Integral Geometry

Question

Consider a smooth curve $\gamma$ of finite length in the unit square $[0,1]\times[0,1]$. Is the following statement correct?

There exists a Lebesgue null set $N$ such that for all $x\in [0,1]\setminus N$ the set $\{y\in[0,1]:(x,y)\in\gamma\}$ has only finitely many points.

Generalization:

Consider a smooth manifold $M$ in the $n$ - cube $[0,1]^n$. Assume that the $n-1$ dimensional Hausdorff measure ${\cal{H}}^{n-1}(M)$ is finite. Is the following statement correct?

There exists a set $N$ with ${\cal{L}}^{n-1}(N)=0$ such that for all $x\in [0,1]^{n-1}\setminus N$ the set $\{y\in[0,1]:(x,y)\in M\}$ has only finitely many points. Here ${\cal{L}}^{n-1}$ denotes the $n-1$ dimensional Lebesgue measure.

Answer 1

By the theorem of Sard (Morse 1939, Sard 1942) the set of all singular values of a $C^k$-mapping $M\to N$ is of Lebesgue measure 0 in $N$ if $k> \max\{0, \dim(M)-\dim(N)$. Here $M$ and $N$ are smooth manifolds. A point $y\in N$ is a regular value of $f$ if $T_xf$ is surjective for all $x\in f^{-1}(y)$. If not, then $y$ is a singular value. Note that $y\in N\setminus f(M)$ is regular. Both statements follow from this.