Proof Check: Multiplication operator on $L^{p}(mathbb R^{d})$ is closed

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 $(f_{n(k)})_{kin mathbb N}$ 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

$$lvert lvert mf-zrvertrvert_{p}^{p}\=intlimits_{mathbb R^{d}}lvert m(x)f(x)-z(x)rvert^{p}dx\=intlimits_{A_{k}^{c}}lvert m(x)f(x)-z(x)rvert^{p}dx\=intlimits_{A_{k}^{c}}limlimits_{kto infty}lvert m(x)f_{n(k)}(x)-z(x)rvert^{p}dx\ =intlimits_{A_{k}^{c}}liminflimits_{kto infty}lvert m(x)f_{n(k)}(x)-z(x)rvert^{p}dx\ leq liminflimits_{kto infty} intlimits_{A_{k}^{c}}lvert m(x)f_{n(k)}(x)-z(x)rvert^{p}dx\leq liminflimits_{kto infty} intlimits_{mathbb R^{d}}lvert m(x)f_{n(k)}(x)-z(x)rvert^{p}dx \ =limlimits_{kto infty} intlimits_{mathbb R^{d}}lvert m(x)f_{n(k)}(x)-z(x)rvert^{p}dx=0.$$ 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?