Let $M_{f}$ be the multiplication operator on a Hilbert space $L^{2}(X,\mu)$ defined as $$M_{f}: D(M_{f})\rightarrow L^{2}(X,\mu),\; M_{f}g=fg$$ where $D(M_{f})=\{g\in L^{2}(X,\mu) : fg\in L^{2}(X,\mu) \}$ and $f$ a measurable function $f:X\rightarrow \mathbb{C}$.
The essential range of $f$ is defined as $$ess\, ran\, f:=\{\lambda \in \mathbb{C}:\mu(f^{-1}(B_{\varepsilon}(\lambda)))>0 \text{ for all } \varepsilon >0\}$$ I want to proof that the following statements are equivalent:
i) $M_{f}$ is self-adjoint
ii) $ess\, ran\, f \subset \mathbb{R}$
iii) $f(x)\in \mathbb{R}$ almost everywhere
$i) \rightarrow ii)$
No idea yet but I'm thankful for suggestions.
$ii) \rightarrow iii)$
We need to show that $ess\, ran\, f$ implies the existence of a set $N$ with measure zero such that $f(x)\in \mathbb{R}$ for all $x \in X\setminus N$.
Set $A:=ess\, ran\, f$. Take $\lambda \in A^{C}$. By the definition of the essential range of $f$ there exists an $\varepsilon_{\lambda} >0$ with $\mu\left(f^{-1}(B_{\varepsilon_{\lambda}}(\lambda)\right)=0$. For $A^{C}$ countable we could conclude that $$ \mu\left( \bigcup\limits_{\lambda \in A^{C}}f^{-1}B_{\varepsilon_{\lambda}}(\lambda)) \right) = 0$$ using the $\sigma-$sub-additivity of measures.
For $A^{C}$ uncountable we cannot uses this. I assume that one could proof $\mu(A^{C})=0$ by finding a compact superset of $A^{C}$ which can be decomposed into a countable number of sets with measure zero. However I was not able to find such a set.
Any suggestions?
$iii) \rightarrow i)$
By assumption there exists a set $N$ wiht $\mu(N)=0$ such that $f(x)\in \mathbb{R}$ for all $x \in X\setminus N$. The self-adjointness then follows using $f(x)=\overline{f(x)}$ for $x \in X\setminus N$: $$\begin{align} \langle M_{f}g,h \rangle &= \int_{X}\overline{(M_{f}g)(x)}h(x) \mathrm{d} \mu(x)\\ &= \int_{X}\overline{f(x)g(x)}h(x) \mathrm{d} \mu(x)\\ &= \int_{X\setminus N}\overline{f(x)g(x)}h(x) \mathrm{d} \mu(x)\\ &= \int_{X\setminus N}\overline{g(x)}f(x)h(x) \mathrm{d} \mu(x)\\ &= \int_{X\setminus N}\overline{g(x)}(M_{f}h)(x) \mathrm{d} \mu(x)\\ &= \int_{X}\overline{g(x)}(M_{f}h)(x) \mathrm{d} \mu(x)\\ &=\langle g,M_{f}h\rangle \end{align}$$
Thus $\langle M_{f}g,h \rangle=\langle g,M_{f}h\rangle$.
I would try proving (i)<->(iii) and (ii)<->(iii).
You can get (i)->(iii) from the same identity you derived in (iii)->(i). If $M_f$ is self-adjoint, then you have $\int \overline{f} \overline{g} h = \int f \overline{g} h$. In other words, $\int (\overline{f} - f) \overline{g} h = 0$ for arbitrary $g,h$, from which it follows that $\overline{f} - f = 0$ almost everywhere, i.e. $f \in \mathbb{R}$ almost everywhere.
For your (ii)->(iii), you have an open cover of $\mathbb{C} \setminus \mathbb{R}$ by open sets of $\mu \circ f^{-1}$-measure zero. Now $\mathbb{C}$, like every separable metric space, has the Lindelöf property, so this open cover admits a countable subcover.
The implication (iii)->(ii) is immediate since $\mathbb{R}$ is a closed set.