The measurability of convex sets

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.

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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):

• The boundary, closure, and interior of a set are always closed, closed, and open respectively, so they are always measurable.
• If $\partial C$ has measure zero, then $\partial C \cap C$ is measurable and has measure zero by completeness of the Lebesgue measure.
• Once you have measurability of $\partial C \cap C$, you have measurability of $C$ since,
$$C=(\partial C \cap C) \cup C^\circ.$$

(3):

• The proof that taking interiors preserves convexity is straightforward from the definitions but a little tedious. See lemma 4 here.

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