[proofplan] Represent the center of $M$ as an abelian von Neumann algebra $L^\infty(X,\mathcal A,\mu)$ over a standard [measure space](/page/Measure%20Space), using separability of the predual to ensure standardness. The [[Central Type Decomposition](/theorems/9316)][citetheorem:9316] gives a countable central partition of the identity into homogeneous type $I_n$ and type $I_\infty$ summands after separability rules out uncountable dimensions. On each homogeneous summand, matrix units over the center identify the algebra with a [tensor product](/page/Tensor%20Product) of its center and a full operator algebra by the [[Matrix Unit Reconstruction Theorem](/theorems/9292)][citetheorem:9292]; this is then read as a direct integral with constant fibre dimension. Assembling these summands gives the measurable field $(K_x)_{x\in X}$, and uniqueness follows from the intrinsic central partition by homogeneous dimension together with the [[Classification of Type I Factors](/theorems/9293)][citetheorem:9293]. [/proofplan]

[step:Represent the center by a standard measure space]Let $M_*$ denote the predual of $M$, the [Banach space](/page/Banach%20Space) of normal linear functionals on $M$. Since $Z(M)$ is a weak-star closed subspace of $M$, its predual is the quotient of $M_*$ by the annihilator of $Z(M)$; hence $Z(M)$ has separable predual because $M_*$ is separable. We use the standard commutative representation theorem in the following precise form: an abelian von Neumann algebra with separable predual is normally $*$-isomorphic to $L^\infty$ of a standard $\sigma$-finite measure space. Therefore there are a standard $\sigma$-finite measure space $(X,\mathcal A,\mu)$ and a normal $*$-isomorphism \begin{align*} \Theta:Z(M)\to L^\infty(X,\mathcal A,\mu). \end{align*} Under this isomorphism, central projections in $Z(M)$ correspond to characteristic functions $\mathbb 1_E$ of measurable subsets $E\in\mathcal A$, modulo $\mu$-null sets. We shall identify $Z(M)$ with $L^\infty(X,\mathcal A,\mu)$ through $\Theta$.[/step]

[guided]The first point is to turn the abstract center into a concrete algebra of scalar multiplication operators. Let $M_*$ denote the predual of $M$, meaning the Banach space of normal linear functionals on $M$. Since $Z(M)$ is weak-star closed in $M$, its predual is obtained from $M_*$ by restriction, equivalently as the quotient of $M_*$ by the closed subspace of normal functionals that vanish on $Z(M)$. Quotients of separable Banach spaces are separable, so the abelian von Neumann algebra $Z(M)$ has separable predual. The standard representation theorem for abelian von Neumann algebras with separable predual then gives a standard $\sigma$-finite measure space $(X,\mathcal A,\mu)$ and a normal $*$-isomorphism \begin{align*} \Theta:Z(M)\to L^\infty(X,\mathcal A,\mu). \end{align*} This is the point where the phrase “standard central decomposition setting” is made precise: the center is not merely abstractly abelian, but is represented as essentially bounded [measurable functions](/page/Measurable%20Functions) on a standard measure space. A projection $z\in Z(M)$ satisfies $z=z^*=z^2$, so $\Theta(z)$ is an idempotent real-valued element of $L^\infty(X,\mathcal A,\mu)$. Hence $\Theta(z)=\mathbb 1_E$ for some measurable set $E\in\mathcal A$, uniquely modulo $\mu$-null sets. Thus central projections in $M$ are exactly measurable central pieces of $X$, up to null sets.[/guided]

[step:Partition the identity into homogeneous type I central summands]We apply the [Central Type Decomposition][citetheorem:9316] to the von Neumann algebra $M$. Its hypotheses are satisfied because $M$ is a von Neumann algebra, and the type I hypothesis means that the type II and type III central summands in that theorem are zero. The refinement of the type I part gives central projections indexed by Hilbert-space dimensions. Since $M$ has separable predual, each central summand $Mz$ with $z\in Z(M)$ also has separable predual: its predual is the quotient of $M_*$ by the normal functionals that vanish on $Mz$. We next rule out uncountable homogeneous dimensions directly. Suppose that $N$ is a nonzero homogeneous type $I_\kappa$ central summand with $\kappa$ uncountable. By the homogeneous type I matrix-unit theorem, $N$ contains matrix units $(e_{ij})_{i,j\in I}$ indexed by a set $I$ of cardinality $\kappa$, with strong diagonal supremum equal to the unit of $N$. Fix $i_0\in I$ and choose a normal state $\rho$ on the nonzero corner $e_{i_0i_0}Ne_{i_0i_0}$. For each $i\in I$, define a normal functional on $N$ by \begin{align*} \rho_i(x)=\rho(e_{i_0i}xe_{ii_0}),\qquad x\in N. \end{align*} Each $\rho_i$ has norm $1$, and if $i\ne j$, then \begin{align*} \rho_i(e_{ii})=1,\qquad \rho_j(e_{ii})=0. \end{align*} Thus $\|\rho_i-\rho_j\|\ge 1$ for distinct $i,j$. The predual $N_*$ therefore contains an uncountable $1$-separated set, so it is nonseparable. This contradicts the separability of every nonzero central summand predual. Hence the only possible homogeneous type I dimensions are $n\in\mathbb N$ and $\aleph_0$. Therefore there is a countable family of pairwise orthogonal central projections \begin{align*} (z_n)_{n\in\mathbb N\cup\{\infty\}}\subset Z(M) \end{align*} such that \begin{align*} \sum_{n\in\mathbb N\cup\{\infty\}} z_n=1 \end{align*} in the strong operator topology, and each summand $Mz_n$ is homogeneous of type $I_n$ for $n\in\mathbb N$ and homogeneous of type $I_\infty$ for $n=\infty$. Here $z_n=0$ is allowed. For each $n\in\mathbb N\cup\{\infty\}$, let $X_n\in\mathcal A$ be a measurable set such that \begin{align*} \Theta(z_n)=\mathbb 1_{X_n} \end{align*} in $L^\infty(X,\mathcal A,\mu)$. The projections are pairwise orthogonal and sum to $1$, so the sets $(X_n)_{n\in\mathbb N\cup\{\infty\}}$ are pairwise disjoint and cover $X$ up to a $\mu$-null set.[/step]

[guided]The [Central Type Decomposition][citetheorem:9316] applies to any von Neumann algebra. We apply it to $M$. Because $M$ is type I, the central projections corresponding to the type II and type III parts must be zero; otherwise a nonzero central summand of $M$ would have type II or type III. The theorem also refines the type I part into homogeneous type $I_\kappa$ pieces, where $\kappa$ is the Hilbert-space dimension of the type I factor on that central piece. The separable-predual hypothesis restricts the possible cardinals. First, any central summand $Mz$ has separable predual because its predual is obtained from $M_*$ by quotienting out the normal functionals that vanish on $Mz$. Now suppose a nonzero homogeneous type I summand $N$ had dimension $\kappa$ with $\kappa$ uncountable. Matrix units $(e_{ij})_{i,j\in I}$ with $|I|=\kappa$ and strong diagonal supremum $1_N$ exist in $N$. Fixing $i_0\in I$ and a normal state $\rho$ on $e_{i_0i_0}Ne_{i_0i_0}$, the formula \begin{align*} \rho_i(x)=\rho(e_{i_0i}xe_{ii_0}) \end{align*} defines normal functionals indexed by $I$. They are pairwise separated because $\rho_i(e_{ii})=1$ while $\rho_j(e_{ii})=0$ for $j\ne i$. Thus $N_*$ contains an uncountable separated family, which is impossible in a separable Banach space. This contradiction rules out uncountable homogeneous dimensions. Thus only finite dimensions and the countably infinite dimension occur. We therefore obtain central projections \begin{align*} (z_n)_{n\in\mathbb N\cup\{\infty\}}\subset Z(M) \end{align*} with strong sum $1$, where $Mz_n$ is homogeneous of type $I_n$ for finite $n$ and homogeneous of type $I_\infty$ for $n=\infty$. The center has already been identified with $L^\infty(X,\mathcal A,\mu)$ by $\Theta$. Each central projection $z_n$ therefore corresponds to an indicator function $\mathbb 1_{X_n}$ for a measurable set $X_n\in\mathcal A$, modulo null sets. Orthogonality of the projections gives disjointness of the sets modulo null sets, and the strong sum condition gives that their union covers $X$ modulo a null set.[/guided]

[step:Reconstruct each homogeneous summand from its center and matrix units]Fix $n\in\mathbb N\cup\{\infty\}$ with $z_n\ne 0$. Define the [Hilbert space](/page/Hilbert%20Space) $K_n$ by \begin{align*} K_n=\mathbb C^n \end{align*} when $n\in\mathbb N$, and by \begin{align*} K_\infty=\ell^2(\mathbb N) \end{align*} when $n=\infty$. Let $I_n=\{1,\dots,n\}$ if $n\in\mathbb N$, and let $I_\infty=\mathbb N$. By the homogeneous type I matrix-unit theorem applied to the homogeneous type $I_n$ von Neumann algebra $Mz_n$ with unit $z_n$, choose a system of matrix units \begin{align*} (e_{ij,n})_{i,j\in I_n}\subset Mz_n \end{align*} such that \begin{align*} e_{ij,n}e_{kl,n}=\delta_{jk}e_{il,n} \end{align*} and \begin{align*} e_{ij,n}^*=e_{ji,n}. \end{align*} The diagonal supports exhaust the central summand, meaning \begin{align*} \bigvee_{i\in I_n} e_{ii,n}=z_n. \end{align*} For $n=\infty$, this is the countable strong operator supremum over $I_\infty=\mathbb N$, and the Hilbert-space tensor factor appearing below is $\mathcal L(\ell^2(\mathbb N))$. Choose and fix an index $i_0\in I_n$, and define the corner von Neumann algebra \begin{align*} N_n=e_{i_0i_0,n}Mz_ne_{i_0i_0,n}. \end{align*} The hypotheses of the [Matrix Unit Reconstruction Theorem][citetheorem:9292] are satisfied for the von Neumann algebra $Mz_n$, whose unit is $z_n$, because the displayed matrix-unit relations hold and the diagonal supremum is $z_n$. Therefore there is a normal $*$-isomorphism \begin{align*} Mz_n\cong N_n\,\overline{\otimes}\,\mathcal L(K_n). \end{align*} Here $\overline{\otimes}$ denotes the von Neumann algebra tensor product. For a homogeneous type I summand, the diagonal projection $e_{i_0i_0,n}$ is an abelian projection with central carrier $z_n$. By the corner-center identification theorem for abelian projections with full central carrier, applied to the von Neumann algebra $Mz_n$ and the projection $e_{i_0i_0,n}$, there is a normal $*$-isomorphism \begin{align*} \Gamma_n:Z(M)z_n&\to N_n \end{align*} defined by \begin{align*} \Gamma_n(a z_n)&=a e_{i_0i_0,n}. \end{align*} Thus the coefficient algebra in the reconstruction theorem is canonically $Z(M)z_n$, and there is a normal $*$-isomorphism \begin{align*} Mz_n\cong Z(M)z_n\,\overline{\otimes}\,\mathcal L(K_n). \end{align*} Using $Z(M)z_n\cong L^\infty(X_n,\mathcal A|_{X_n},\mu|_{X_n})$, this becomes \begin{align*} Mz_n\cong L^\infty(X_n,\mathcal A|_{X_n},\mu|_{X_n})\,\overline{\otimes}\,\mathcal L(K_n). \end{align*} For a constant measurable field with fibre $K_n$, the direct integral is by definition the von Neumann algebra of essentially bounded measurable functions from $X_n$ into $\mathcal L(K_n)$, which is exactly the spatial tensor product below: \begin{align*} L^\infty(X_n,\mathcal A|_{X_n},\mu|_{X_n})\,\overline{\otimes}\,\mathcal L(K_n)\cong \int_{X_n}^{\oplus}\mathcal L(K_n)\,d\mu(x). \end{align*}[/step]

[guided]The purpose of the matrix units is to turn the homogeneous summand into a matrix algebra over one diagonal corner. Fix $n\in\mathbb N\cup\{\infty\}$ with $z_n\ne 0$, and set $K_n=\mathbb C^n$ for finite $n$ and $K_\infty=\ell^2(\mathbb N)$ for $n=\infty$. Also set $I_n=\{1,\dots,n\}$ for finite $n$ and $I_\infty=\mathbb N$. The homogeneous type I matrix-unit theorem applies to the homogeneous type $I_n$ von Neumann algebra $Mz_n$ with unit $z_n$, and gives matrix units $(e_{ij,n})_{i,j\in I_n}\subset Mz_n$ satisfying \begin{align*} e_{ij,n}e_{kl,n}=\delta_{jk}e_{il,n} \end{align*} and \begin{align*} e_{ij,n}^*=e_{ji,n}. \end{align*} Their diagonal projections fill the unit of the summand: \begin{align*} \bigvee_{i\in I_n} e_{ii,n}=z_n. \end{align*} This last equality is the precise meaning of the diagonal support condition; it is what allows the reconstruction theorem to recover all of $Mz_n$. In the case $n=\infty$, the index set is $I_\infty=\mathbb N$, so the displayed supremum is a countable supremum in the strong operator topology, and the reconstruction theorem gives the tensor factor $\mathcal L(\ell^2(\mathbb N))$. Choose an index $i_0\in I_n$ and define \begin{align*} N_n=e_{i_0i_0,n}Mz_ne_{i_0i_0,n}. \end{align*} We apply the [Matrix Unit Reconstruction Theorem][citetheorem:9292] to $Mz_n$ with unit $z_n$. Its hypotheses are exactly the matrix-unit identities above together with the diagonal supremum condition $\bigvee_{i\in I_n} e_{ii,n}=z_n$. Therefore \begin{align*} Mz_n\cong N_n\,\overline{\otimes}\,\mathcal L(K_n) \end{align*} by a normal $*$-isomorphism. It remains to identify the coefficient algebra $N_n$. In a homogeneous type I summand, the chosen diagonal projection $e_{i_0i_0,n}$ is abelian and has central carrier $z_n$. Therefore the corner-center identification theorem for abelian projections with full central carrier applies to the von Neumann algebra $Mz_n$ and the projection $e_{i_0i_0,n}$. Its hypotheses are exactly that $e_{i_0i_0,n}$ is abelian and has central carrier equal to the unit $z_n$ of $Mz_n$. It gives a normal $*$-isomorphism \begin{align*} \Gamma_n:Z(M)z_n&\to N_n \end{align*} defined by \begin{align*} \Gamma_n(a z_n)&=a e_{i_0i_0,n}. \end{align*} Substituting this corner identification into the reconstruction isomorphism gives \begin{align*} Mz_n\cong Z(M)z_n\,\overline{\otimes}\,\mathcal L(K_n). \end{align*} Finally, the center representation sends $Z(M)z_n$ onto $L^\infty(X_n,\mathcal A|_{X_n},\mu|_{X_n})$, so \begin{align*} Mz_n\cong L^\infty(X_n,\mathcal A|_{X_n},\mu|_{X_n})\,\overline{\otimes}\,\mathcal L(K_n). \end{align*} For a constant measurable field with fibre $K_n$, this tensor product is precisely the algebra of essentially bounded measurable operator fields on $X_n$ with values in $\mathcal L(K_n)$, namely \begin{align*} \int_{X_n}^{\oplus}\mathcal L(K_n)\,d\mu(x). \end{align*}[/guided]

[step:Assemble the constant homogeneous fields into one measurable field]Define a measurable field of Hilbert spaces $(K_x)_{x\in X}$ by setting \begin{align*} K_x=K_n \end{align*} for $x\in X_n$, after modifying the sets $X_n$ on a common null set so that they form an actual partition of $X$. Since the index set $\mathbb N\cup\{\infty\}$ is countable and each $K_n$ is separable, this is a measurable field in the standard direct-integral sense. The central direct-sum decomposition \begin{align*} M=\bigoplus_{n\in\mathbb N\cup\{\infty\}} Mz_n \end{align*} means the von Neumann algebra of bounded families $(x_n)$ with $x_n\in Mz_n$, represented inside $M$ by the strong sum $\sum_n x_n$ along the countable central partition. This decomposition therefore gives a normal $*$-isomorphism \begin{align*} \Phi:M\to \int_X^\oplus \mathcal L(K_x)\,d\mu(x) \end{align*} by combining the isomorphisms on the summands $Mz_n$. This assembly is compatible with the measurable partition because multiplication by $z_n$ corresponds under $\Theta$ to multiplication by $\mathbb 1_{X_n}$, and the direct integral over the disjoint union $X=\bigcup_n X_n$ is the countable direct sum of the direct integrals over the pieces $X_n$, modulo the common null set. Under this isomorphism, an element $a\in Z(M)$ corresponds to the decomposable operator whose fibre at $x\in X$ is \begin{align*} \Theta(a)(x)I_{K_x}. \end{align*} Thus \begin{align*} \Phi(Z(M))=L^\infty(X,\mathcal A,\mu) \end{align*} acting fibrewise by scalar multiplication.[/step]

[guided]We now glue the homogeneous pieces along the measurable partition of the center. After changing the sets $X_n$ on one common null set, the family $(X_n)_{n\in\mathbb N\cup\{\infty\}}$ is an actual countable partition of $X$. Define the fibre Hilbert space by $K_x=K_n$ for $x\in X_n$. Since the partition is countable and each $K_n$ is separable, this is a measurable field of separable Hilbert spaces. The central projection $z_n$ corresponds to multiplication by $\mathbb 1_{X_n}$. Hence the central direct sum $M=\bigoplus_n Mz_n$ matches the decomposition of the direct integral over the disjoint union $X=\bigcup_n X_n$: an element of $M$ is a bounded family of elements in the summands $Mz_n$, and its image is the corresponding essentially bounded measurable operator field whose restriction to $X_n$ lies in $\mathcal L(K_n)$. Combining the normal $*$-isomorphisms on the summands therefore defines a normal $*$-isomorphism \begin{align*} \Phi:M\to \int_X^\oplus \mathcal L(K_x)\,d\mu(x). \end{align*} If $a\in Z(M)$, then on the central piece $X_n$ it acts through $\Theta(a)|_{X_n}$ times the identity on $K_n$. Therefore the fibre of $\Phi(a)$ at $x\in X$ is \begin{align*} \Theta(a)(x)I_{K_x}. \end{align*} This proves that $\Phi(Z(M))=L^\infty(X,\mathcal A,\mu)$ and that the center acts by scalar multiplication on the fibres.[/guided]

[step:Identify the fibre dimension as the unique central invariant]Define the dimension function \begin{align*} d:X\to \mathbb N\cup\{\aleph_0\},\qquad x\mapsto \dim K_x. \end{align*} By construction, $d(x)=n$ on $X_n$ for $n\in\mathbb N$ and $d(x)=\aleph_0$ on $X_\infty$, so $d$ is measurable because each level set is measurable. The central projection $z_n$ is intrinsically characterized as the largest central projection for which $Mz_n$ is homogeneous of type $I_n$. Hence, after the center is identified with $L^\infty(X,\mathcal A,\mu)$, the measurable set $X_n$ is determined up to a $\mu$-null set. Conversely, the fibre over $x\in X_n$ is the type $I_n$ factor $\mathcal L(K_n)$. By the [Classification of Type I Factors][citetheorem:9293], factors $\mathcal L(K)$ and $\mathcal L(K')$ are isomorphic exactly when $\dim K=\dim K'$. Therefore any other decomposition over the same identified center has the same level sets of the fibre-dimension function up to null sets. This proves that $x\mapsto \dim K_x$ is uniquely determined almost everywhere after the center has been identified. If a different standard measure-space representation of the same abelian von Neumann algebra $Z(M)$ is used, the representation theorem identifies the two measure algebras by a measure-class isomorphism, so the same central projections $z_n$ correspond to matching level sets under that isomorphism. This gives the stated equivalence up to measure-class isomorphism and completes the proof.[/step]

[guided]The dimension function records which homogeneous central summand contains a point of the base space. Define \begin{align*} d:X\to \mathbb N\cup\{\aleph_0\},\qquad x\mapsto \dim K_x. \end{align*} Because $K_x=K_n$ on $X_n$, we have $d(x)=n$ on $X_n$ for finite $n$ and $d(x)=\aleph_0$ on $X_\infty$. Each set $X_n$ is measurable, so every level set of $d$ is measurable. The uniqueness is central, not a consequence of the particular matrix units chosen above. For each $n\in\mathbb N\cup\{\infty\}$, the projection $z_n$ is the largest central projection such that the corresponding summand is homogeneous of type $I_n$. This characterization depends only on $M$ and its center. Once the center is identified with $L^\infty(X,\mathcal A,\mu)$, the projection $z_n$ corresponds to the measurable set $X_n$ modulo null sets, so the level set $\{x\in X:d(x)=n\}$ is determined modulo null sets. Finally, the [Classification of Type I Factors][citetheorem:9293] says that a factor $\mathcal L(K)$ determines the Hilbert-space dimension of $K$. Thus no alternative direct-integral decomposition over the same identified center can replace the fibres on a positive-measure subset of $X_n$ by factors of a different dimension. The fibre-dimension function is therefore uniquely determined almost everywhere once the center is identified. If the same abelian center is represented as $L^\infty(Y,\mathcal B,\nu)$ instead, the standard representation theorem gives a measure-class isomorphism between the measure algebras of $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ implementing the same central projections. Under that isomorphism, the level sets of the two dimension functions agree modulo null sets, which is the asserted uniqueness up to measure-class isomorphism.[/guided]