Averages of a Function Let $(X,M,\mu)$ a measure space and $f:X \rightarrow \mathbb{C}$ a function in $L^{\infty}(\mu)$. Define $A_f$ as the set of all averages
\begin{equation}
\frac{1}{\mu(E)} \int_{E} f d \mu
\end{equation}
where $E \in M$ with $ 0 < \mu(E) < \infty$. An exercise in Rudin, Real and Complex Analysis, asks whether there exists a
measure space such that the set $A_f$ is convex for all $f \in L^{\infty}(\mu)$.
Obviously, a space $X$ containing only one point gives an affirmative answer.
But does there exists a non trivial example? That is, a measure space $X$ in which
$A_f$ contains more than one point for some $f$?
Even in the special case in which $X=\mathbb{R}$ and $\mu$ is the Lebesgue measure,
I don't know whether $A_f$ is convex for all $f \in L^{\infty}(\mu)$ or not.
Do you have some idea?
Thank you very very much for your help.
My Best Regards,
Maurizio Barbato
PS If we define the essential range of $f \in L^{\infty}(\mu)$ as the set of all complex numbers $w$
such that 
\begin{equation}
\mu({x: |f(x) - w| < \epsilon}) > 0
\end{equation}
for all $\epsilon > 0$, then it is easy to prove that $R_f$ is compact. Moreover,
the closure of $A_f$ is contained in the convex hull of $R_f$, and in the special
case in which $X=\mathbb{R}$ and $\mu$ is the Lebesgue measure, we have 
$cl(A_f)=conv(R_f)$. This was proved by prof. David Ullrich. See the posts
http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1465722&messageID=5234768
and 
http://mathforum.org/kb/thread.jspa?forumID=253&threadID=559133&messageID=1677532

When $\mu$ is nonatomic and $f$ is essentially bounded, $A_{f}$ is always convex. This follows painlessly from Liapunoff's convexity theorem, which states that the range of a nonatomic vector measure is convex (and closed).
In particular, the set $E$ in $\mathbf R^{3}$ of triples of the form $(\mu(A),\int_{A} f\:d\mu)$ is convex. Let $C$ be the set in $\mathbf R^{3}$ consisting of nonnegative scalar multiples of members of $E$; note that $C$ is convex. The plane $P$ in $\mathbf R^{3}$ of points of the form $(1,y,z)$ is also convex, so the intersection $P\cap C$ is convex. But $P\cap C$ is a copy of $A_{f}$, and so $A_{f}$ is convex.
This is Lemma 3.3 of this paper:

Zhou, Lin. Strictly Fair Allocations in Large Exchange Economies. 1991

https://cowles.yale.edu/sites/default/files/files/pub/d09/d0972.pdf
