[proofplan] We first use ergodicity to show that every non-zero eigenfunction has constant modulus and that each eigenspace is one-dimensional. Orthogonality of distinct eigenspaces inside the separable Hilbert space $L^2(X,\mathcal B,\mu)$ gives countability, and products and conjugates give the subgroup property. We then choose the normalized eigenfunctions multiplicatively, use them as coordinate functions for a measurable map into the compact dual group $G=\widehat{\Lambda(T)}$, and prove by Fourier coefficients that the pushforward of $\mu$ is Haar measure. Finally, the pullback by this map sends the character basis of $L^2(G)$ onto the eigenfunction basis of $L^2(X)$, so the factor map is an isomorphism; the classification follows by comparing both systems with the same compact rotation model. [/proofplan]

[step:Normalize eigenfunctions and make each eigenspace one-dimensional] Let \begin{align*} U_T:L^2(X,\mathcal B,\mu;\mathbb C)&\to L^2(X,\mathcal B,\mu;\mathbb C)\\ [f]&\mapsto [f\circ T] \end{align*} be the Koopman operator of $T$. Since $T$ is measure-preserving, $U_T$ is an isometry. Define the eigenvalue set \begin{align*} \Lambda(T):=\{\lambda\in \mathbb T:\ker(U_T-\lambda I)\neq \{0\}\}. \end{align*} For $\lambda\in \Lambda(T)$, write \begin{align*} E_\lambda:=\ker(U_T-\lambda I) \end{align*} for the corresponding eigenspace. Let $\lambda\in\Lambda(T)$ and let $f\in E_\lambda$ be non-zero. Then \begin{align*} |f|\circ T=|f\circ T|=|\lambda f|=|f| \end{align*} as elements of $L^2(X,\mathcal B,\mu)$. By the invariant-function form of the [Equivalence of Ergodicity Conditions](/theorems/3444), every $T$-invariant function in $L^2(X,\mathcal B,\mu)$ is constant almost everywhere. Hence there is a constant $c_\lambda>0$ such that $|f|=c_\lambda$ almost everywhere. Therefore $c_\lambda^{-1}f$ is a $\mathbb T$-valued eigenfunction with eigenvalue $\lambda$. Now let $u,v\in E_\lambda$ be $\mathbb T$-valued eigenfunctions. Define \begin{align*} h:X&\to \mathbb T\\ x&\mapsto u(x)\overline{v(x)} \end{align*} as an element of $L^\infty(X,\mathcal B,\mu;\mathbb T)$. Since $u\circ T=\lambda u$ and $v\circ T=\lambda v$ almost everywhere, \begin{align*} h\circ T=(u\circ T)\overline{(v\circ T)} =\lambda u\,\overline{\lambda v} =u\overline v =h \end{align*} almost everywhere. Applying the invariant-function criterion again, $h=a$ almost everywhere for some $a\in\mathbb T$. Thus $u=av$ almost everywhere. Consequently every eigenspace $E_\lambda$ is one-dimensional. [/step]

[step:Prove that the eigenvalue set is a countable subgroup of $\mathbb T$] The constant function $\mathbf 1_X:X\to\mathbb C$, $x\mapsto 1$, belongs to $E_1$, so $1\in\Lambda(T)$. Let $\lambda,\eta\in\Lambda(T)$. Choose $\mathbb T$-valued eigenfunctions $u_\lambda\in E_\lambda$ and $u_\eta\in E_\eta$, whose existence was proved in the previous step. The product \begin{align*} u_\lambda u_\eta:X&\to \mathbb T\\ x&\mapsto u_\lambda(x)u_\eta(x) \end{align*} satisfies \begin{align*} (u_\lambda u_\eta)\circ T =(u_\lambda\circ T)(u_\eta\circ T) =(\lambda u_\lambda)(\eta u_\eta) =(\lambda\eta)(u_\lambda u_\eta) \end{align*} almost everywhere. Hence $\lambda\eta\in\Lambda(T)$. Also \begin{align*} \overline{u_\lambda}\circ T =\overline{u_\lambda\circ T} =\overline{\lambda u_\lambda} =\lambda^{-1}\overline{u_\lambda} \end{align*} almost everywhere, so $\lambda^{-1}\in\Lambda(T)$. Thus $\Lambda(T)$ is a subgroup of $\mathbb T$. For countability, choose for each $\lambda\in\Lambda(T)$ a $\mathbb T$-valued eigenfunction $u_\lambda\in E_\lambda$. Since $\mu$ is a probability measure, \begin{align*} \|u_\lambda\|_{L^2(X,\mu)}^2 =\int_X |u_\lambda(x)|^2\,d\mu(x) =\int_X 1\,d\mu(x) =1. \end{align*} If $\lambda\neq\eta$, then \begin{align*} \langle u_\lambda,u_\eta\rangle_{L^2(X,\mu)} &=\langle U_Tu_\lambda,U_Tu_\eta\rangle_{L^2(X,\mu)}\\ &=\lambda\overline{\eta}\,\langle u_\lambda,u_\eta\rangle_{L^2(X,\mu)}. \end{align*} Because $\lambda\overline{\eta}\neq 1$, it follows that \begin{align*} \langle u_\lambda,u_\eta\rangle_{L^2(X,\mu)}=0. \end{align*} Thus $\{u_\lambda:\lambda\in\Lambda(T)\}$ is an orthonormal set. By the [Separability of $L^p$ Spaces](/theorems/548), applied with $p=2$ to the standard probability space $(X,\mathcal B,\mu)$, the Hilbert space $L^2(X,\mathcal B,\mu)$ is separable. A separable Hilbert space has at most countable orthonormal sets, so $\Lambda(T)$ is countable. [/step]

[step:Choose eigenfunctions that multiply according to the eigenvalue group] Let $\mathcal E$ be the abelian group, under pointwise multiplication of equivalence classes, of all $\mathbb T$-valued eigenfunctions of $U_T$. Define \begin{align*} q:\mathcal E&\to \Lambda(T)\\ u&\mapsto \lambda \quad\text{where } U_Tu=\lambda u. \end{align*} The map $q$ is a surjective group homomorphism. Its kernel consists exactly of the constant $\mathbb T$-valued functions, by ergodicity and the one-dimensionality of $E_1$. We construct a group homomorphism \begin{align*} s:\Lambda(T)&\to \mathcal E \end{align*} such that $q\circ s=\operatorname{id}_{\Lambda(T)}$. Suppose $H$ is a subgroup of $\Lambda(T)$ and \begin{align*} s_H:H&\to\mathcal E \end{align*} is a homomorphism with $q\circ s_H=\operatorname{id}_H$. Let $\alpha\in\Lambda(T)$ and set $K:=\langle H,\alpha\rangle$. If no positive integer $n$ satisfies $\alpha^n\in H$, then each element of $K$ has a unique form $\beta\alpha^k$ with $\beta\in H$ and $k\in\mathbb Z$. Choose $u\in\mathcal E$ with $q(u)=\alpha$, and define \begin{align*} s_K(\beta\alpha^k):=s_H(\beta)u^k. \end{align*} The uniqueness of the representation makes $s_K$ well-defined, and the definition gives a homomorphism extending $s_H$ with $q\circ s_K=\operatorname{id}_K$. If there is a positive integer $n$ with $\alpha^n\in H$, let $n$ be the least such integer. Choose $u\in\mathcal E$ with $q(u)=\alpha$. Then \begin{align*} u^n s_H(\alpha^n)^{-1}\in \ker q, \end{align*} so there is a constant $c\in\mathbb T$ such that \begin{align*} u^n s_H(\alpha^n)^{-1}=c. \end{align*} Choose $a\in\mathbb T$ with $a^n=c^{-1}$, and set $v:=au$. Then \begin{align*} v^n=s_H(\alpha^n). \end{align*} Every element of $K$ has a unique form $\beta\alpha^k$ with $\beta\in H$ and $0\leq k\leq n-1$. Define \begin{align*} s_K(\beta\alpha^k):=s_H(\beta)v^k. \end{align*} The relation $v^n=s_H(\alpha^n)$ proves that $s_K$ is multiplicative when exponents cross $n$, so $s_K$ is a homomorphism extending $s_H$ and satisfying $q\circ s_K=\operatorname{id}_K$. Since $\Lambda(T)$ is countable, choose an enumeration $(\lambda_j)_{j\in\mathbb N}$ of $\Lambda(T)$ with $\lambda_1=1$. Start with $H_1=\{1\}$ and $s_{H_1}(1)=\mathbf 1_X$. Applying the extension construction inductively to \begin{align*} H_j:=\langle \lambda_1,\ldots,\lambda_j\rangle \end{align*} gives compatible homomorphic sections $s_{H_j}:H_j\to\mathcal E$. Their union is the required section $s:\Lambda(T)\to\mathcal E$. For each $\lambda\in\Lambda(T)$, set \begin{align*} e_\lambda:=s(\lambda). \end{align*} Then \begin{align*} e_1=\mathbf 1_X,\qquad e_{\lambda\eta}=e_\lambda e_\eta,\qquad U_Te_\lambda=\lambda e_\lambda \end{align*} for all $\lambda,\eta\in\Lambda(T)$. [/step]

[step:Realize the multiplicative eigenfunctions as a measurable map into the dual group] For each $\lambda\in\Lambda(T)$, choose a measurable representative \begin{align*} \widetilde e_\lambda:X&\to\mathbb T \end{align*} of the equivalence class $e_\lambda$. Since $\Lambda(T)$ is countable, there is a measurable set $X_0\in\mathcal B$ with $\mu(X_0)=1$ such that $T(X_0)\subseteq X_0$ and, for every $x\in X_0$ and every $\lambda,\eta\in\Lambda(T)$, \begin{align*} \widetilde e_\lambda(Tx)=\lambda \widetilde e_\lambda(x), \qquad \widetilde e_{\lambda\eta}(x)=\widetilde e_\lambda(x)\widetilde e_\eta(x), \qquad \widetilde e_1(x)=1. \end{align*} Indeed, take the union of the countably many null sets on which one of these identities fails, and then remove all of its inverse images under the maps $T^k$ for $k\in\mathbb N\cup\{0\}$. Let \begin{align*} G:=\widehat{\Lambda(T)} \end{align*} be the compact abelian group of all homomorphisms $\gamma:\Lambda(T)\to\mathbb T$, equipped with the product topology inherited from $\mathbb T^{\Lambda(T)}$. Let $\mathcal B_G$ be its Borel $\sigma$-algebra. Define \begin{align*} \Phi:X&\to G \end{align*} by \begin{align*} \Phi(x)(\lambda)= \begin{cases} \widetilde e_\lambda(x),&x\in X_0,\\ 1,&x\notin X_0. \end{cases} \end{align*} For $x\in X_0$, the identity $\widetilde e_{\lambda\eta}(x)=\widetilde e_\lambda(x)\widetilde e_\eta(x)$ shows that $\lambda\mapsto\Phi(x)(\lambda)$ is a homomorphism, so $\Phi(x)\in G$. Since the coordinate maps generate $\mathcal B_G$ and each coordinate $\Phi(\cdot)(\lambda)$ is $\mathcal B$-measurable, $\Phi$ is measurable. Define \begin{align*} g:\Lambda(T)&\to\mathbb T\\ \lambda&\mapsto \lambda \end{align*} and \begin{align*} R_g:G&\to G\\ \gamma&\mapsto g\gamma, \end{align*} where $(g\gamma)(\lambda)=g(\lambda)\gamma(\lambda)$. Since $\Lambda(T)$ is a subgroup of $\mathbb T$, the map $g$ is a character of $\Lambda(T)$, hence $g\in G$. For $x\in X_0$ and $\lambda\in\Lambda(T)$, \begin{align*} \Phi(Tx)(\lambda) =\widetilde e_\lambda(Tx) =\lambda \widetilde e_\lambda(x) =g(\lambda)\Phi(x)(\lambda) =(R_g(\Phi(x)))(\lambda). \end{align*} Therefore \begin{align*} \Phi\circ T=R_g\circ \Phi \end{align*} almost everywhere. [/step]

[step:Identify the pushforward measure as Haar measure by Fourier coefficients] Let \begin{align*} \nu:=\Phi_*\mu \end{align*} be the pushforward probability measure on $(G,\mathcal B_G)$, and let $m_G$ denote Haar probability measure on $G$. For each $\lambda\in\Lambda(T)$, define the continuous character \begin{align*} \chi_\lambda:G&\to\mathbb T\\ \gamma&\mapsto \gamma(\lambda). \end{align*} Then \begin{align*} \int_G \chi_\lambda(\gamma)\,d\nu(\gamma) &=\int_X \chi_\lambda(\Phi(x))\,d\mu(x)\\ &=\int_X \widetilde e_\lambda(x)\,d\mu(x). \end{align*} If $\lambda=1$, this integral equals $1$. If $\lambda\neq 1$, then $e_\lambda$ is orthogonal to the constant eigenspace $E_1$, so \begin{align*} \int_X \widetilde e_\lambda(x)\,d\mu(x) =\langle e_\lambda,\mathbf 1_X\rangle_{L^2(X,\mu)} =0. \end{align*} Thus \begin{align*} \int_G \chi_\lambda(\gamma)\,d\nu(\gamma) = \begin{cases} 1,&\lambda=1,\\ 0,&\lambda\neq 1. \end{cases} \end{align*} The same identities hold for $m_G$. For $\lambda=1$, $\chi_1$ is the constant function $1$. For $\lambda\neq 1$, characters on the discrete abelian group separate points, so there is some $\gamma_0\in G$ such that $\chi_\lambda(\gamma_0)\neq 1$. If \begin{align*} I_\lambda:=\int_G \chi_\lambda(\gamma)\,d m_G(\gamma), \end{align*} then left-invariance of Haar measure gives \begin{align*} I_\lambda =\int_G \chi_\lambda(\gamma_0\gamma)\,d m_G(\gamma) =\chi_\lambda(\gamma_0) I_\lambda. \end{align*} Since $\chi_\lambda(\gamma_0)\neq 1$, $I_\lambda=0$. The linear span of $\{\chi_\lambda:\lambda\in\Lambda(T)\}$ is a unital self-adjoint subalgebra of $C(G;\mathbb C)$ and separates points of $G$. By the [Stone-Weierstrass Theorem](/theorems/886), it is uniformly dense in $C(G;\mathbb C)$. Since $\nu$ and $m_G$ are regular Borel probability measures on the compact Hausdorff space $G$ and agree on a uniformly dense subspace of $C(G;\mathbb C)$, the [Riesz-Markov-Kakutani Representation Theorem](/theorems/976) implies \begin{align*} \nu=m_G. \end{align*} [/step]

[step:Use spectral density to make the factor map an isomorphism] Define the pullback operator \begin{align*} C_\Phi:L^2(G,\mathcal B_G,m_G;\mathbb C)&\to L^2(X,\mathcal B,\mu;\mathbb C)\\ [F]&\mapsto [F\circ\Phi]. \end{align*} Since $\Phi_*\mu=m_G$, for every $F\in L^2(G,\mathcal B_G,m_G;\mathbb C)$, \begin{align*} \|C_\Phi F\|_{L^2(X,\mu)}^2 &=\int_X |F(\Phi(x))|^2\,d\mu(x)\\ &=\int_G |F(\gamma)|^2\,d(\Phi_*\mu)(\gamma)\\ &=\int_G |F(\gamma)|^2\,d m_G(\gamma)\\ &=\|F\|_{L^2(G,m_G)}^2. \end{align*} Thus $C_\Phi$ is an isometry. For each $\lambda\in\Lambda(T)$, \begin{align*} C_\Phi\chi_\lambda=e_\lambda \end{align*} as elements of $L^2(X,\mathcal B,\mu)$. The linear span of $\{\chi_\lambda:\lambda\in\Lambda(T)\}$ is dense in $L^2(G,\mathcal B_G,m_G)$ because it is uniformly dense in $C(G;\mathbb C)$ and continuous functions are dense in $L^2(G,m_G)$. The linear span of $\{e_\lambda:\lambda\in\Lambda(T)\}$ is dense in $L^2(X,\mathcal B,\mu)$ by the pure discrete spectrum hypothesis and the one-dimensionality of the eigenspaces. Hence the range of $C_\Phi$ contains a dense subspace of $L^2(X,\mathcal B,\mu)$. Since the range of an isometry is closed, $C_\Phi$ is surjective. Therefore $C_\Phi$ is unitary. For standard probability systems, a measure-preserving factor map whose pullback operator is unitary has pullback $\sigma$-algebra equal to the whole target $L^2$ space, hence is an isomorphism modulo null sets. Hence \begin{align*} \Phi:(X,\mathcal B,\mu,T)\to (G,\mathcal B_G,m_G,R_g) \end{align*} is a measure-theoretic isomorphism. The rotation is ergodic as well. If $\lambda\neq 1$, then \begin{align*} \chi_\lambda(g)=g(\lambda)=\lambda\neq 1, \end{align*} so no non-constant character is fixed by $R_g$. Since characters are dense in $L^2(G,m_G)$, every $R_g$-invariant $L^2$ function is constant. Thus $R_g$ is ergodic. Equivalently, the cyclic subgroup $\{g^n:n\in\mathbb Z\}$ is dense in $G$. [/step]

[step:Classify pure discrete ergodic systems by their eigenvalue group] Let $(X_1,\mathcal B_1,\mu_1,T_1)$ and $(X_2,\mathcal B_2,\mu_2,T_2)$ be ergodic measure-preserving systems with pure discrete spectrum. First suppose \begin{align*} \Lambda(T_1)=\Lambda(T_2)=:\Lambda. \end{align*} Applying the construction above to each system gives measure-theoretic isomorphisms \begin{align*} \Phi_i:(X_i,\mathcal B_i,\mu_i,T_i)&\to (G,\mathcal B_G,m_G,R_g), \qquad i\in\{1,2\}, \end{align*} where \begin{align*} G:=\widehat{\Lambda} \end{align*} and \begin{align*} g:\Lambda&\to\mathbb T\\ \lambda&\mapsto \lambda. \end{align*} Therefore $\Phi_2^{-1}\circ\Phi_1$ is a measure-theoretic isomorphism from $(X_1,\mathcal B_1,\mu_1,T_1)$ to $(X_2,\mathcal B_2,\mu_2,T_2)$. Conversely, suppose \begin{align*} \Psi:(X_1,\mathcal B_1,\mu_1,T_1)&\to (X_2,\mathcal B_2,\mu_2,T_2) \end{align*} is a measure-theoretic isomorphism. Define \begin{align*} C_\Psi:L^2(X_2,\mathcal B_2,\mu_2;\mathbb C)&\to L^2(X_1,\mathcal B_1,\mu_1;\mathbb C)\\ [F]&\mapsto [F\circ\Psi]. \end{align*} The operator $C_\Psi$ is unitary and intertwines the Koopman operators: \begin{align*} C_\Psi U_{T_2}F &=(F\circ T_2)\circ\Psi\\ &=F\circ(T_2\circ\Psi)\\ &=F\circ(\Psi\circ T_1)\\ &=(F\circ\Psi)\circ T_1\\ &=U_{T_1}C_\Psi F. \end{align*} If $F\neq 0$ satisfies $U_{T_2}F=\lambda F$, then $C_\Psi F\neq 0$ and \begin{align*} U_{T_1}(C_\Psi F)=C_\Psi(U_{T_2}F)=\lambda C_\Psi F. \end{align*} Hence $\Lambda(T_2)\subseteq \Lambda(T_1)$. Applying the same argument to $\Psi^{-1}$ gives $\Lambda(T_1)\subseteq\Lambda(T_2)$. Therefore \begin{align*} \Lambda(T_1)=\Lambda(T_2). \end{align*} This proves that ergodic measure-preserving systems with pure discrete spectrum are classified, up to measure-theoretic isomorphism, by their eigenvalue group. [/step]