Let $M \subset \mathbb{R}^{p}$ be a surface of class $C^{2}$. If $2\dim M < p$ then the union $X = \bigcup_{x \in M}T_{x}M$ of tangent vetorial spaces to $M$ in all points is a null measure set in $\mathbb{R}^{p}$.
I don't know if it is correct, but I tried to use Sard Theorem. For this, I have to show that $X$ is the set of singular points of $f$ where $f:M \to N$ is a $C^{1}$ function between same dimensions surfaces. This seems strange to me, I'm probably not on the right way. Can anybody help me?
EDIT (Second idea). We know that $M$ can be covered by coordinated neighborhoods $U_{i}$. Let $U = \bigcup U_{i}$ where $U_{i}$ is an open of each parameterization $g_{i}$. Define $f: U \times R^{m} \to \mathbb{R}^{p}$ by $f(q,v) = Dg_{i}q(v)$ (the differential of $g_{i}$ in $q$ applied in $v$). So, $U\times \mathbb{R}^{m}$ is open and has dimension $2m<p$. Since $f \in C^{1}$, $f(U \times \mathbb{R}^{2m})$ is a null set and $f(U \times \mathbb{R}^{m}) = \bigcup T_{x}M$. But, the problem is: the $U_{i}$ need not be disjoint so, $f$ is not well-defined.
I have no more ideas.
In $\mathbb{R}^p$, being a null-measure set is a local property (that's actually how you define null measure in an arbitrary manifold !) : if $X$ is a set such that each $x\in X$ has a neighbourhood $U$ in $\mathbb{R}^p$ such that $U\cap X$ has measure $0$, then $X$ has measure $0$.
Here that's what you have: locally $TM$ is the set of singular points of a $C^1$ map and the dimensions allow you to apply Sard's theorem locally. By the above claim, this suffices.
To prove the above claim, note that if $U$ is such an open neighbourhood for $x\in X$,then one may find a rational vector $q_x \in U$ and a rational number $r_x\in \mathbb{Q}$ such that $x\in B(q_x, r_x)\subset U$. Then $X\cap B(q_x, r_x)$ also has measure $0$ and $X\subset \displaystyle\bigcup_{x\in X}B(q_x, r_x)\cap X$, and the right hand side is actually a countable union of measure $0$ sets (countable because we chose the $q_x, r_x$ rational), so $X$ also has measure $0$