Is my proof of closedness of multiplication operator corect?

Question

I am considering an operator $A: L^2(\mathbb R , d \mu) \supset D(A)\to L^2 (\mathbb R, d\mu)$ defined by $(Af)(x)=a(x)f(x)$ for known measurable function $a$. Domain is of course all those functions for which result is square integrable. I want to (directly) show that it is closed, that is its graph is closed subset of $L^2 \oplus L^2$. I want to make sure my proof is correct. I consider an element of closure $(f,g)$. Then there is a sequence $ f_n \in D(A)$ converging to $f$ such that $af_n $ converges to g. I choose subsequence of $f_n$ that converges pointwise a.e. (and denote it like original sequence from now on). Then $af_n$ converges pointwise a.e. to $af$. However we also know that its subsequence converges pointwise a.e. to $g$ which is square integrable. Since sequence can have only one limit point we deduce that those functions are equal almost everywhere, which means that $f \in D(A)$. I use similar arguments to show that $A$ is closable, which combined with statement above shows it's closed. If above is not correct I will appreciate hints how to fix it. I'm also curious about alternative methods.

Answer 1

$(f,af) \in \mathcal{G}(A)$ iff $\sqrt{1+|a|^{2}}f \in L^{2}_{\mu}$. Furthermore, for any such $f$, $$ \|(f,af)\|_{\mathcal{G}(A)}=\|\sqrt{1+|a|^{2}}f\|_{L^{2}_{\mu}}=\|f\|_{L^{2}_{d\nu}}, $$ where $d\nu = \sqrt{1+|a|^{2}}d\mu$. So the graph is a Hilbert space.