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I need the following theorem: Let $A$ and $B$ be two $n-1$ dimensional bounded smooth manifolds in $R^{n}$. Let $C$ be the set of all points in $A \cap B$ where the normals to $A$ and $B$ are unequal. Then $C$ has measure zero, in the $n-1$ dimensional Hausdorff measure.

I actually only need the cases $n \leq 4$, and where the normals are actually orthogonal.

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This is Thom-Boardman transversality, and is an elementary exercise, since the intersection is a submanifold of dimension $n-2$.

A pretty complete reference: https://en.wikipedia.org/wiki/Transversality_(mathematics)

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    $\begingroup$ Once you use the words "Thom-Boardman transversality", the claims of elementarity (?) become a little cloudy, since I, for one, had never heard of it before today. $\endgroup$
    – Igor Rivin
    Commented Feb 1, 2017 at 16:32
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    $\begingroup$ The intersection need not be of dimension $n-2$ if the intersection is not transversal. Take for example two smooth functions $f_1,f_2:\mathbb{R}^2\to\mathbb{R}$ such that, for $i=1,2$, $f_i$ is equal to $0$ on the disk of radius $i$ and is equal to $-\sqrt{(i+5)^2-x^2-y^2}$ outside the disk of radius $i+1$. Note that the graph of $f_i$ can be capped to a sphere of radius $i+5$. The resulting hypersurfaces in $\mathbb{R}^3$ have the a planar disk of radius $1$ in common. $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 1, 2017 at 19:24

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