# 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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What do you know about the topological properties of convex sets and the Lebesgue Measure on $\mathbb{R}^n$? – Tim Duff Oct 5 '12 at 6:51
There is the following way to prove it, which is probably not the most elementary. One can assume that the convex set has non empty interior. Then the projection onto the closure can be shown to be Lipschitz and differentiable everywhere, except at the boundary. By Rademacher's theorem, it implies that the boundary is Lebesgue negligible, and the measurability follows easily. – Ahriman Oct 5 '12 at 8:08
I suspect that if the interior is empty, then you can show that the interior of the affine hull is empty, and from this show that the set is contained in a hyperplane, and hence has measure zero. – copper.hat Oct 5 '12 at 8:10
If you can show that the boundary of your set $\partial C$ has measure zero by squeezing it and using convexity, then $\partial C \cap C$ is measurable by completeness of the Lebesgue measure. You can also reduce the problem to the bounded case by cutting off your set with a countable collection of larger and larger balls. – Nick Alger Oct 5 '12 at 9:59
Note that the answer depends on what you mean by "measurable".  A convex set need not be Borel measurable.  (Take the open unit ball together with a non-Borel subset of the unit sphere.) – Nate Eldredge Oct 5 '12 at 12:48

## 2 Answers

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.

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.

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So: proving that the boundary has measure zero shows not only Lebesgue measurability, but even that the Jordan content exists. – GEdgar Oct 5 '12 at 14:05
Yeah, for convex sets the interior coincides with the interior of the closure, so it makes sense that it would have Jordan measure as well. – Nick Alger Oct 6 '12 at 12:57

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.

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