How to prove the measurability of convex sets in $R^n$ ? I have seen a proof, but too long and not very intuitive.If you have seen any, please post it here.
Let $C$ be your convex set, and assume without loss of generality(1) that it contains zero as an interior point and is bounded.
The question boils down to showing that $\partial C$ has measure zero(2), which can be shown by squeezing the boundary between the interior $C^\circ$, and a slightly expanded version of the interior, $\frac{1}{1-\epsilon}C^\circ$.
Let $p \in \partial C$. Since $0$ is an interior point, by convexity the point $q:=(1-\epsilon)p$ lies in the interior of the cone $K:=\{sp + (1-s)x: x \in B_r(0) \}$, and therefore $q \in C^\circ$. But then $p=\frac{1}{1-\epsilon}q \in \frac{1}{1-\epsilon}C^\circ$.
Thus
$$\partial C \subset \frac{1}{1-\epsilon}C^\circ.$$
Since for any set the boundary and the interior are disjoint,
$$\partial C \subset \frac{1}{1-\epsilon}C^\circ \setminus C^\circ.$$
Since the interior of a convex set is convex(3) and $C^\circ$ contains zero, $C^\circ$ is contained in it’s dilation:
$$C^\circ \subset \frac{1}{1-\epsilon}C^\circ.$$
Finally, since we have assumed $C^\circ$ is bounded, the measure of the boundary,
$$\lambda(\partial C) \le \lambda(\frac{1}{1-\epsilon}C^\circ \setminus C^\circ) = (\frac{1}{1-\epsilon})^n\lambda(C^\circ)-\lambda(C^\circ),$$
can be made as small as desired by taking $\epsilon \rightarrow 0$.
Tying up loose ends:
(1):
If the set is not bounded, cut it off with a countable collection of successively larger balls. Since the countable union of measurable sets is measurable, this suffices.
If the set $C$ contains some interior point, translate the set so that the interior point is at zero. Since the Lebesgue measure is translation invariant, this suffices.
If the set $C$ contains no interior points, then all it’s point must lie within a $n-1$ dimensional plane, otherwise $C$ would contain a n-tetrahedron (simplex), and a simplex contains interior points. Thus $C$ would lie within a measure zero set and the result is trivial.
(2):
(3):
Edit: To add, the approach in the answer here:
Why does a convex set have the same interior points as its closure?
is similar to the reasoning in my post, and shines some light onto what’s going on. The technique there could be adapted easily to prove the result here as well, and you would get a similar proof.
A relatively simple proof of a more general result (measurability with respect to every complete product measure of $\sigma$-finite Borel measures) can be found in
Lang, Robert A note on the measurability of convex sets. Arch. Math. (Basel) 47 (1986), no. 1, 90–92.