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](http://arxiv.org/pdf/0705.4200v1.pdf).



Is an integral against a probability measure in the convex hull of the range?
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 $f$.


