Let $M\subset \Bbb{R}^{p}$ a surface of class $C^2$. I need to prove that, if $2\dim M < p$, so the set $$X=\bigcup_{x\in M} T_x M$$
has measure zero in $\Bbb{R}^{p}$.
My definition of "measure zero in $\Bbb{R}^{p}$" is the following:
A set $X$ has measure zero in $\Bbb{R}^{p}$ if, for all $\varepsilon > 0$, exists a covering of $X$ by rectangles $Q_1,Q_2,\dots$ such that $$ \sum_{i\in\Bbb{N}}v(Q_i)<\varepsilon.$$
Here $v(Qi)$ denotes the volume of the rectangle $Q_i$.
Let $m=\dim M$. So, for all $x\in M$, exists an open set $U_x\subset \Bbb{R}^{m}$ and a parametrization $\varphi_x:U_x\rightarrow V_x\subset M$ of class $C^2$.
So one can parameterize $T_xM$ by the following function:
$$\psi_x: \Bbb{R}^m \rightarrow T_x M,\quad \psi_x(v)=\varphi'(x)\cdot v.$$
Particularly, $\dim T_x M = m<p/2$.
Ok... and that is all that I know. Given $\varepsilon >0$, how can I construct the covering by rectangles whose volumes sums to a number less than $\varepsilon$ ?