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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.

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up vote 2 down vote accepted

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 $f$.

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Note: The link in my questions to the affirmative answer in the continuous case will go to a theorem by Jankovic and Merkle that the integral is a convex combination of $n$ points in the range when $f$ is continuous. By Caratheodory's theorem, without assuming continuity we will have a combination of $n+1$ points in the range. So continuity lets one simplify the convex combination from $n+1$ to $n$ points. That helps me to understand the significance of the Jankovic-Merkle theorem. –  Alexander Pruss May 1 at 15:21

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