# Spectral Theorem for unbounded self adjoint operators

So I've been working on the Spectral Theorem for self adjoint unbounded Operators with the Book by Rudin and got to a problem: Let $$(X,\mathcal{A})$$ be a measure space, $$H$$ a complex Hilbert space and $$P:\mathcal{A}\rightarrow B(H)$$ a resolution of the identity. Then, to every measurable function $$f:X\rightarrow\mathbb{C}$$ there exists a densely defined operator $$\Psi(f)$$ in $$H$$, with domain $$D(\Psi(f)) = \{x\in H: \int_{X} \vert f(\lambda) \vert^{2} \ d\langle P(\lambda)x,x\rangle < \infty\}$$, which is characterized by $$\langle \Psi(f)x,y\rangle = \int_{X} f(\lambda) \ d\langle P(\lambda)x,y\rangle$$ for all $$x\in D(\Psi(f))$$ and $$y\in H$$.
My problem is the following theorem: In the above situation, if $$D(\Psi(f)) = H$$ then $$f$$ is essentially bounded.
$$\textbf{Proof:}$$ Since $$\Psi(f)$$ is a closed operator, the closed graph theorem implies $$\Psi(f)\in B(H)$$. If $$f_{n} = f\chi_{A_{n}}$$ for $$A_{n} = \{x\in X : \vert f(x)\vert \leq n\}$$ and $$n\in\mathbb{N}$$, then it follows $$\Vert f_{n}\Vert_{\infty} = \Vert \Psi(f_{n})\Vert = \Vert \Psi(f)\Psi(\chi_{A_{n}}) \Vert \leq\Vert \Psi(f) \Vert,$$ since $$\Vert\Psi(\chi_{A_{n}})\Vert = \Vert \chi_{A_{n}} \Vert_{\infty}\leq 1$$. Thus $$\Vert f \Vert_{\infty} \leq \Vert\Psi(f)\Vert$$ and $$f$$ is essentially bounded.
I don't know how we can conclude that $$\Vert f \Vert_{\infty} \leq \Vert \Psi(f)\Vert$$, since $$f_{n}\rightarrow f$$ only pointwise. I also know that $$\Vert \Psi(f_{n}) \Vert$$ converges to $$\Vert\Psi(f)\Vert$$ but I can't see how that could help me. I would appreciate any help.

$$|f_n(x)| \leq \|\Psi (f)\|$$ for all $$x$$ outside some set $$E_n$$ of measure $$0$$. If $$x \notin \cup_n E_n$$ and $$f_n(x) \to f(x)$$ then $$|f_n(x)| \leq \|\Psi (f)\|$$ for all $$n$$ so $$|f(x)| \leq \|\Psi (f)\|$$. This proves that $$\|f\|_{\infty} \leq \|\Psi (f)\|$$.