[proofplan] First we prove the result for positive increasing nets, where every scalar coefficient $(T_i\xi,\xi)_H$ is an increasing net converging to $(T\xi,\xi)_H$ by strong convergence. A positive normal functional on a concrete von Neumann algebra admits a summable vector-state decomposition, so the problem reduces to interchanging an increasing directed limit with a nonnegative countable series. The general self-adjoint case is reduced to the positive case by fixing one index $i_0$ and applying the positive argument to the tail $(T_i-T_{i_0})_{i\ge i_0}$. [/proofplan]

[step:Reduce positive normal functionals to summable vector states]We use the standard vector-state decomposition theorem for positive normal functionals on concrete von Neumann algebras, obtained from normal positive extension to $\mathcal{L}(K)$ and the trace-class description of the predual of $\mathcal{L}(K)$ in [citetheorem:9274]: if $N\subseteq \mathcal{L}(K)$ is a von Neumann algebra on a [Hilbert space](/page/Hilbert%20Space) $K$ and $\psi:N\to\mathbb C$ is a positive normal linear functional, then there exists a sequence $(\eta_n)_{n\ge 1}$ in $K$ such that \begin{align*} \sum_{n=1}^{\infty}\|\eta_n\|_K^2<\infty \end{align*} and, for every $R\in N$, \begin{align*} \psi(R)=\sum_{n=1}^{\infty}(R\eta_n,\eta_n)_K. \end{align*} This theorem applies with $N=M$, $K=H$, and $\psi=\varphi$, because $M\subseteq \mathcal{L}(H)$ is a von Neumann algebra and $\varphi:M\to\mathbb C$ is positive and normal. Hence there exists a sequence $(\xi_n)_{n\ge 1}$ in $H$ such that \begin{align*} \sum_{n=1}^{\infty}\|\xi_n\|_H^2<\infty \end{align*} and, for every $S\in M$, \begin{align*} \varphi(S)=\sum_{n=1}^{\infty}(S\xi_n,\xi_n)_H. \end{align*} For each $S\in M$, the displayed series is absolutely convergent because \begin{align*} \sum_{n=1}^{\infty}|(S\xi_n,\xi_n)_H|\le \|S\|_{\mathrm{op}}\sum_{n=1}^{\infty}\|\xi_n\|_H^2<\infty. \end{align*}[/step]

[guided]The point of this step is to replace an abstract normal functional by concrete scalar coefficients in the Hilbert space representation. We use the standard vector-state decomposition theorem for positive normal functionals on concrete von Neumann algebras; its standard proof extends the functional normally and positively to $\mathcal{L}(K)$ and then uses the trace-class predual description of $\mathcal{L}(K)$ in [citetheorem:9274]. Its hypotheses are: a von Neumann algebra $N\subseteq \mathcal{L}(K)$ on a Hilbert space $K$, and a positive normal linear functional $\psi:N\to\mathbb C$. Its conclusion is that there is a sequence $(\eta_n)_{n\ge 1}$ in $K$ with \begin{align*} \sum_{n=1}^{\infty}\|\eta_n\|_K^2<\infty \end{align*} such that \begin{align*} \psi(R)=\sum_{n=1}^{\infty}(R\eta_n,\eta_n)_K \end{align*} for every $R\in N$. We verify the hypotheses in the present setting. The theorem statement gives that $H$ is a Hilbert space, that $M\subseteq\mathcal{L}(H)$ is a von Neumann algebra, and that $\varphi:M\to\mathbb C$ is a positive normal linear functional. Therefore the theorem applies with $N=M$, $K=H$, and $\psi=\varphi$. We obtain a sequence $(\xi_n)_{n\ge 1}$ in $H$ satisfying \begin{align*} \sum_{n=1}^{\infty}\|\xi_n\|_H^2<\infty \end{align*} and \begin{align*} \varphi(S)=\sum_{n=1}^{\infty}(S\xi_n,\xi_n)_H \end{align*} for every $S\in M$. We also need to know that the series can be evaluated safely on every bounded operator $S\in M$. For each $n\ge 1$, the [Cauchy-Schwarz inequality](/theorems/432) in $H$ gives \begin{align*} |(S\xi_n,\xi_n)_H|\le \|S\xi_n\|_H\|\xi_n\|_H. \end{align*} The definition of the operator norm gives \begin{align*} \|S\xi_n\|_H\le \|S\|_{\mathrm{op}}\|\xi_n\|_H. \end{align*} Combining these inequalities, \begin{align*} |(S\xi_n,\xi_n)_H|\le \|S\|_{\mathrm{op}}\|\xi_n\|_H^2. \end{align*} Summing over $n$ yields \begin{align*} \sum_{n=1}^{\infty}|(S\xi_n,\xi_n)_H|\le \|S\|_{\mathrm{op}}\sum_{n=1}^{\infty}\|\xi_n\|_H^2<\infty. \end{align*} Thus the representation of $\varphi(S)$ is an absolutely convergent scalar series.[/guided]

[step:Prove the theorem for positive increasing nets] Assume first that $0\le T_i\le T_j$ whenever $i\le j$ and that $T_i\to T$ strongly. Fix $n\ge 1$. Since $T_j-T_i\ge 0$ for $i\le j$, the scalar net $((T_i\xi_n,\xi_n)_H)_{i\in I}$ is increasing. Strong convergence gives \begin{align*} \lim_{i\in I}(T_i\xi_n,\xi_n)_H=(T\xi_n,\xi_n)_H. \end{align*} Also $T-T_i\ge 0$ for every $i\in I$, because for each $\eta\in H$, \begin{align*} ((T-T_i)\eta,\eta)_H=\lim_{j\in I,\,j\ge i}((T_j-T_i)\eta,\eta)_H\ge 0. \end{align*} Therefore \begin{align*} 0\le (T_i\xi_n,\xi_n)_H\le (T\xi_n,\xi_n)_H \end{align*} for every $i\in I$ and $n\ge 1$. Define $a_{i,n}:=(T_i\xi_n,\xi_n)_H$ and $a_n:=(T\xi_n,\xi_n)_H$ for $i\in I$ and $n\ge 1$. Then $0\le a_{i,n}\uparrow a_n$ in $i$ for each $n$. The next claim is the directed-net version of monotone convergence for a countable series of nonnegative terms; the finite-subset directedness argument replaces the sequential [monotone convergence theorem](/theorems/509). We claim that \begin{align*} \sup_{i\in I}\sum_{n=1}^{\infty}a_{i,n}=\sum_{n=1}^{\infty}a_n. \end{align*} The inequality $\le$ follows from $a_{i,n}\le a_n$. For the reverse inequality, let $N\ge 1$. Since the finite set $\{1,\dots,N\}$ is finite and $a_{i,n}\uparrow a_n$ for each $1\le n\le N$, the directedness of $I$ gives \begin{align*} \sup_{i\in I}\sum_{n=1}^{N}a_{i,n}=\sum_{n=1}^{N}a_n. \end{align*} Hence \begin{align*} \sup_{i\in I}\sum_{n=1}^{\infty}a_{i,n}\ge \sum_{n=1}^{N}a_n. \end{align*} Taking the supremum over $N$ gives the reverse inequality. Using the vector-state decomposition, \begin{align*} \sup_{i\in I}\varphi(T_i)=\sup_{i\in I}\sum_{n=1}^{\infty}(T_i\xi_n,\xi_n)_H=\sum_{n=1}^{\infty}(T\xi_n,\xi_n)_H=\varphi(T). \end{align*} Since $\varphi$ is positive, $T_i\le T_j$ implies $\varphi(T_i)\le\varphi(T_j)$, so $\varphi(T_i)\uparrow\varphi(T)$ in the positive case. [/step]

[step:Pass from positive increasing nets to arbitrary self-adjoint increasing nets] Now let $(T_i)_{i\in I}$ be an arbitrary increasing net of [self-adjoint operators](/page/Self-Adjoint%20Operators) in $M$ with strong limit $T\in M$. Since $I$ is directed and nonempty, fix $i_0\in I$. Let \begin{align*} I_{i_0}:=\{i\in I:i\ge i_0\} \end{align*} with the order inherited from $I$. This is a directed set. Define a net $(A_i)_{i\in I_{i_0}}$ in $M$ by \begin{align*} A_i:=T_i-T_{i_0}. \end{align*} Then each $A_i$ is self-adjoint, $A_i\ge 0$, and $A_i\le A_j$ whenever $i\le j$ in $I_{i_0}$. Strong convergence of $T_i$ to $T$ gives strong convergence of $A_i$ to \begin{align*} A:=T-T_{i_0}. \end{align*} By the positive case, \begin{align*} \varphi(A_i)\uparrow\varphi(A). \end{align*} Since $\varphi$ is linear, \begin{align*} \varphi(T_i)=\varphi(A_i)+\varphi(T_{i_0}) \end{align*} for every $i\in I_{i_0}$, and \begin{align*} \varphi(T)=\varphi(A)+\varphi(T_{i_0}). \end{align*} Therefore \begin{align*} \lim_{i\in I,\,i\ge i_0}\varphi(T_i)=\varphi(T). \end{align*} The full net $(\varphi(T_i))_{i\in I}$ is increasing because $T_i\le T_j$ and $\varphi$ is positive. The subset $I_{i_0}$ is cofinal in $I$: for every $k\in I$, directedness gives an index $l\in I$ with $l\ge k$ and $l\ge i_0$, hence $l\in I_{i_0}$. Therefore the tail indexed by $I_{i_0}$ has the same supremum as the full increasing net, and we obtain \begin{align*} \sup_{i\in I}\varphi(T_i)=\varphi(T). \end{align*} Thus \begin{align*} \varphi(T_i)\uparrow\varphi(T). \end{align*} This proves the theorem. [/step]