Suppose $(X, \mu)$ is a measure space. If we additionally assume that $X$ is $\sigma$-finite, then for $f\in L^\infty(X, \mu)$, the spectrum of the multiplication operator $M_f$ on $L^2$ equals the essential range of $f$.

Although $\sigma$-finiteness plays a role in the proof that I know for this statement, I'm not sure if the assumption can be dropped with a different proof. I wonder if anyone can provide a proof or a counterexample for this. (My feeling is that a counterexample should exist if $X$ is not $\sigma$-finite.)


  • $\begingroup$ Do you mean $\sigma$-finite? For a general measure space, there is no topology, so it doesn't quite make sense to talk about $\sigma$-compactness. $\endgroup$
    – Mike F
    Commented Aug 26, 2016 at 4:08
  • $\begingroup$ @MikeF Sure! Thanks for catching the typo! $\endgroup$
    – EPS
    Commented Aug 26, 2016 at 4:16

1 Answer 1


You can screw things up a bit if you allow, for instance, points which have infinite measure. Consider, e.g., $X = \{a,b\}$ and $\mu$ given by $\mu(\{a\}) = \infty$, $\mu(\{b\})=1$. Note that, for this measure space, the L-2 functions must vanish on $a$.

Consider the function $f \in L^\infty(X)$ with $f(a) = 1, f(b)=0$. The corresponding multiplication operator $M_f$ is the zero operator on $L^2(X)$, so the spectrum of $M_f$ is just $\{0\}$. However, the essential range of $f$ is $\{0,1\}$.


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