From Chapter X. of John B. Conway's textbook A Course in Functional Analysis:
1.10 Example Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space and let $\phi:X\to\mathbb C$ be a $\Omega$-measurable function. Let $\mathscr D=\{f\in L^2(\mu)|\phi f\in L^2(\mu)\}$ and define $Af=\phi f$ for all $f$ in $\mathscr D$. Then $A\in\mathscr C(L^2(\mu))$, $\operatorname{dom} A^*=\mathscr D$ and $A^* f=\bar \phi f$ for $f$ in $\mathscr D$.
From definition 1.3, $\mathscr C(L^2(\mu)) = \mathscr C(L^2(\mu),L^2(\mu)) =$ the collection of all closed densely defined operators from $L^2(\mu)$ to itself.
Neither I nor my professor find out why $\sigma$-finite matters here. I tried analyzing details:
By Theorem 3.11 of Papa Rudin, $L^p(\mu)$ is a complete metric space, for every $1<p<\infty$ and every positive measure $\mu$. So $L^2(\mu)$ is a complete Hilbert space.
$A$ is a well-defined linear operator $\mathscr D\to L^2(\mu)$.
$\mathscr D\subset L^2(\mu)$ is dense. In fact, if $f\in L^2(\mu)$, let $E_n = \{x\in X\mid |f(x)|<n\}$ and $f_n=\chi_{E_n} f$ for each integer $n$. Then $\int_X|\phi f_n|^2 \mathrm d\mu = \int_{E_n}|\phi f|^2 \mathrm d\mu < n\|f\|_2$, so $f_n\in\mathscr D$. But $f_n\to f$ in $L^2(\mu)$ as $n\to\infty$.
Since $A$ is densely defined, $A^*$ is well defined. $\langle Ak, h\rangle = \int k\phi \bar h = \int k \overline{\bar \phi h} = \langle k, \bar\phi h\rangle$. So $A^* f= \bar \phi f$ wherever definable. Its domain is
$$\{f\in L^2(\mu)|\bar\phi f\in L^2(\mu)\} = \{f\in L^2(\mu)|\phi f\in L^2(\mu)\} = \mathscr D \text.$$
Throughout the proof, $\sigma$-finite is never used. But I do understand that if the condition is dropped, under certain circumstances, $A$ can be $0$ for nonzero $\phi$ (which is sort of uncanny).
But I am not very confident with my argument. Is this correct, i.e. the statements in this example remains valid without the condition?