A convex subset of a set has 'smaller' boundary than the set? Let $A\subseteq B$ be subsets of the real plane. Show that if $A$ is convex and $B$ is bounded, then the length of the border of $A$ $\leq$ the length of the border of $B$.
 A: *

*Given any convex closed subset $A$ of a Euclidean space $\mathbb R^n$, we can define the nearest-point projection $p_A\colon\mathbb R^n\to A$ by sending each point $x\in\mathbb R^n$ into $y\in A$ that attains the minimum $\min_{y\in A}|x-y|$. Due to the convexity of $A$, $p_A$ is well-defined. This concept generalizes the orthogonal projection onto a linear subspace.

*A notable property of orthogonal projections is shared by $p_A$: it is a contraction in the sense that $|p_A(x)-p_A(y)|\le |x-y|$ for all $x,y\in \mathbb R^n$. 

*It remains to show that $p_A$ maps $\partial B$ onto $\partial A$. The claim then follows from the fact that contractions do not increase length (or area, etc). 
The proofs of 2 and 3 are easier to carry out yourself than to watch/read someone else do it.
A: First, show that for a general half-space $H$, $B\cap H$ has smaller boundary than $B$.
If $A$ is sufficiently smooth, you can write $A$ as the intersection of $B$ with a countable family of half-spaces, say 
$$A=B\cap(\bigcap_{k=1}^\infty H_k).$$
From this (and lower semicontinuity of the perimeter functional, see any text of sets of finite perimeter), it follows that $A$ has smaller boundary than $B$.
