I would like to prove that for a normal operator T on a Hilbert space H there exists a measure space $(\Omega, \Sigma, \mu)$, a unitary operator $U: H \to L^2(\Omega)$ and a bounded measurable function $f$ on $\Omega$ so that $UTU^*\varphi = f \cdot \varphi \ \forall \varphi$ in $L^2(\Omega).$. I would like to use the corresponding theorem for bounded self-adjoint Operators, and have written $T$ as $$T = S_1 + iS_2 \text{ where } S_1 := \frac{T + T^*}{2}, \quad S_2 = \frac{T - T^*}{2i}.$$ $S_1$ and $S_2$ are self-adjoint, so there exist measure spaces $(\Omega_1, \Sigma_1, \mu_1)$ and $(\Omega_2, \Sigma_2, \mu_2)$, measurable functions $f_1$ and $f_2$ and unitary operators $U_1: H \to L^2(\Omega_1), \ U_2: H \to L^2(\Omega_2)$, so that $U_1S_1U_1^* = M_{f_1}$ and $U_2S_2U_2^* = M_{f_2}.$ Chat GPT suggests $\Omega = \Omega_1 \times \Omega_2$
$U: H \to L^(\Omega)$
$Ux(\lambda_1, \lambda_2) = U_1x(\lambda_1) U_2x(\lambda_2) $ $f: \Omega \to C$
$f(\lambda_1, \lambda_2) = f_1(\lambda_1) + if_2(\lambda_2)$
I think it should be easy to prove the theorem with all these tools, but I'm having problems, and want to confirm with more knowledgable people whether this approach even works before I spend too much time on it. I have gotten this far: $UTU^*\varphi(\lambda_1, \lambda_2)$
$= (US_1U^*\varphi)(\lambda_1, \lambda_2) + i(US_2U^*\varphi)(\lambda_1, \lambda_2) = U_1(S_1U^*\varphi)(\lambda_1)\cdot U_2(S_1U^*\varphi)(\lambda_2) + iU_1(S_2U^*\varphi)(\lambda_1)\cdot U_2(S_2U^*\varphi)(\lambda_2)$
This is supposed to equal $f_1(\lambda_1)\varphi(\lambda_1, \lambda_2) + if_2((\lambda_1)\varphi(\lambda_1, \lambda_2).$ Can someone help me prove this or tell me why the approach doesn't work (ChatGPT isn't the best at maths).