The centroid of a set lies in its convex hull I've come across the following claim on a Wikipedia page about the centroid:

The geometric centroid of a convex object always lies in the object



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*How should I prove this claim for convex subsets of $\mathbb{R}^{n}$, where the centroid is usually located by using the integral formula?

*Is it true that more generally, the centroid of a (not necessarily convex) subset of $\mathbb{R}^{n}$ lies in its convex hull?
 A: Assume $X$ is a closed, convex subset of $\mathbb{R}^n$ with positive volume. The centroid of $X$, is by definition the normalized vector-valued integral:
$$c_X=\frac{1}{\text{Vol}(X)}\int_{X}x\,dx$$
The integral here is with respect to the Lebesgue measure in $\mathbb{R}^n$. Now, as with any integration process, you can find a sequence of partitions of $X$ into small pieces, say, $\{X_{i,m}\}_{i=1}^{N_m}$,
where $X_{i,m}$ are subsets of $X$ with positive volume, such that 
$$\lim_{m\to\infty}\sum_{i=1}^{N_m}x_{i,m}\text{Vol}(X_{i,m})=\int_X x\,dx$$
where $x_{i,m}$ are points in $X_{i,m}$. Dividing both sides by the volume of $X$ you get
$$\frac{1}{\text{Vol}X}\lim_{m\to\infty}\sum_{i=1}^{N_m}x_{i,m}\text{Vol}(X_{i,m})= c_X$$
Observe that for each $m$, the sum on the left hand side is a convex combination of the points $x_{i,m}$, because the volumes of $X_{i,m}$ sum up to the volume of $X$. Thus, since $X$ is convex, $c_X$ is a limit of points in $X$, and since $X$ is closed, $c_X$ belongs to $X$.
It is not true that in general the centroid of a subset $X$ of $\mathbb{R}^n$ belongs to $X$. Take, for instance the unit disc in the plane, punctured at its center. The centroid of this set is of course the removed center.
A: Actually the correct result is that the centroid of a convex object lies in the closure of the object.
Let $A \subseteq \mathbb{R}^n$ be convex set with positive Lebesgue measure and let $\overline{x}$ be the centroid of $A$. Assume that $\overline{x}\notin \overline{A}$. 
Recall this theorem:

Hahn-Banach separation theorem 
  Let $A$ and $B$ be nonempty disjoint convex subsets of a real normed space $X$. If $A$ is open, there exists a bounded linear functional $f \in X'$ and $t\in \mathbb{R}$ such that $$f(a) < t \le f(b) \quad\text{ for all } a \in A, b \in B$$

Then there exits $r > 0$ such that $A$ and $B(\overline{x},r)$ are disjoint. The theorem implies there exists a linear functional on $\mathbb{R}^n$ and $t \in\mathbb{R}$ such that $f(\overline{x}) < t \le f(x)$ for all $x \in A$. Then we have
$$f(\overline{x}) = \frac1{\lambda(A)} f\left(\int_Ax\,dx\right) =  \frac1{\lambda(A)} \int_Af(x)\,dx \ge  \frac1{\lambda(A)} \int_At\,dx = t$$
which contradicts $f(\overline{x}) < t$. Therefore $\overline{x} \in \overline{A}$.
