Definitions: Let $H$ be a Hilbert space over the field $\mathbb{C}$. Call a linear map $A:H\to H$ unbounded if its domain $D(A)$ is dense in $H$ and $\not\exists C>0:\forall \psi\in D(A):||A\psi|| \leq C||\psi||$. Call such an operator $A$ closable if the closure of the graph $G(A):=\{(\psi, A\psi): \psi\in D(A)\}$ in $H^2$ is the graph of some other function, i.e. $\forall \psi\in D(A)\exists!\phi\in H:(\psi,\phi)\in G(A)$. Denote this operator whose graph is equal to the closure of $G(A)$ by $A^{\text{cl}}$. Then $A^{\text{cl}}$ is an extension of $A$, i.e. $\forall \psi\in D(A):A\psi = A^{\text{cl}}\psi$.
Problem: I am trying to understand one specific point in a proof which shows that $A^* = (A^\text{cl})^*$. The author claims that if $\psi\in D(A^\text{cl})$ then there exists a sequence $(\psi_n)_n$ with $\psi_n\in D(A)$ such that $\lim_{n\to\infty}\psi_n = \psi, \lim_{n\to\infty}A\psi_n = A^{\text{cl}}\psi$. I don't quite follow why $\lim_{n\to\infty}A\psi_n = A^{\text{cl}}\psi$. If $A$, and $A^\text{cl}$ for that matter, were bounded on $D(A)$, then this would be a simple consequence of the completeness of $H$. However no such assumption is present at this moment.
This is because of the definition of $A^{cl}\psi$. Indeed, $A^{cl}$ is defined so that its graph is the closure of the graph of $A$. Therefore, if $\psi \in D(A^{cl})$, then it must be because there are some $(\psi_n, A \psi_n) \to (\psi, y)$ for some $y \in H$. We then define $A^{cl}\psi := y$.
As you can see, talking about closures with respect to graphs allows us to define extensions of $A$ without requiring continuity from $A$, simply because we ask that both $\psi_n$ and $A \psi_n$ converge at the same time. In the continuous setting, this is not required, by the closed graph theorem.