Let $m : \mathbb R^{d} \to \mathbb K$ measurable and $A: \operatorname{dom}(A) \to L^{p}(\mathbb R^{d})$
such that $Af(x)=m(x)f(x)$ and $\operatorname{dom}(A):=\{f \in L^{p}(\mathbb R^{d}):mf \in L^{p}(\mathbb R^{d})\}$. Show that $A$ is a closed operator.
My proof:
Let $(f_{n})_{n \in \mathbb N}\subseteq \operatorname{dom}(A)$ with $f_{n} \xrightarrow{L^{p}} f$ and $Af_{n} \xrightarrow{ L^{p}} z$. Then by $L^{p}$ convergence there exists a measurable $A_{k}$ and a subsequence $(f_{n(k)})_{k\in \mathbb N}$ such that $\mu(A_{k})=0$ and $f_{n(k)}(x)\xrightarrow{k \to \infty} f(x),\; \forall x \in A_{k}^{c}$.
Now we want to show that $mf=z$ a.e., thus consider
$$\lvert \lvert mf-z\rvert\rvert_{p}^{p}\\=\int\limits_{\mathbb R^{d}}\lvert m(x)f(x)-z(x)\rvert^{p}dx\\=\int\limits_{A_{k}^{c}}\lvert m(x)f(x)-z(x)\rvert^{p}dx\\=\int\limits_{A_{k}^{c}}\lim\limits_{k\to \infty}\lvert m(x)f_{n(k)}(x)-z(x)\rvert^{p}dx\\ =\int\limits_{A_{k}^{c}}\liminf\limits_{k\to \infty}\lvert m(x)f_{n(k)}(x)-z(x)\rvert^{p}dx\\ \leq \liminf\limits_{k\to \infty} \int\limits_{A_{k}^{c}}\lvert m(x)f_{n(k)}(x)-z(x)\rvert^{p}dx\\\leq \liminf\limits_{k\to \infty} \int\limits_{\mathbb R^{d}}\lvert m(x)f_{n(k)}(x)-z(x)\rvert^{p}dx \\ =\lim\limits_{k\to \infty} \int\limits_{\mathbb R^{d}}\lvert m(x)f_{n(k)}(x)-z(x)\rvert^{p}dx=0.$$ Thus $mf = z$ a.e. and since $z \in L^{p}$, it follows that $f \in \operatorname{dom}(T)$ and hence $A$ is closed.
The critical part of my proof was to use Fatou's Lemma. Are there any other alternative ways to prove this and is my proof at all correct?