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Suppose that we have given any measure space $(\Omega, \Sigma, \mu)$ (such that $L^2(\mu)$ is not trivial) and consider the multiplication operator $M_g : L^2(\mu) \rightarrow L^2(\mu)$ given by $M_g (\phi) (x) = g(x) \phi (x)$. Here $g$ is a bounded, measurable function from $\Omega$ to $\mathbb{C}$.

Now suppose that $M_g$ is injective, i.e. we have that for every $f \in L^2(\mu)$

$$ g \cdot f = 0 $$

implies

$$ f = 0. $$

Now we look at the set $M=\{ \omega \in \Omega : g(\omega)=0 \}$ where $g$ vanishes.

Question: Does $M$ always have $\mu$-measure zero?

Partial Answer:

Suppose that $\mu$ is $\sigma$-finite and suppose that $M$ does not have measure zero. Then we can find a subset $M_0$ of $M$ which has positive but finite measure, i.e. the characteristic function $\chi_{M_0}$ is in $L^2(\mu)$. But we have $M_g \chi_{M_0} = 0$, so by the assumption $\chi_{M_0}$ is the zero function. This is a contradiction.

The problem with this is that in general we do not have $\sigma$-finiteness of $\mu$. So this argument won't work.

Maybe some context: This statement is needed in a proof of the spectral theorem for unbounded self-adjoint operators on a (not necessarily separable) Hilbert space. The function $g$ is given by the spectral theorem for bounded, normal operators, i.e. we are given a non-trivial Hilbert space $H$ and a normal operator $A$ on it which then is unitarily equivalent to multiplication by $g$ on $L^2(\mu)$. If the Hilbert space were separable, the function $g$ would live on a $\sigma$-finite measure space; but this is not assumed.

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  • $\begingroup$ What if you take a one point space with infinite measure? $\endgroup$ – Jonas Meyer Jan 19 '17 at 16:48
  • $\begingroup$ You're right. That would be a counterexample. In this case we'd have $L^2(\mu)=\{ 0 \}$. One should exclude this case because in the proof on which this question is based it is assumed that $L^2(\mu)$ is isomorphic to an abstract non-trivial Hilbert space. $\endgroup$ – NiU Jan 19 '17 at 23:08
  • $\begingroup$ Your partial answer works under the assumption that $\mu$ is semifinite (that is, for every $S\in \Sigma$ with $\mu(S) = +\infty$ there is an $S'\subset S$, $S'\in \Sigma$, with $0 < \mu(S') < +\infty$). $\endgroup$ – Daniel Fischer Jan 21 '17 at 17:33

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