Let $I=[0,1]^d$ be the closed $d$-dimensional unit cube, and let $f:I \to \mathbb{R}$ be a $C^2$ function. Its graph $f(I)$ can be embedded naturally as a manifold in $\mathbb{R}^{d+1}$.
Denote by $B_h ({\bf x})$ the ball of radius $h>0$ centered at ${\bf x} \in \mathbb{R}^{d+1}$, and denote by $\sigma $ the $d$-dimensional surface measure on $f(I)$ induced by the Lebesgue measure on $\mathbb{R}^{d+1}$.
My question: Is it true that $\sigma \left( f(I) \cap B_h (x) \right) = O(h^d)$?
In other words: If $f(I)$ is the graph of a "nice" function, it seems reasonable that it has a finite curvature, and so a "small" $d+1$ dimensional ball will only contain something similar to a $d$-dimensional ball on the manifold. I have a sketch of a proof, but a reference for a textbook/paper book would be much better.