integral over the pre-image of a convex set

I really don't know how to start with this problem:

Let $\mu$ be a measure on $(\mathbb{C},\mathcal{B}(\mathbb{C}))$, where $\mathcal{B}(\mathbb{C})$ denotes the Borel $\sigma-$algebra of $\mathbb{C}$, and $f:\mathbb{C}\longrightarrow\mathbb{C}$ an integrable function. Let $C$ be a convex set (and possibly closed) such that $\mu(D)\in(0,\infty)$, where $D=f^{-1}(C)$. Prove that $$\dfrac{1}{\mu(D)}\int_{D}f\;d\mu \in C.$$

I thought it would be a good idea to start with simple functions, but $\mu(D)$ varies with $f$, and I don't know how to deal with it.

• You say "and possibly closed". If $C$ is not closed this is false. – Kabo Murphy May 7 '18 at 5:30
• Do you have a counterexample? – Ángela Flores May 7 '18 at 5:30
• @KaviRamaMurthy: I believe the result is true in finite dimensions for arbitrary convex $C$, with a rather laborious inductive proof. It is straightforward in one dimension. – copper.hat May 7 '18 at 14:49

To simplify notation, let $\nu A = \dfrac{1}{\mu(D)} \mu (A \cap D)$.

It is straightforward to show that $c= \int f d \nu \in \overline{C}$, the closure of $C$.

In the following I will treat $\mathbb{C}$ as $\mathbb{R}^2$.

If $\phi$ is a linear functional, then $\phi(c) = \int \phi(f(z)) d\nu(z) \le \int_{D}(\sup_{x \in C} \phi(x)) d\nu(z) = \sup_{x \in C} \phi(x)$, from which the inclusion follows (for two closed convex sets $A,B$ then $A \subset B$ iff $\sigma_A(h) \le \sigma_B(h)$ for all $h$, where $\sigma_C$ is the support functions of $C$).

It takes a little more work to show that, in fact, $c \in C$.

Suppose $c \in \overline{C} \setminus C$, then there is a non zero linear functional $\phi$ such that $\phi(c) \le \phi(x)$ for all $x \in C$. Since $\phi(c) = \int \phi(f(z)) d \nu(z)$, then we must have $\phi(f(z)) = \phi(c)$ for ae. $z \in D$. Let $L =\ker \phi$, then $f(z) \in L+\{c\}$ for ae. $z \in D$. There is an affine bijection $i:L+\{c\} \to \mathbb{R}$ such that $i(c) = 0$ and by choosing $i$ appropriately, we must have $0 \le i(f(z))$ for ae. $z \in D$. In a similar manner, we have $i(c) = 0 = \int i(f(z)) d \nu(z)$, and hence $i(f(z)) = 0$ for ae. $z \in D$. In particular, there is some $z_0 \in D$ such that $f(z_0) = c$, which contradicts the initial assumption. Hence $c \in C$.

This line of reasoning can be extended to $\mathbb{R}^n$ valued functions.

Note on the separation theorem:

One nice version is from Rockafellar's "Convex Analysis": Theorem 11.6. Let $C$ be a convex set, and let $D$ be a non-empty convex subset of $C$ (for instance, a subset consisting of a single point). In order that there exist a non-trivial supporting hyperplane to $C$ containing $D$, it is necessary and sufficient that $D$ be disjoint from $\operatorname{ri} C$.

• Honestly, I didn't understand. First, why can a linear functional "enter" under the integral sign? And second, why this inequality proves what I want? – Ángela Flores May 7 '18 at 5:13
• Are you familiar with the Hahn Banach separation theorem? – copper.hat May 7 '18 at 5:16
• Not really. I've never heard of that theorem. – Ángela Flores May 7 '18 at 5:19
• I don't know how to get the above result without it :-(. – copper.hat May 7 '18 at 5:21
• Anyway, thank you – Ángela Flores May 7 '18 at 5:24