Borel Functional Calculus. A Question regarding some basic facts.

Question

In section 20.3 on "The Borel Functional Calculus" in Kehe Zhu's "An introduction to operator algebra" one finds the following densely written passage:

"Suppose $A$ is a von Neumann algebra and $T$ is normal in $A$. If E is the resolution of identity for $T$, then for each Borel set $S$ in $\sigma(T)$ the projection $E(S)=\chi_{S}(T)$ belongs to $A$. In fact, $f(T)$ belongs to $A$ for every $f$ in $B_{\infty}(\sigma(T))$ because

$$\langle f(T)x,y\rangle =\int_{\sigma(T)}f dE_{x,y} $$

and so $f(T)$ is in the weak-operator closure of the C*-subalgebra generated by $T$".

Questions:

1) It is clear, that $E(S)=\chi_{S}(T)$ is a projection. But why does it belong to $A$?

2) It is also clear from the definitions, that
$$\langle f(T)x,y\rangle = \int_{\sigma(T)}fdE_{x,y}.$$ But why does it follow, that $f(T)$ is in the weak-operator closure of the C*-subalgebra generated by $T$?

[Since stepfunctions made up of Borel measurable sets are dense in $B_{\infty}(\sigma(T))$ it does not seem too difficult to show, that part (1) implies that $f(T)$ belongs $A$. But how do we conclude, that f(T) belongs to the weak-operator closure of the C*-subalgebra generated by $T$?]

Update: There is a related question, the answer to which might trivialize the above questions: Given some normal operator $T \in B(H)$, what is the range of the Borel functional calculus $f \mapsto f(T)$, $B_{\infty}(\sigma(T)) \to B(H)$? Does $f(T)$ always belong to some nice subset of $B(H)$ related to $T$? And why? Kehe Zhu does not mention this, but I guess there is some simple answer.

Answer 1

The answer is that $f(T)$ always belongs to the von Neumann algebra generated by $T$. You can see that precisely via the two questions you are asking:

1. $E(S)$ belongs to $A$: It is easy to see that $E(S)$ is in the double commutant of $T$; from the fact that $T$ is normal, the commutant of $T$ is a von Neumann algebra (the only non-trivial part is that of adjoints, which follows from the Fuglede-Putnam Theorem). So $E(S)$ belongs to any von Neumann algebra that contains $T$.

2. This is the same as question 1, only with $f$ instead of $\chi_S$. Since the integral $\langle f(T)x,y\rangle$ is a limit of integrals of simple functions, this gives you $f(T)$ as a wot limit of linear combinations of projections, and by part 1 these linear combinations are in the von Neumann algebra generated by $T$.