Properties of the multiplication operator, self-ajointness Let $(\Omega, \Sigma,\mu)$ a measurable space, $f:\Omega\to \mathbb{R}$ $\mu$-measurable. a.My first question: What does "f $\mu$-measurable" mean?I only know, what it means that "f is measurable" but including the measure "$\mu$", $\mu$-measurability?Now consider the operator $$M_f:D(M_f)=\{u\in L^2(\Omega); f\cdot u\in L^2(\Omega)\}\subseteq L^2(\Omega)\to L^2(\Omega),\; u\mapsto f\cdot u$$
$D(M_f)$ means the domain of the multiplicationoperator $M_f$. $M_f$ is always linear and if $f\in L^{\infty}(\Omega)$, $M_f$ is bounded. Now, let $f:\Omega\to \mathbb{R}$ be only measurable. I want to know, why $M_f$ is self-adjoint. Therefore I have to check the following properties:
$$1. M_f\; \text{ is densely defined}$$
$$2. M_f\; \text{ is symmetric}$$
$$3. Im(M_f+iId)=L^2(\Omega);\; Im(M_f-iId)=L^2(\Omega)\; \text{(I think, it is enough to prove $Im(M_f+iId)=L^2(\Omega))$}$$
The second property is no problem. I stuck with 1. and 3. We had this proof in lecture and we said if we proved 
1.:It is $\Omega=\bigcup_{n\in\mathbb{N}}\Omega_n$ with $\Omega_n=\{x\in\Omega; |f(x)|\le n\}$ and $\{u\in L^2(\Omega);\exists  n\in\mathbb{N}:$ such that $u=0$ almost everywhere in $\Omega\setminus\Omega_n$ $\}\subseteq D(M_f)$. And the set $\{u\in L^2(\Omega);\exists  n\in\mathbb{N}:$ such that $u=0$ almost everywhere in $\Omega\setminus\Omega_n$ $\}$ is dense in $L^2(\Omega)$.
b. I don't understand this. Why is the set $\{u\in L^2(\Omega);\exists  n\in\mathbb{N}:$ such that $u=0$ almost everywhere in $\Omega\setminus\Omega_n$ $\}$ dense in $L^2(\Omega)$?  And the proof of the 3. property we had in lecture: We only want to prove $Im(M_f+iId)=L^2(\Omega)$ (in other words: $M_f+iId$ is surjective). Let $g\in L^2(\Omega)$. It is $u:=\frac{g}{f+i}\in L^2(\Omega)$ (auxiliary calculation: $(M_f+iId)(u)=g\iff fu+iu=g\iff u=\frac{g}{f+i}$) and $f\cdot u\in L^2(\Omega)$. Therefore it is $u\in D(M_f)$ and $M_fu+iu=g$, so it is $g\in Im(M_f+iId)$. My questions: c. Why is  $u:=\frac{g}{f+i}\in L^2(\Omega)$ and $f\cdot u\in L^2(\Omega)$?  Any help would be appreciated. Regards
 A: a. "$f$ is $\mu$-measurable" means that $f^{-1}(X)\in\Sigma$ for every measurable set $X\subseteq\mathbb R.$
b. For $g\in L^2(\Omega)$ consider the restriction $g_n$ of $g$ onto $\Omega_n.$ Use the Dominated convergence theorem to show that $g_n\to g,\ n\to\infty.$
c. If $f$ is real-valued, then $\frac1{f+i}$ and $\frac f{f+i}$ are bounded. Hence for $g\in L^2(\Omega)$ we have $g\cdot \frac1{f+i},g\cdot \frac f{f+i}\in L^2(\Omega)$
A: Properties (2) and (3) are all you need in order to prove that $M_{f}$ is densely-defined. This is a point that seems not to be commonly known for some reason. So, focus on (3), because you've already proved (2). To see that $(M_{f}\pm iI)$ are surjective, let $g \in L^{2}$ and observe that $g/(f\pm i)$ are in $L^{2}$ because these are measurable functions and
$$
          \left|\frac{g}{f\pm i}\right|^{2}=\frac{|g|^{2}}{f^{2}+1} \le |g|^{2}.
$$
Therefore, $h_{\pm}=g/(f\pm i)$ are in $\mathcal{D}(M_{f})$ and $(M_{f}\pm iI)g_{\pm}=g$, which proves $M_{f}\pm iI$ are surjective.
Proving that $\mathcal{D}(M_{f})$ is dense in $L^{2}$ is equivalent to showing that $\mathcal{D}(M_{f})^{\perp}=\{0\}$. So, assume $g \in \mathcal{D}(M_{f})^{\perp}$, and show that $g=0$. By the above, there exists $h \in \mathcal{D}(M_{f})$ such that
$$
                   g = (M_{f}+iI)h.
$$
And, by assumption, $h \perp g$, which gives
$$
                0=(g,h) = (M_{f}h,h)+i(h,h).
$$
Because $M_{f}$ is symmetric, then $(M_{f}h,h)$ is real. Thus $0 = \Im(g,h)=\|h\|^{2}$, which proves that $h = 0$ and $g=(M_{f}+iI)h=0$. So $\mathcal{D}(M_{f})^{\perp}=\{0\}$.
Note: To prove that $M_{f}$ is selfadjoint, it is not sufficient to just show that $M_{f}+i I$ is surjective or that $M_{f}-iI$ is surjective. Both are required.
