|\nabla f|}\operatorname{area}(f^{-1}(t))\,dt $$ (copypaste from the link, you can translate easily the notation)