# Integral Geometry

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.

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.