Proof of the spectral theorem I am currently going to through my proof of the spectral theorem that we had in class, but I feel that I have copied some nonsense from the board.
So we defined the Cayley transform $U= (T-i)(T+i)^{-1} = (T+i-2i)(T+i)^{-1}=id-2i(T+i)^{-1}$ and therefore $(T+i)^{-1}= \frac{1}{2i}(id-U)$. Then we showed earlier that any unitary operator can be written as $e^{iA}$ for some self-adjoint bounded $A$ such that $\sigma(A) \subset [-\pi,\pi]$.
Now $-U$ is definitely unitary, as the Cayley transform is unitary, so we have 
$$\langle -U_t x,x \rangle =  \int e^{it}d\mu_x,$$
where $\mu_x$ is the Borel measure from the functional calculus of $A$.
Thus by using  $(T+i)^{-1}= \frac{1}{2i}(id-U)$ we can write
$$\langle (T+i)^{-1}x,x\rangle  = \frac{1}{2i} \int(1+e^{it}) d\mu_x$$
and now comes the problem, somehow at the end we made a $arctan$ substitution and I wrote 
$$\langle (T+i)^{-1}x,x\rangle  = \frac{1}{2i} \int_{\mathbb{R}} (1+e^{i \arctan(s)}) ds$$
and then we rewrote the integrand by Euler's identity so that we ende up with (this I undestand again) 
$$\langle (T+i)^{-1}x,x\rangle  = \frac{1}{2i} \int_{\mathbb{R}} \frac{1}{s+i} ds$$
The problem is that I don't get this substitution, is there anybody who could make this substitution of the arctan function rigorous, i.e. I don't get what this $ds$ now means, what kind of measure is this?
If anything is unclear, please let me know.
 A: Any mention of $A$ is a non-sequitur for the purposes of this discussion. The operators $T$ and $U$ are the only operators involved. The operator $U=(T-iI)(T+iI)^{-1}$ is a unitary operator with its spectrum $\sigma(U)\subseteq\mathbb{T}$, where $\mathbb{T}$ is the unit circle in the complex plane. Note that $1$ may be in the spectrum of $U$, but it cannot be in the discrete spectrum because $Ux=x$ iff
$$
\begin{align}
        x =Ux & = (T-iI)(T+iI)^{-1}x \\
              & = (T+iI)(T+iI)^{-1}x-2i(T+iI)^{-1}x \\
              & = x-2i(T+iI)^{-1}x \\
       \iff & (T+iI)^{-1}x = 0 \\
       \iff & x = 0.
\end{align}
$$
So $-U$ does not have $-1$ in its point spectrum, allowing you to write
$$
          (-Ux,x) = \int_{\sigma(-U)}zd(E(z)x,x)=\int_{-\pi+0}^{\pi-0}e^{it}d\mu_{x}(t).
$$
The function $\arctan(u)$ is a bijection between $(-\infty,\infty)$ and $(-\pi,\pi)$. This leads to a new finite Borel measure $\nu_{x}(S)=\mu_{x}(\arctan(S))$ such that
$$
               (-Ux,x) = \int_{-\infty}^{\infty}e^{i\arctan(u)}d\nu_{x}(u).
$$
The notation $ds$ in your notes is not a correct one because it suggests Lebesgue measure, due to the fact that the variable of integration is also $s$.
