Let $B_\varphi f =\varphi .f$ for $ f \in L^2$, be a multiplicative operator with a measurable symbol $\varphi$. The domain of $B$ are those $f\in L^2$ s.t $\varphi . f \in L^2$. Prove that $B$ is closed and domain of $B$ is dense in $L^2$.
I tried closed graph theorem to prove closeness of B but I messed up. No idea for density.
Any hint would be appreciated.
Note that $\varphi f \in L^2$ for some $f\in L^2$ iff $|\varphi|f\in L^2$ because $|\varphi|f$ is measurable and $|\varphi f|^2=||\varphi|f|^2$. Let $$ \mathcal{D} = \{ f \in L^2 : \varphi f \in L^2 \} = \{ f \in L^2 : |\varphi|f \in L^2 \}. $$ The operator $Kg = (1+|\varphi|)^{-1}g$ is bounded and selfadjoint with $\|K\|\le 1$. Every $h \in \mathcal{R}(K)$ has the property that $|\varphi|h \in L^2$. So $\mathcal{R}(K) \subseteq \mathcal{D}$, which gives $$ L^2 = \{0\}^{\perp}=\mathcal{N}(K)^{\perp}=\overline{\mathcal{R}(K)}\subseteq\overline{D} \implies \mathcal{D}\mbox{ is dense}. $$ To show that $B$ is closed, suppose $\{ f_n \} \subset \mathcal{D}$ converges in $L^2$ to $f\in L^2$, and suppose that $\{ \varphi f_n \}$ converges in $L^2$ to $g\in L^2$. Then it must be shown that $\varphi f=g$ a.e.. One way to do this is to obtain subsequences of $\{ f_n \}$ and $\{ \varphi f_n \}$ that converge pointwise a.e. to $f$ and $g$ respectively. Then $\varphi f_n$ also converges to $\varphi f$ pointwise a.e., forcing $\varphi f=g$ a.e..