I have a question about a part of some proof.
Let $\lambda$ be any number on the unit circle in the complex plane.
Then, we know that there exists a real bounded Borel function $g(\lambda)$ such that $\lambda=\exp(ig(\lambda))$, where $g(\lambda)\in(-\pi,\pi]$.
Now, let $H=M_g$ be the multiplication operator defined by $g$, which is a self-adjoint operator on some Hilbert space, the detail of which I omit for simplicity.
Then, could we prove easily exp$(iH)=M_\lambda$ using something like Borel functional calculus or anything else?
Borel functional calculus can be defined (or otherwise characterized) in such a way that if a normal operator $T$ is unitarily equivalent to the multiplication operator $M_\phi$ and $f$ is a Borel function, then $f(T)$ is unitarily equivalent to $M_{f\circ\phi}$ (through the same unitary).
In your case $T=M_g$ and $f=\exp(i\,\cdot\,)$, hence $f(T)=M_{f\circ g}=M_{\mathrm{id}}$.
However, the exponential of a bounded operator (normal or not) can also be defined via the exponential series. One can show that both definitions coincide for normal operators. Using that $g\mapsto M_g$ is a $\ast$-homomorphism, this definition also implies $\exp(iM_g)=M_{\mathrm{id}}$ immediately.