= Solution
Let $K=\sigma(T)$ for a bounded <normal operator> on a nonzero complex <Hilbert space>. The zero Hilbert space has the trivial calculus. Take the <inner product> linear in its first argument. The <continuous functional calculus> is an isometric <unital> star-representation $\pi:C(K)\to\mathcal B(H)$ with $\pi(z)=T$. The <Borel calculus> will act on $B_b(K)$, the algebra of actual bounded <Borel measurable functions>. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.
Here is a construction that also covers a nonseparable <Hilbert space>. For $v\in H$, the positive functional $f\mapsto\langle\pi(f)v,v\rangle$ has a finite <regular Borel measure> $\mu_v$ by the <Riesz-Markov-Kakutani representation theorem>. On <continuous functions>,
$$
\|\pi(f)v\|^2=\int_K|f|^2\,d\mu_v.
$$
<Continuous functions> are dense in $L^2(K,\mu_v)$. Hence $f\mapsto\pi(f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi(C(K))v}$. This subspace reduces $T$ and $T^*$, which correspond to multiplication by $z$ and $\overline z$. Decompose $H$ as an <orthogonal direct sum> of such cyclic <reducing subspaces>, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the <cyclic multiplication model for a normal operator>.
On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a <unital> star-homomorphism
$$
\Phi:B_b(K)\longrightarrow\mathcal B(H),\qquad\Phi(z)=T,\qquad\|\Phi(f)\|\le\|f\|_\infty.
$$
Thus $(fg)(T)=f(T)g(T)$, $\overline f(T)=f(T)^*$, and positivity of $f$ gives positivity of $f(T)$. The <continuous functional calculus> is recovered on $C(K)$.
Set $E(S)=\Phi(\mathbf1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an <orthogonal projection>, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the <strong operator topology>. Indeed on every multiplication summand the indicators add pointwise, and the tail squared <norm> is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a <projection-valued measure>, and
$$
\boxed{f(T)=\int_K f(z)\,dE(z),\qquad T=\int_Kz\,dE(z)}.
$$
For $v,w\in H$, polarization gives scalar measures $\mu_{v,w}(S)=\langle E(S)v,w\rangle$ satisfying $\langle f(T)v,w\rangle=\int f\,d\mu_{v,w}$. Regularity of these finite scalar measures and uniqueness in the <Riesz-Markov-Kakutani representation theorem> make the <projection-valued measure> unique: its values on <continuous functions> determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.
The correct convergence assertion is bounded pointwise convergence in the <strong operator topology>. If $f_n\to f$ pointwise and $\sup_n\|f_n\|_\infty<\infty$, then
$$
\|(f_n(T)-f(T))v\|^2=\int_K|f_n-f|^2\,d\mu_v\longrightarrow0
$$
by the <dominated convergence theorem>. This need not be convergence in <operator norm>. For multiplication by $t$ on $L^2[0,1]$, the functions $\mathbf1_{[0,1/n]}$ converge pointwise to $\mathbf1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.
A <Borel measurable function> can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\varepsilon\})=0$ for every $\varepsilon>0$, and
$$
\|f(T)\|=\inf\{C\ge0:E(\{|f|>C\})=0\}.
$$
The upper norm bound follows by integrating $|f|^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\|f(T)\|\ge C$. The induced map on the quotient by these spectral-null <Borel measurable functions> is isometric. Every nonempty relatively <open> subset of $K$ has nonzero spectral projection, since otherwise a nonzero <continuous function> supported there would contradict faithfulness of the <continuous functional calculus>. This full support gives the usual continuous <spectral mapping theorem>, but for a <Borel measurable function> the exact statement is instead
$$
\boxed{\sigma(f(T))=\{\lambda:E(\{|f-\lambda|<\varepsilon\})\ne0\text{ for every }\varepsilon>0\}}.
$$
If an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\|(f(T)-\lambda)v\|$ arbitrarily small, contradicting invertibility. This is <spectral essential range in Borel functional calculus>. For example $\mathbf1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.
Spectral projections provide <reducing subspaces>, and $E(\{\lambda\})H=\ker(T-\lambda I)$: the squared <norm> of $(T-\lambda)v$ is $\int|z-\lambda|^2\,d\mu_v$. In finite dimension the calculus is simply $f(T)=\sum_jf(\lambda_j)P_j$, the sum over <eigenspaces>. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z|^{1/2}e^{i\operatorname{Arg}(z)/2}$ with $s(0)=0$, giving a <normal operator> $s(T)$ with $s(T)^2=T$ even when a continuous scalar branch is unavailable.
Any <bounded operator> commuting with both $T$ and $T^*$ commutes with their <continuous functional calculus> and with all $E(S)$. To justify the latter, regularity approximates $\mathbf1_S$ by uniformly bounded <continuous functions> in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the <Borel calculus> lies in the <Von Neumann algebra> generated by $T$. On a <separable Hilbert space>, a countable dense list of vectors gives a controlling measure $\mu=\sum_j2^{-j}\mu_{v_j}/(1+\|v_j\|^2)$; spectral-null sets are exactly its <null sets>, and the range is the represented $L^\infty(K,\mu)$, equal to that <Von Neumann algebra>. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in[0,1]$, the generated <Von Neumann algebra> has arbitrary bounded diagonal functions, while the original <Borel calculus> has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the <Borel functional calculus for a normal operator> from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.
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