Uniqueness of the Spectral Measure

Question

I am studying the spectral theorem for unbounded s.a. operators. The existence of a spectral measure E sucht that the Operator $T=\int_{\sigma(t)}\lambda dE(\lambda):=E(id)$, is already shown. What i want to understand now is the uniqueness of the spectral measure.

The Idea in my script is to show that any other spectral measure $F$ with $T=\int_{\sigma(t)}\lambda dF(\lambda)$ has to satisfy:

$\int_{\sigma(T)} \frac{1}{\lambda-z}dE(\lambda) = \int_{\sigma(T)} \frac{1}{\lambda-z}dF(\lambda)$ for $z \in \mathbb{C}/\mathbb{R}$,

and then to use the Theorem of Stone-Weierstraß to approximate the continuous functions by functions of the form $p_N(\lambda)=\sum_{i=1}^{N}\frac{1}{\lambda - z_k}$ uniformly.

And Finally to approximate the bounded measurable functions by the continuous functions pointwise in order to show that $E=F$.

My problem is to understand these approximation procedures. If $\phi = \lim_{n \rightarrow \infty} p_n$ is a continuous function then in which sense can i take the limit $E(\phi)=\lim_{n \rightarrow \infty} E(p_n)$?

Finally: why is it enough to proof $E(f)=F(f)$ only for bounded measurable $f$?

Answer 1

To answer the question in which sense one would understand $E(\phi) = \lim\limits_{n\rightarrow\infty} E(p_n)$ you should look at the definition of a spectral measure. A spectral measure is a map (with some properties) $E:\mathcal{A}\rightarrow L(H)$ for some $\sigma$-algebra $\mathcal{A}$ and Hilbert space $H$, where $L(H)$ is the space of bounded linear operators $A:H\rightarrow H$. Hence, I would understand $E(\phi) = \lim\limits_{n\rightarrow\infty} E(p_n)$ as convergence in $H$.
For your second question, I guess it is enough to show the equality for bounded measurable functions, since we can approximate an unbounded measurable function $f$ with $f_n = f\boldsymbol{1}_{A_n}$, where $A_n = \{x: \vert f(x) \vert < n \}$, and then one usually uses some convergence theorem (e.g. monotone convergence) to obtain equality even for unbounded measurable functions.
Hope this makes all sense and helps, it's been a while since I've been in touch with the topic.

Answer 2

I know how to obtain convergence in the strong operator topology know:

Since uniform convergence implies $L^2$-convergence, we obtain by using the measure $\nu_x(A):= \langle x, E(A)x \rangle $ with $x \in H$, that

$||E(p_n)x-E(\phi)x||=\int_X |p_n - \phi|^2d \nu_x \rightarrow 0$

For the approximation of the bounded measurable functions, using the dominated convergence theorem, again convergence in the sense of the strong operator topology follows.

Then as @stange pointed out, one obtains uniqueness for all measurable funktions, since simple functions are dense in the measurable functions.