TITLE: Spectrum and point spectrum of a multiplication operator
QUESTION [5 upvotes]: I am considering an arbitrary measure space $(X, \mathcal{A}, \mu)$ and the Hilbert space $H = L^2(\mu)$. The space $B = L^{\infty}(\mu)$ of essentially bounded functions can then act on $H$ as bounded linear operator by multiplication, i.e. if $g \in B$, then $M_g(f) = gf \in H$ for all $f \in H$.
I've managed to prove the following if $\mu$ is either $\sigma$-finite or locally finite.

For any $g \in B$, the spectrum of $M_g$ is equal to the essential range of $g$ and the point spectrum of $M_g$ is equal to $\{ \lambda \in \mathbb{C} \mid \exists A \in \mathcal{A} \text{ with } \mu(A)>0: f(A) = \lambda\}$.

I also know this doesn't hold for general measure spaces, because there is the following counter example:
 If we have $X = \{0,1\}$, $\mu(0) = 1$, $\mu(1) = \infty$, $g(0) = 0$ and $g(1) = 1$. Then $M_g=0$ but the essential range and point spectrum of $g$ is $\{0,1\}$.
My questions are the following then: 

Is there a more general class of measures for which this would hold? I know locally finite doesn't imply $\sigma$-finite and vice versa, so preferably it would hold them both.
If such a class doesn't exist, are there maybe counterexamples less trivial than the one I mentioned? 
Are there any non-trivial counterexamples where a part of the statement still holds, meaning the essential range isn't the spectrum but the point spectrum is still the same set described in the statement? Or maybe the other way around.

REPLY [3 votes]: Let $(X', \mathscr M', \mu')$ be an arbitary measure space and $x_\infty\notin X'$.
Set $X=X'\cup \{x_\infty\}, \mathscr M=\mathscr M'\cup \{A\cup \{x_\infty\}|A\in \mathscr M'\}$, 
and $\mu(A)=\mu'(A)$ if $A\in\mathscr M'$ and $ \mu(A)=\infty$ if $x_\infty\in A$.
Let $g(x)=1\forall x\in X'$ and $g(x_\infty)=0$. Then $0\in\operatorname{ess-ran}(g)$ but $M_g $ is invertible.
$\sigma_p(M_g)=\{\lambda\in \mathbb{C}|\exists A\in X~ \text{with} ~ \mu(A)>0:f(A)=\lambda\} $ for an arbitary measure space.