Usual convex combination and the one with measure Let $X$ be a Borel measurable subset of $\Bbb R^n$ and let $\nu$ be a probability measure on $X$. Can we always find an integer $m$, points  $x_1,\dots,x_m\in X$ and coefficients $a_1,\dots,a_m \geq 0$ such that
$$
  \sum_{i=1}^m a_i = 1,\qquad \sum_{i=1}^m a_ix_i = \int_X x \;\nu(\mathrm dx).
$$
If not, is that true for closed or compact $X$?
 A: I believe this is true without putting any extra assumptions on $X$: 
Since $\nu$ is a probability measure, there exists a sequence of finite convex combinations of delta measures
$\{ \sum_{i=1}^{m_k} a_{k,i} \delta_{x_{k,i}} \}_k$ converging weakly to $\nu$,
with each $x_{k,i} \in X$. Therefore if $p = \int_X x~\nu(dx)$, we have
$$
p = \lim_{k \to \infty} \int_X x~d(\sum_{i=1}^{m_k} a_{k,i} \delta_{x_{k,i}}) = \lim_{k \to \infty} \sum_{i=1}^{m_k} a_{k,i} x_{k,i}.
$$
We know that $\sum_{i=1}^{m_k} a_{k,i} x_{k,i}$ belongs to the convex hull
of $X$ for each $k$, so $p$ belongs to the closed convex hull. If we already
know the convex hull of $X$ to be closed (for instance if $X$ is compact), then we're done.
Otherwise:
To see that $p$ must also belong to the convex hull, let $I$ be the interior of the convex hull.
If $p \notin I$, then one can choose $v \in \mathbb{R}^n$
and $\alpha \in \mathbb{R}$ so
that $v \cdot x < \alpha$ for all $x \in I$ and $v \cdot p \geq \alpha$. Then
the function
$$
x \mapsto v \cdot p - v \cdot x = \alpha - v \cdot x
$$
is nonnegative on $X$. But $\int_X (p-x)~\nu(dx) = 0$, so
$\int_X (\alpha-v\cdot x)~\nu(dx) = 0$. Therefore $v \cdot x = \alpha$ for
$\nu$-almost every $x \in X$, which means that $\nu$ is supported in the
hyperplane $H = \{ x \in X : v \cdot x = \alpha \}$.
By induction on the dimension, $p$ belongs to the convex hull of $X \cap H$
(which itself is contained in the convex hull of $X$).
The base case $n=1$ for the induction holds because the convex hull of $X$ is just
an interval. If $I$ denotes the interior of the interval and $p \notin I$, then
either $p-x$ or $x-p$ is nonnegative for all $x \in X$. Assume the former. Then
$\int_X (p-x)~\nu(dx) = 0$, which implies that $\nu = \delta_p$, and
this is impossible unless $p \in X$.
