[proofplan] Write $K:=\sigma(T)$ and view the [Borel functional calculus](/theorems/2696) as the spectral integral $\Phi_E(u)=\int_K u\,dE$ for bounded Borel functions $u:K\to\mathbb C$. The algebraic rules are first checked on bounded Borel simple functions by refining the measurable partitions and using the projection identities $E(A)E(B)=E(A\cap B)$ and $E(A)^*=E(A)$. The passage from simple functions to arbitrary bounded Borel functions is made through scalar matrix coefficients $\mu_{h,k}(A)=(E(A)h,k)_H$ and dominated convergence. The final strong convergence statement follows by applying dominated convergence to the positive scalar measure $\mu_h=\mu_{h,h}$ and the identity $\|u(T)h\|_H^2=\int_K |u|^2\,d\mu_h$. [/proofplan]

[step:Fix the spectral-integral notation and scalar measures] Set \begin{align*} K:=\sigma(T). \end{align*} For a bounded Borel function $u:K\to\mathbb C$, write \begin{align*} u(T):=\Phi_E(u):=\int_K u\,dE. \end{align*} For $h,k\in H$, define the scalar set function $\mu_{h,k}:\mathcal B(K)\to\mathbb C$ by \begin{align*} \mu_{h,k}(A):=(E(A)h,k)_H. \end{align*} By Countable Additivity in Matrix Coefficients, applied to the projection-valued measure $E$ on the compact [metric space](/page/Metric%20Space) $K$, $\mu_{h,k}$ is a finite complex measure on $K$, and $\mu_h:=\mu_{h,h}$ is a finite positive measure satisfying \begin{align*} \mu_h(K)=\|h\|_H^2. \end{align*} By the defining matrix-coefficient formula for spectral integrals, for every bounded Borel function $u:K\to\mathbb C$ and every $h,k\in H$, \begin{align*} (u(T)h,k)_H=\int_K u\,d\mu_{h,k}. \end{align*} [/step]

[step:Verify the algebraic rules for simple Borel functions]Let $s,t:K\to\mathbb C$ be bounded Borel simple functions. Choose a finite Borel partition $(A_i)_{i=1}^m$ of $K$ and complex numbers $(a_i)_{i=1}^m$ such that \begin{align*} s=\sum_{i=1}^m a_i\mathbb 1_{A_i}. \end{align*} Choose a finite Borel partition $(B_j)_{j=1}^n$ of $K$ and complex numbers $(b_j)_{j=1}^n$ such that \begin{align*} t=\sum_{j=1}^n b_j\mathbb 1_{B_j}. \end{align*} By the definition of the spectral integral for simple functions, \begin{align*} s(T)=\sum_{i=1}^m a_iE(A_i) \end{align*} and \begin{align*} t(T)=\sum_{j=1}^n b_jE(B_j). \end{align*} Linearity for simple functions follows immediately from the linearity of finite sums after passing to the common finite Borel partition $(A_i\cap B_j)_{1\le i\le m,\ 1\le j\le n}$. For multiplicativity, use the projection-valued measure identity \begin{align*} E(A_i)E(B_j)=E(A_i\cap B_j). \end{align*} Then \begin{align*} s(T)t(T)=\sum_{i=1}^m\sum_{j=1}^n a_ib_jE(A_i\cap B_j). \end{align*} Since \begin{align*} st=\sum_{i=1}^m\sum_{j=1}^n a_ib_j\mathbb 1_{A_i\cap B_j}, \end{align*} we obtain \begin{align*} (st)(T)=s(T)t(T). \end{align*} For the adjoint rule, use $E(A_i)^*=E(A_i)$ for each Borel set $A_i$. Thus \begin{align*} s(T)^*=\sum_{i=1}^m \overline{a_i}E(A_i). \end{align*} Since \begin{align*} \overline{s}=\sum_{i=1}^m \overline{a_i}\mathbb 1_{A_i}, \end{align*} we get \begin{align*} (\overline{s})(T)=s(T)^*. \end{align*}[/step]

[guided]The point of starting with simple functions is that the spectral integral is then only a finite sum of projections. Let \begin{align*} s=\sum_{i=1}^m a_i\mathbb 1_{A_i} \end{align*} and \begin{align*} t=\sum_{j=1}^n b_j\mathbb 1_{B_j}, \end{align*} where $(A_i)_{i=1}^m$ and $(B_j)_{j=1}^n$ are finite Borel partitions of $K$, and where $a_i,b_j\in\mathbb C$. The spectral integral of a [simple function](/page/Simple%20Function) is defined by replacing each indicator $\mathbb 1_A$ with the projection $E(A)$, so \begin{align*} s(T)=\sum_{i=1}^m a_iE(A_i) \end{align*} and \begin{align*} t(T)=\sum_{j=1}^n b_jE(B_j). \end{align*} For addition, we must put $s$ and $t$ on a common partition. The intersections $A_i\cap B_j$ are Borel sets and form a finite Borel partition of $K$ after omitting empty sets. On $A_i\cap B_j$, the function $\alpha s+\beta t$ has value $\alpha a_i+\beta b_j$. Therefore the finite-sum definition gives \begin{align*} (\alpha s+\beta t)(T)=\alpha s(T)+\beta t(T). \end{align*} For multiplication, the essential projection-valued identity is \begin{align*} E(A_i)E(B_j)=E(A_i\cap B_j). \end{align*} Using finite distributivity in $\mathcal L(H)$, \begin{align*} s(T)t(T)=\sum_{i=1}^m\sum_{j=1}^n a_ib_jE(A_i)E(B_j). \end{align*} Substituting the projection identity gives \begin{align*} s(T)t(T)=\sum_{i=1}^m\sum_{j=1}^n a_ib_jE(A_i\cap B_j). \end{align*} But the pointwise product satisfies \begin{align*} st=\sum_{i=1}^m\sum_{j=1}^n a_ib_j\mathbb 1_{A_i\cap B_j}. \end{align*} Applying the simple-function definition of the spectral integral to this last expression yields \begin{align*} (st)(T)=s(T)t(T). \end{align*} For adjoints, each $E(A_i)$ is an [orthogonal projection](/theorems/437), so $E(A_i)^*=E(A_i)$. Since the Hilbert-space adjoint is conjugate-linear with respect to scalar multiplication, \begin{align*} s(T)^*=\sum_{i=1}^m \overline{a_i}E(A_i). \end{align*} The conjugate simple function is \begin{align*} \overline{s}=\sum_{i=1}^m \overline{a_i}\mathbb 1_{A_i}. \end{align*} Hence \begin{align*} (\overline{s})(T)=s(T)^*. \end{align*}[/guided]

[step:Pass from simple functions to bounded Borel functions]Let $u:K\to\mathbb C$ be a bounded Borel function. The standard bounded simple approximation theorem gives a sequence $(u_m)_{m=1}^{\infty}$ of bounded Borel simple functions $u_m:K\to\mathbb C$ such that $u_m(z)\to u(z)$ for every $z\in K$ and \begin{align*} |u_m(z)|\le \|u\|_{\infty} \end{align*} for every $m\in\mathbb N$ and every $z\in K$. One obtains such a sequence by applying the usual dyadic simple approximations separately to the real and imaginary parts of $u$ and then truncating within the closed disk of radius $\|u\|_{\infty}$. Applying the [dominated convergence theorem](/theorems/4) to the finite complex measure $\mu_{h,k}$ gives, for every $h,k\in H$, \begin{align*} \int_K u_m\,d\mu_{h,k}\to \int_K u\,d\mu_{h,k}. \end{align*} Equivalently, \begin{align*} (u_m(T)h,k)_H\to (u(T)h,k)_H. \end{align*} Choose uniformly bounded Borel simple functions $s_m,t_m:K\to\mathbb C$ such that $s_m(z)\to f(z)$ and $t_m(z)\to g(z)$ for every $z\in K$, with $|s_m(z)|\le \|f\|_{\infty}$ and $|t_m(z)|\le \|g\|_{\infty}$. Then $\alpha s_m+\beta t_m$ is a bounded Borel simple function, $\alpha s_m(z)+\beta t_m(z)\to \alpha f(z)+\beta g(z)$ for every $z\in K$, and it is uniformly bounded by $|\alpha|\|f\|_{\infty}+|\beta|\|g\|_{\infty}$. Applying the simple-function identity \begin{align*} (\alpha s_m+\beta t_m)(T)=\alpha s_m(T)+\beta t_m(T) \end{align*} and passing to scalar matrix coefficients by dominated convergence gives \begin{align*} ((\alpha f+\beta g)(T)h,k)_H=(\alpha f(T)h+\beta g(T)h,k)_H \end{align*} for every $h,k\in H$. Hence $(\alpha f+\beta g)(T)=\alpha f(T)+\beta g(T)$. For multiplicativity, first fix a bounded Borel function $v:K\to\mathbb C$ and a bounded Borel simple function $s:K\to\mathbb C$ with representation \begin{align*} s=\sum_{i=1}^m a_i\mathbb 1_{A_i}. \end{align*} For every $h,k\in H$, \begin{align*} (s(T)v(T)h,k)_H=\sum_{i=1}^m a_i(E(A_i)v(T)h,k)_H. \end{align*} Since $E(A_i)^*=E(A_i)$ and $E(B)E(A_i)=E(B\cap A_i)$ for Borel sets $B\subset K$, the scalar measure $B\mapsto (E(B)h,E(A_i)k)_H$ is the restriction of $\mu_{h,k}$ to $A_i$. Hence \begin{align*} (E(A_i)v(T)h,k)_H=\int_K \mathbb 1_{A_i}v\,d\mu_{h,k}. \end{align*} Summing over $i$ gives \begin{align*} (s(T)v(T)h,k)_H=\int_K sv\,d\mu_{h,k}=((sv)(T)h,k)_H. \end{align*} Now choose uniformly bounded simple approximations $(s_m)_{m=1}^{\infty}$ to $f$. Applying the preceding identity with $s=s_m$ and $v=g$, then using dominated convergence for the matrix coefficients of $s_m(T)g(T)$ and of $(s_mg)(T)$, gives for every $h,k\in H$ \begin{align*} (f(T)g(T)h,k)_H=((fg)(T)h,k)_H. \end{align*} Thus $(fg)(T)=f(T)g(T)$. For the adjoint rule, take uniformly bounded simple approximations $(s_m)_{m=1}^{\infty}$ to $f$. The simple-function adjoint rule gives $(\overline{s_m})(T)=s_m(T)^*$ for every $m\in\mathbb N$. Passing to scalar matrix coefficients by dominated convergence gives, for every $h,k\in H$, \begin{align*} ((\overline f)(T)h,k)_H=(f(T)^*h,k)_H. \end{align*} Therefore $(\overline f)(T)=f(T)^*$.[/step]

[guided]The limiting argument has two separate jobs. First, it must say why arbitrary bounded Borel functions can be reached from simple functions. Second, for products, it must justify passing limits through an operator product, which is not a consequence of [weak convergence](/page/Weak%20Convergence) alone. Let $u:K\to\mathbb C$ be a bounded Borel function. The standard bounded simple approximation theorem provides bounded Borel simple maps $u_m:K\to\mathbb C$ with $u_m(z)\to u(z)$ for every $z\in K$ and $|u_m(z)|\le \|u\|_{\infty}$ for every $m\in\mathbb N$ and $z\in K$. Concretely, approximate $\operatorname{Re}u$ and $\operatorname{Im}u$ by dyadic step functions and truncate the resulting complex values to stay in the closed disk of radius $\|u\|_{\infty}$. Since $\mu_{h,k}$ is a finite complex measure, dominated convergence applies to the dominating constant $\|u\|_{\infty}$. Thus \begin{align*} \int_K u_m\,d\mu_{h,k}\to \int_K u\,d\mu_{h,k}. \end{align*} By the matrix-coefficient definition of the spectral integral, this is exactly \begin{align*} (u_m(T)h,k)_H\to (u(T)h,k)_H. \end{align*} Linearity is now immediate at the level of scalar matrix coefficients: approximate the three bounded Borel functions $f$, $g$, and $\alpha f+\beta g$ by uniformly bounded simple functions and pass the simple-function identity to the limit using dominated convergence. Multiplicativity requires more care because weak convergence of $s_m(T)$ to $f(T)$ does not by itself allow us to multiply by a varying or fixed operator without justification. We avoid that problem by proving a mixed identity with one simple factor and one bounded Borel factor. Fix a bounded Borel function $v:K\to\mathbb C$ and write the simple function $s:K\to\mathbb C$ as \begin{align*} s=\sum_{i=1}^m a_i\mathbb 1_{A_i}. \end{align*} Then, for $h,k\in H$, \begin{align*} (s(T)v(T)h,k)_H=\sum_{i=1}^m a_i(E(A_i)v(T)h,k)_H. \end{align*} The projection $E(A_i)$ is self-adjoint, so \begin{align*} (E(A_i)v(T)h,k)_H=(v(T)h,E(A_i)k)_H. \end{align*} For a Borel set $B\subset K$, the measure appearing in this last matrix coefficient satisfies \begin{align*} (E(B)h,E(A_i)k)_H=(E(A_i)E(B)h,k)_H=(E(A_i\cap B)h,k)_H. \end{align*} Therefore integration against this measure is the same as integrating over $A_i$ with respect to $\mu_{h,k}$, and hence \begin{align*} (E(A_i)v(T)h,k)_H=\int_K \mathbb 1_{A_i}v\,d\mu_{h,k}. \end{align*} After summing over $i$, we obtain \begin{align*} (s(T)v(T)h,k)_H=\int_K sv\,d\mu_{h,k}=((sv)(T)h,k)_H. \end{align*} Now let $(s_m)_{m=1}^{\infty}$ be uniformly bounded simple approximations to $f$ and set $v=g$. The mixed identity gives \begin{align*} (s_m(T)g(T)h,k)_H=((s_mg)(T)h,k)_H. \end{align*} The left side converges to $(f(T)g(T)h,k)_H$ by the matrix-coefficient convergence applied to the fixed vector $g(T)h$. The right side converges to $((fg)(T)h,k)_H$ because $s_mg\to fg$ pointwise and $|s_mg|\le \|f\|_{\infty}\|g\|_{\infty}$. Hence \begin{align*} (f(T)g(T)h,k)_H=((fg)(T)h,k)_H. \end{align*} Since this holds for all $h,k\in H$, the operators are equal. For the adjoint rule, choose uniformly bounded simple approximations $s_m\to f$. The simple-function rule gives $(\overline{s_m})(T)=s_m(T)^*$ for every $m$. Taking matrix coefficients and passing to the limit gives \begin{align*} ((\overline f)(T)h,k)_H=(f(T)^*h,k)_H \end{align*} for every $h,k\in H$. Hence $(\overline f)(T)=f(T)^*$.[/guided]

[step:Derive the norm identity for one vector] Let $u:K\to\mathbb C$ be a bounded Borel function and let $h\in H$. By the adjoint and multiplicative rules already proved, \begin{align*} \|u(T)h\|_H^2=(u(T)h,u(T)h)_H. \end{align*} Using the definition of the Hilbert-space adjoint, \begin{align*} (u(T)h,u(T)h)_H=(u(T)^*u(T)h,h)_H. \end{align*} The adjoint and multiplicative rules give \begin{align*} u(T)^*u(T)=(\overline u)(T)u(T)=(|u|^2)(T). \end{align*} Therefore, by the matrix-coefficient formula with the positive measure $\mu_h$, \begin{align*} \|u(T)h\|_H^2=\int_K |u|^2\,d\mu_h. \end{align*} [/step]

[step:Apply dominated convergence to prove strong convergence]Assume that $(f_n)_{n=1}^{\infty}$ is uniformly bounded by a constant $M>0$ and that $q:K\to\mathbb C$ is a bounded Borel function satisfying $f_n(z)\to q(z)$ for every $z\in K$. Fix $h\in H$. Define $u_n:K\to\mathbb C$ by \begin{align*} u_n(z):=f_n(z)-q(z). \end{align*} Then $u_n$ is bounded and Borel, $u_n(z)\to 0$ for every $z\in K$, and \begin{align*} |u_n(z)|^2\le (M+\|q\|_{\infty})^2 \end{align*} for every $n\in\mathbb N$ and every $z\in K$. Since $\mu_h$ is a finite positive measure on $K$, the [dominated convergence theorem](/theorems/7529) gives \begin{align*} \int_K |u_n|^2\,d\mu_h\to 0. \end{align*} By the norm identity from the previous step, \begin{align*} \|f_n(T)h-q(T)h\|_H^2=\int_K |f_n-q|^2\,d\mu_h. \end{align*} Thus \begin{align*} \|f_n(T)h-q(T)h\|_H\to 0. \end{align*} Since $h\in H$ was arbitrary, $f_n(T)h\to q(T)h$ in $H$ for every $h\in H$. This is exactly strong convergence of $f_n(T)$ to $q(T)$ on each vector.[/step]

[guided]The statement now names the pointwise limit as a bounded Borel map $q:K\to\mathbb C$, so the operator $q(T)$ is defined by the Borel functional calculus. Fix a vector $h\in H$ and define the bounded Borel difference function $u_n:K\to\mathbb C$ by \begin{align*} u_n(z):=f_n(z)-q(z). \end{align*} The pointwise convergence hypothesis says precisely that $u_n(z)\to 0$ for each $z\in K$. The uniform bound on the sequence and boundedness of $q$ give the pointwise estimate \begin{align*} |u_n(z)|^2\le (M+\|q\|_{\infty})^2. \end{align*} The dominating function is constant and integrable with respect to $\mu_h$ because $\mu_h$ is a finite positive measure. Therefore the dominated convergence theorem applies and yields \begin{align*} \int_K |u_n|^2\,d\mu_h\to 0. \end{align*} The norm identity from the previous step converts this scalar integral convergence into Hilbert-space norm convergence: \begin{align*} \|f_n(T)h-q(T)h\|_H^2=\int_K |f_n-q|^2\,d\mu_h\to 0. \end{align*} Taking square roots gives \begin{align*} \|f_n(T)h-q(T)h\|_H\to 0. \end{align*} Because the vector $h\in H$ was arbitrary, $f_n(T)h\to q(T)h$ for every $h\in H$.[/guided]