[proofplan] We prove first that normality implies complete additivity by replacing an arbitrary orthogonal family of projections with the directed net of its finite partial sums. These finite sums form an increasing bounded net of projections whose strong limit is the join, and normality preserves this bounded monotone limit. Conversely, the assumed complete additivity is exactly the projection-additivity condition in the normality criterion for positive functionals, so applying that criterion to the positive functional $\phi$ gives normality. [/proofplan] [step:Replace the orthogonal family by its finite partial-sum net] Assume first that $\phi$ is normal. Let $I$ be an index set, let $(p_i)_{i\in I}$ be a pairwise orthogonal family in $\mathcal P(M)$, where $\mathcal P(M)$ denotes the set of projections in $M$, and define \begin{align*} p=\bigvee_{i\in I}p_i. \end{align*} Let $\mathcal F(I)$ denote the directed set of finite subsets of $I$, ordered by inclusion. For each $F\in\mathcal F(I)$, define \begin{align*} q_F=\sum_{i\in F}p_i\in M. \end{align*} Because the projections $p_i$ are pairwise orthogonal, each $q_F$ is a projection. If $F\subset G$, then \begin{align*} q_G-q_F=\sum_{i\in G\setminus F}p_i \end{align*} is a projection, hence positive, so $q_F\le q_G$. Thus $(q_F)_{F\in\mathcal F(I)}$ is an increasing net of [self-adjoint operators](/page/Self-Adjoint%20Operators) in $M$, bounded by $1$. The supremum of this net in the projection order is $p$. Indeed, every $q_F$ is bounded above by $p$, and if $r\in\mathcal P(M)$ satisfies $q_F\le r$ for all finite $F\subset I$, then $p_i\le r$ for every $i\in I$, so $p\le r$ by the definition of $p$ as the join. The net $(q_F)_{F\in\mathcal F(I)}$ is increasing and bounded in the self-adjoint part of $M\subseteq\mathcal L(H)$, and its supremum in the projection order is $p$. By [citetheorem:9262], the monotone strong limit is this supremum, so $q_F\to p$ strongly. [guided] The reason for introducing finite subsets is that an uncountable sum of projections is not interpreted as an algebraic sum. The statement defines the right-hand side as the supremum of finite partial sums, so we build exactly that directed system. Let $\mathcal F(I)$ be the set of finite subsets of $I$, ordered by inclusion. This is directed: if $F,G\in\mathcal F(I)$, then $F\cup G\in\mathcal F(I)$ and both $F\subset F\cup G$ and $G\subset F\cup G$. For each finite subset $F\subset I$, define \begin{align*} q_F=\sum_{i\in F}p_i. \end{align*} Since the projections $p_i$ are pairwise orthogonal, the finite sum $q_F$ is again a projection: it is self-adjoint because each $p_i$ is self-adjoint, and its square is \begin{align*} q_F^2=\sum_{i\in F}p_i^2+\sum_{\substack{i, j\in F, i\ne j}}p_ip_j=\sum_{i\in F}p_i=q_F. \end{align*} If $F\subset G$, then the difference is \begin{align*} q_G-q_F=\sum_{i\in G\setminus F}p_i. \end{align*} This is a projection, hence a positive operator, so $q_F\le q_G$. Therefore $(q_F)_{F\in\mathcal F(I)}$ is an increasing net of projections, hence an increasing bounded net of self-adjoint elements of $M$. Now we identify its supremum. Since $p=\bigvee_{i\in I}p_i$, each $p_i\le p$, and therefore every finite sum $q_F$ satisfies $q_F\le p$. Conversely, suppose $r\in\mathcal P(M)$ is a projection with $q_F\le r$ for every finite $F\subset I$. Taking $F=\{i\}$ gives $p_i\le r$ for every $i\in I$. Since $p$ is the least projection dominating all the $p_i$, this implies $p\le r$. Thus $p$ is exactly the supremum of the net $(q_F)_{F\in\mathcal F(I)}$. For increasing bounded nets of self-adjoint operators in a von Neumann algebra, the supremum is the strong operator limit by [citetheorem:9262]. Applying that result to the net $(q_F)$ gives strong convergence $q_F\to p$. [/guided] [/step] [step:Use normality to pass from finite additivity to complete additivity] Since $\phi$ is a positive normal functional and $(q_F)_{F\in\mathcal F(I)}$ is an increasing bounded net of self-adjoint elements of $M$ with strong limit $p$, [citetheorem:9263] gives \begin{align*} \phi(q_F)\uparrow \phi(p). \end{align*} For each finite $F\subset I$, linearity of $\phi$ gives \begin{align*} \phi(q_F)=\phi\left(\sum_{i\in F}p_i\right)=\sum_{i\in F}\phi(p_i). \end{align*} Hence \begin{align*} \phi(p)=\sup\left\{\sum_{i\in F}\phi(p_i):F\subset I\text{ is finite}\right\}. \end{align*} This proves complete additivity on arbitrary pairwise [orthogonal projection](/theorems/437) families. [/step] [step:Recognize complete projection additivity as the normality criterion] Conversely, assume that for every index set $I$ and every pairwise orthogonal family $(p_i)_{i\in I}$ in $\mathcal P(M)$, with \begin{align*} p=\bigvee_{i\in I}p_i, \end{align*} one has \begin{align*} \phi(p)=\sup\left\{\sum_{i\in F}\phi(p_i):F\subset I\text{ is finite}\right\}. \end{align*} Since $\phi$ is a state, it is a positive functional on $M$. The displayed condition is precisely complete additivity on arbitrary pairwise orthogonal projection families, with the sum interpreted as the supremum of finite partial sums. By [citetheorem:9272], this condition is equivalent, for positive functionals on a von Neumann algebra, to normality. Therefore $\phi$ is normal. [/step] [step:Conclude the equivalence] The first two steps prove that every normal state is completely additive on arbitrary orthogonal families of projections. The third step proves that complete additivity on arbitrary orthogonal projection families forces the state to be normal. Hence the two conditions are equivalent. [/step]