Is an integral against a probability measure in the convex hull of the range? This may be really obvious but I don't see it.  Let $f:\Omega \to \mathbb R^n$ be integrable with respect to a probability measure $\mu$.  Does it follow that $\int_\Omega f \, d\mu$ is in the convex hull (not just the closed convex hull) of the range of $f$?  
If the answer is yes, does this remain true if $\mu$ is merely finitely additive?
If $f$ is continuous and $\Omega$ is an interval, the answer is affirmative.
 A: Sorry!  It was rather easy, so perhaps it should be closed.
The second question clearly gets a negative.  Let $\mu$ be a finitely additive measure on $\mathbb Z^+$ that assigns zero to all finite subsets.  Let $n=1$.  Let $f(k)=1/k$.  Then $\int_{\mathbb Z^+} fd\mu =0$, but $0$ is not in the convex hull of the range.
The first question gets a positive.  It's clear for $n=1$.  Suppose $n>1$ and it's true for smaller dimensions.  Let $e=\int_\Omega fd\mu$.  Let $C$ be the convex hull of the range of $f$.  If $e\notin C$, there is a hyperplane $H$ through $e$ such that $C$ lies to one side of it.  Let $p$ be a normal of $H$ such that $p\cdot f(x) \le p\cdot e$ for all $x\in\Omega$.  Then $\int_\Omega (p\cdot f) d\mu = p\cdot e$ but $p\cdot f(x) \le p\cdot e$ for all $x$, so $p\cdot f = p\cdot e$ almost everywhere.  We can modify $f$ without changing its range so as to ensure $p\cdot f = p\cdot e$ everywhere.  But then the range of $f$ lies in the $(n-1)$-dimensional hyperplane $H$, and so by the $(n-1)$-dimensional case we have $e$ in the convex hull of the range of $f$.
