I'm trying to prove the following claim but I really don't know where to start:
Let $\left(X,\mathcal{F},\mu\right)$ be a finite measure space and let $f:\left(X,\mathcal{F}\right)\to\mathbb{R}$ be a measurable function such that $\int_{X}\left|f\right|d\mu<\infty$ . Suppose $C\subseteq\mathbb{R}$ is a closed set such that $\frac{1}{\mu\left(E\right)}\cdot\int_{E}fd\mu\in C$ for all $E\in\mathcal{F}$ with $\mu\left(E\right)>0$ . Show that $f\left(x\right)\in C$ for almost all $x\in X$ .
Help would be appreciated!
If $\mu(f^{-1}(\mathbb{R}\setminus C)) > 0$, then, since $\mathbb{R}\setminus C$ is the disjoint union of countably many open intervals, there is an open interval $(a,b)$ in the complement of $C$ with $\mu(f^{-1}(a,b)) > 0$. Then there is a compact interval $[\alpha,\beta] \subset (a,b)$ with $\mu(f^{-1}([\alpha,\beta])) > 0$. For $E = f^{-1}([\alpha,\beta])$, we then have
$$\frac{1}{\mu(E)}\cdot\int_E f\,d\mu \in [\alpha,\beta],$$
which contradicts the assumption that the averages always lie in $C$.