[proofplan] We use the definition that a self-adjoint operator $T$ is affiliated with $M$ when every unitary operator in $M'$ commutes with $T$ in the domain-preserving sense. In one direction, affiliation implies commutation with the bounded [Borel functional calculus](/theorems/2696) of $T$, hence with the spectral projections $E_T(B)$; the bicommutant theorem then puts these projections in $M$. Conversely, if all spectral projections lie in $M$, then every unitary in $M'$ commutes with the spectral measure, so it preserves the spectral-[integral domain](/page/Integral%20Domain) of $T$ and commutes with the spectral integral defining $T$. [/proofplan] [step:Fix the affiliation convention and the spectral-integral domain] We write $M' := \{a \in \mathcal{L}(H) : ax = xa \text{ for all } x \in M\}$ for the commutant of $M$. The self-adjoint operator $T$ is affiliated with $M$, written $T\eta M$, if for every unitary $u\in M'$ one has $uT\subseteq Tu$; equivalently, $u(\operatorname{Dom}(T))\subseteq \operatorname{Dom}(T)$ and \begin{align*} T(u\xi)=u(T\xi) \end{align*} for every $\xi\in\operatorname{Dom}(T)$. Since $u^{-1}\in M'$ is also unitary, this condition implies $u(\operatorname{Dom}(T))=\operatorname{Dom}(T)$. For $\xi\in H$, define the finite positive Borel measure $\mu_\xi$ on $\mathbb{R}$ by \begin{align*} \mu_\xi(B)=(E_T(B)\xi,\xi)_H,\qquad B\in\mathcal{B}(\mathbb{R}). \end{align*} By the [spectral theorem for self-adjoint operators](/theorems/6911), the domain of $T$ is \begin{align*} \operatorname{Dom}(T)=\left\{\xi\in H:\int_{\mathbb{R}}\lambda^2\,d\mu_\xi(\lambda)<\infty\right\}. \end{align*} Moreover, if $f:\mathbb{R}\to\mathbb{C}$ is a bounded Borel function, the bounded Borel functional calculus gives an operator \begin{align*} f(T)=\int_{\mathbb{R}}f(\lambda)\,dE_T(\lambda)\in\mathcal{L}(H). \end{align*} [/step] [step:Derive membership of the spectral projections from affiliation] Assume $T\eta M$. Let $u\in M'$ be unitary. By affiliation, $u$ commutes with $T$ in the domain-preserving self-adjoint sense. The commutation theorem for the Borel functional calculus of a self-adjoint operator then gives \begin{align*} u f(T)=f(T)u \end{align*} for every bounded Borel function $f:\mathbb{R}\to\mathbb{C}$. Taking $f=\mathbb{1}_B$ for a Borel set $B\in\mathcal{B}(\mathbb{R})$, we obtain \begin{align*} uE_T(B)=E_T(B)u. \end{align*} Thus $E_T(B)$ commutes with every unitary in $M'$. The unitaries of a unital $C^*$-algebra generate it as a linear span, so $E_T(B)$ commutes with every element of $M'$. Hence \begin{align*} E_T(B)\in (M')'. \end{align*} Since $M$ is a von Neumann algebra, the bicommutant theorem gives $(M')'=M$. Therefore $E_T(B)\in M$ for every Borel set $B\subseteq\mathbb{R}$. [guided] Assume that $T$ is affiliated with $M$. This means that every unitary $u\in M'$ preserves the domain of $T$ and satisfies \begin{align*} T(u\xi)=u(T\xi) \end{align*} for every $\xi\in\operatorname{Dom}(T)$. The point of using unitaries is that the spectral theorem converts this domain-level commutation relation into commutation with the bounded Borel functional calculus. More precisely, the standard commutation theorem for [self-adjoint operators](/page/Self-Adjoint%20Operators) says: if a unitary operator $u\in\mathcal{L}(H)$ commutes with a self-adjoint operator $T$ in the domain-preserving sense, then $u$ commutes with every bounded Borel function of $T$. Therefore, for every bounded Borel map $f:\mathbb{R}\to\mathbb{C}$, \begin{align*} u f(T)=f(T)u. \end{align*} Now choose the bounded Borel function $f:\mathbb{R}\to\mathbb{C}$ given by $f=\mathbb{1}_B$, where $B\in\mathcal{B}(\mathbb{R})$. The functional calculus identifies this operator with the spectral projection: \begin{align*} \mathbb{1}_B(T)=E_T(B). \end{align*} Hence \begin{align*} uE_T(B)=E_T(B)u \end{align*} for every unitary $u\in M'$. We now translate commutation with all unitaries of $M'$ into membership in $M$. Since $M'$ is a unital $C^*$-algebra, its elements are linear combinations of unitaries. Thus an operator commuting with every unitary in $M'$ commutes with every element of $M'$. Therefore \begin{align*} E_T(B)\in (M')'. \end{align*} Finally, because $M$ is a von Neumann algebra, the bicommutant theorem gives \begin{align*} (M')'=M. \end{align*} Consequently $E_T(B)\in M$ for every Borel set $B\subseteq\mathbb{R}$. [/guided] [/step] [step:Show spectral projections in $M$ force domain preservation] Assume now that $E_T(B)\in M$ for every $B\in\mathcal{B}(\mathbb{R})$. Let $u\in M'$ be unitary. Since each $E_T(B)$ belongs to $M$, and $u$ commutes with every element of $M$, we have \begin{align*} uE_T(B)=E_T(B)u \end{align*} for every Borel set $B\subseteq\mathbb{R}$. Let $\xi\in\operatorname{Dom}(T)$. For every Borel set $B\subseteq\mathbb{R}$, \begin{align*} \mu_{u\xi}(B)=(E_T(B)u\xi,u\xi)_H=(uE_T(B)\xi,u\xi)_H=(E_T(B)\xi,\xi)_H=\mu_\xi(B). \end{align*} Hence $\mu_{u\xi}=\mu_\xi$. Since $\xi\in\operatorname{Dom}(T)$, \begin{align*} \int_{\mathbb{R}}\lambda^2\,d\mu_\xi(\lambda)<\infty. \end{align*} The equality of measures gives \begin{align*} \int_{\mathbb{R}}\lambda^2\,d\mu_{u\xi}(\lambda)<\infty. \end{align*} Therefore $u\xi\in\operatorname{Dom}(T)$. Thus \begin{align*} u(\operatorname{Dom}(T))\subseteq\operatorname{Dom}(T). \end{align*} Applying the same argument to $u^{-1}\in M'$ gives equality of domains. [/step] [step:Commute the unitary with the spectral integral defining $T$] For each $n\in\mathbb{N}$, define the bounded Borel function \begin{align*} f_n:\mathbb{R}&\to\mathbb{R} \end{align*} \begin{align*} \lambda&\mapsto \lambda\,\mathbb{1}_{[-n,n]}(\lambda). \end{align*} Let \begin{align*} T_n:=f_n(T)=\int_{\mathbb{R}}f_n(\lambda)\,dE_T(\lambda)\in\mathcal{L}(H). \end{align*} Since $u$ commutes with every spectral projection $E_T(B)$, it commutes with every bounded Borel function of $T$. Hence \begin{align*} uT_n=T_nu \end{align*} for every $n\in\mathbb{N}$. Let $\xi\in\operatorname{Dom}(T)$. By the spectral-integral construction of $T$, \begin{align*} T_n\xi\to T\xi \end{align*} in $H$. Since $u\xi\in\operatorname{Dom}(T)$, also \begin{align*} T_nu\xi\to Tu\xi \end{align*} in $H$. Using $T_nu=uT_n$ and the boundedness of $u$, we get \begin{align*} Tu\xi=\lim_{n\to\infty}T_nu\xi=\lim_{n\to\infty}uT_n\xi=uT\xi. \end{align*} Therefore $uT\subseteq Tu$ for every unitary $u\in M'$. [/step] [step:Conclude the equivalence] The first implication showed that affiliation of $T$ with $M$ forces every spectral projection $E_T(B)$ to belong to $M$. The converse showed that if every spectral projection $E_T(B)$ belongs to $M$, then every unitary $u\in M'$ preserves $\operatorname{Dom}(T)$ and satisfies \begin{align*} T(u\xi)=u(T\xi) \end{align*} for every $\xi\in\operatorname{Dom}(T)$. This is exactly the affiliation condition. Hence $T$ is affiliated with $M$ if and only if $E_T(B)\in M$ for every Borel set $B\subseteq\mathbb{R}$. [/step]