Essential range and integral means I am looking at the notion of essential range, which is defined for a function $f:(X,\mu)\rightarrow \mathbb{C}$ as the set $R_f$ of points $z\in\mathbb{C}$ satisfying
$$\forall\epsilon>0, \mu(\{x\in X:|z-f(x)|<\epsilon\})>0.$$
I consider the case where $f\in L^{\infty}(\mu)$. In this case, let's consider the set $A_f$ of complex numbers of the form
$$\frac{1}{\mu(E)}\int_E f\ d\mu,$$
where $E$ is a measurable set satisfying $0<\mu(E)<\infty$.
I conjecture that 

$A_f$ is included in the convex hull of $R_f$.

I have first proved that $\mu(X-f^{-1}(R_f))=0$, so that each element of $A_f$ is actually a ''mean'' over a set $E\cap f^{-1}(R_f)$, comprised of elements mapped into $R_f$. This convinces me that $A_f$ is included into the convex hull of $R_f$, but I can't manage to prove it. Is there a simple or elegant way to come to the result?
NB: I also proved that $R_f$ is compact with supremum $\|f\|_{\infty}$, may it be useful...
 A: Let's prove that there exists $\alpha\in R_f$ such that $\alpha\geq A_f$.
Suppose that this does not happen. Let $b=\|f\|_{\infty}$, then $A_f\leq b$. Consider the closed interval $[A_f,b]$. For any $z$ in it, there exists $\varepsilon_z>0$ such that the set $$\{x\in X\big| |z-f(x)|<\varepsilon_z\}=f^{-1}\left(B(z,\varepsilon_z)\right)$$ has measure zero, since any $z\in[A_f,b]$ is not in $R_f$. Then the balls $B(z,\varepsilon_z)$ form an open cover of $[A_f,b]$, which is compact, therefore there exist finitely many $z_1,\dots z_n\in[A_f,b]$ such that the balls $B(z_i,\varepsilon_{z_i})$ cover $[A_f,b]$. Therefore, $$[A_f,b]\subseteq\bigcup_{i=1}^nB(z_i,\varepsilon_{z_i})\Rightarrow f^{-1}\left([A_f,b]\right)\subseteq\bigcup_{i=1}^nf^{-1}\left(B(z_i,\varepsilon_{z_i})\right),$$ from which you see that $f^{-1}\left([A_f,b]\right)$ has measure zero. Therefore, $f(x)<A_f$ almost everywhere, so $$0=A_f-\frac{1}{\mu(E)}\int_Ef\,d\mu=\frac{1}{\mu(E)}\int_E(A_f-f)\,d\mu,$$ and since $A_f-f>0$ almost everywhere, you get a contradiction. So, there exists $z\geq A_f$ such that $z\in R_f$.
If $z=A_f$, you are done. If not, then $z>A_f$, and you can show similarly that there exists $w\leq A_f$ such that $w\in R_f$. If $w=A_f$, you are done; if not, then $w<A_f$ and $A_f$ is a convex combination of $z$ and $w$, so $A_f$ is in the convex hull of $R_f$.
