Proof. (1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By Urysohn’s Lemma, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B}= 1$. By Goldstine’s Theorem, there exists a net $\angles{g_\gamma}_{\gamma \in C}\subset C(\sigma_{B(H)}(T); \complex)$ such that $g_{\gamma} \to \one_{B}$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$. As $f\one_{B} = \one_{B}$, $fg_{\gamma} \to \one_{B}$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as well.

By the Stone-Weierstrass Theorem, $h(T) \in A$ for all $h \in C(\sigma_{B(H)}(T); \complex)$ with $h(0) = 0$. In particular, $fg_{\gamma}(T) \in A$ for all $\gamma \in C$. Thus the $L^{\infty}$ functional calculus implies that $\one_{B}(T) \in A$ as well.

If $B$ is arbitrary, then $\one_{B \setminus B_\complex(0, r)}\to \one_{B}$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as $r \downto 0$. As $\one_{B \setminus B_{\complex}(0, r)}(T) \in A$ for all $r > 0$, $\one_{B}(T) \in A$ as well.

By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By Lemma 26.6.4, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ is uniformly dense in $\bracsn{f \in B^\infty(\sigma_{B(H)(T)}; \complex)|f(0) = 0}$. Therefore $f(T) \in A$ for all $f \in B^{\infty}(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$.$\square$