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.