Proof. Let $\Omega $ be the spectrum of N, and we shall regard f as a continuous function on $\Omega $ . Then there exists an open neighborhood $O \subseteq \Omega $ of $\mathcal {U}$ s.t. $|f(t) - \mathcal {U}(f)| \leq \epsilon $ whenever $t \in O$ . As clopen sets form a basis of topology for $\Omega $ , O can be chosen to be clopen. Letting $p = 1_O$ concludes the proof.

Proof. That $\|\cdot \|_{E, 2, \infty }$ defines a norm is an easy exercise. To prove completeness, we let $\{\varphi _i\}_{i \in I}$ be a maximal collection of normal states on N with disjoint supports. Then $\sum _{i \in I} \mathrm {supp}(\varphi _i) = 1$ , so the direct sum of the GNS representations of M, associated with $\varphi _i \circ E$ ,

$$ \begin{align*} \pi = \bigoplus_{i \in I} \pi_{\varphi_i \circ E} \end{align*} $$

is a faithful normal representation. Thus, a uniformly dense subset of positive linear functionals on M is of the form

$$ \begin{align*} \phi(x) = \sum_{i \in I_0} \varphi_i(E(y_ix)), \end{align*} $$

where $I_0 \subseteq I$ is a finite subset and $y_i \in M_+$ is supported on $\mathrm {supp}(\varphi _i)$ .

Now, let $(x_n) \subseteq (M)_1$ be a Cauchy sequence under $\|\cdot \|_{E, 2, \infty }$ . For any finite subset $I_0 \subseteq I$ and $y_i \in M_+$ supported on $\mathrm {supp}(\varphi _i)$ , we have

$$ \begin{align*} \begin{split} \sum_{i \in I_0} \varphi_i(E(y_i(x_n - x_m)^\ast (x_n - x_m))) &\leq \sum_{i \in I_0} \|y_i\|_\infty \varphi_i(E((x_n - x_m)^\ast (x_n - x_m)))\\ &\leq \sum_{i \in I_0} \|y_i\|_\infty \|x_n - x_m\|_{E, 2, \infty}^2\\ &\to 0 \end{split} \end{align*} $$

as $n, m \to \infty $ . Hence, as $\|x_n\|_\infty \leq 1$ for all n, we see that $x_n \to x$ in the strong $^\ast $ topology for some $x \in (M)_1$ .

We claim that $\|x_n - x\|_{E, 2, \infty }^2 \to 0$ to conclude the proof. For any $\epsilon> 0$ , there exists $N> 0$ s.t. $\|x_n - x_m\|_{E, 2, \infty }^2 \leq \epsilon $ whenever $n, m \geq N$ ; that is, $\|E((x_n - x_m)^\ast (x_n - x_m))\|_\infty \leq \epsilon $ . As $x_m \to x$ in the strong $^\ast $ topology, $(x_n - x_m)^\ast (x_n - x_m) \to (x_n - x)^\ast (x_n - x)$ in the strong $^\ast $ topology. As E is normal, we have $E((x_n - x_m)^\ast (x_n - x_m)) \to E((x_n - x)^\ast (x_n - x))$ in the strong $^\ast $ topology, whence $\|E((x_n - x)^\ast (x_n - x))\|_\infty \leq \epsilon $ ; that is, $\|x_n - x\|_{E, 2, \infty }^2 \leq \epsilon $ whenever $n \geq N$ . The concludes the proof.

Proof of Proposition 2.2.

Let $q: M \to M^{/E, \mathcal {U}} = M/I_{E, \mathcal {U}}$ be the natural quotient map. We claim that $q((M)_1)$ is complete under $\|\cdot \|_{\tau _{E, \mathcal {U}}, 2}$ . Granted the claim, then as $q((M)_1)$ is contained in $(M^{/E, \mathcal {U}})_1$ and contains all elements of the latter space of operator norm strictly smaller than 1, consequently dense in the latter space under $\|\cdot \|_{\tau _{E, \mathcal {U}}, 2}$ , we must have $q((M)_1) = (M^{/E, \mathcal {U}})_1$ and it is complete. Whence, it is a tracial von Neumann algebra with $\tau _{E, \mathcal {U}}$ a faithful, normal, tracial state, by [Reference Anantharaman and PopaAP16, Proposition 2.6.4].

To prove the claim, we let $(x_n) \subseteq q((M)_1)$ be a Cauchy sequence under $\|\cdot \|_{\tau _{E, \mathcal {U}}, 2}$ . By taking a subsequence if necessary, we may assume $\|x_n - x_{n+1}\|_{\tau _{E, \mathcal {U}}, 2} \leq 2^{-(n+1)}$ . We shall construct by induction a sequence , with and . We start with an arbitrary with . Now, assume up to some n have been constructed, let be arbitrarily chosen with . Since,

by Lemma 2.3, there exists a projection $p \in N$ s.t. $\mathcal {U}(p) = 1$ and

Let . Since both and have operator norms bounded by $1$ , as well. We have

We also have

where in the final equality, we used that $\mathcal {U}(p) = 1$ and therefore $\mathcal {U}(1-p) = 0$ . Thus, and . This concludes the inductive construction.

By Lemma 2.4, for some in $\|\cdot \|_{E, 2, \infty }$ . Let . We shall show that $x_n \to x$ in $\|\cdot \|_{\tau _{E, \mathcal {U}}, 2}$ to conclude the proof of the claim. Indeed,

This proves the claim.

Proof. First, we show that $Z(M^{/E,\mathcal {U}}) = \pi _{E,\mathcal {U}}(Z(M))$ . It is immediate that $\pi _{E,\mathcal {U}}(Z(M)) \subseteq Z(M^{/E,\mathcal {U}})$ since $\pi _{E,\mathcal {U}}$ is a $*$ -homomorphism. Conversely, suppose that $x \in Z(M^{/E,\mathcal {U}})$ , and fix with . Let $E_0: M \to Z(M)$ be the center-valued trace. By the generalized Dixmier averaging theorem, there is a sequence

of convex combinations of unitary conjugates of such that in norm. Since and , we see that . Hence, , and so , which implies that $x \in \pi _{E,\mathcal {U}}(Z(M))$ as desired.

Next, we give an isomorphism $\pi _{E,\mathcal {U}}(Z(M)) \cong Z(M)^{/E|_{Z(M)},\mathcal {U}}$ . First, note that $E|_{Z(M)}$ is a faithful normal conditional expectation from $Z(M)$ onto N, and so $Z(M)^{/E|_{Z(M)},\mathcal {U}}$ is well defined. Two elements and in $Z(M)$ represent the same element of $Z(M)^{/E|_{Z(M)},\mathcal {U}}$ if and only if if and only if and represent the same element in $M^{/E,\mathcal {U}}$ . Hence, there is a well-defined injective $*$ -homomorphism $Z(M)^{/E|_{Z(M)},\mathcal {U}} \to M^{/E,\mathcal {U}}$ , and its image is clearly equal to $\pi _{E,\mathcal {U}}(Z(M))$ .

In the case that $N = Z(M)$ , then $Z(M)^{/E|_{Z(M)},\mathcal {U}} \cong \mathbb {C}$ since $\mathcal {U}$ is a pure state on $Z(M)$ , and hence $Z(M^{/E,\mathcal {U}}) \cong \mathbb {C}$ , so $M^{/E,\mathcal {U}}$ is a factor.

Proof of Proposition 2.9 / Corollary B.

Since $A \mathbin {\overline {\otimes }} M \equiv B \mathbin {\overline {\otimes }} N$ , the Keisler–Shelah theorem implies that $A \mathbin {\overline {\otimes }} M \equiv B \mathbin {\overline {\otimes }} N$ admit isomorphic ultrapowers. In particular, there exists a character $\mathcal {U}$ on some abelian von Neumann algebra C s.t. there exists a trace-preserving isomorphism $\pi : (A \mathbin {\overline {\otimes }} M)^{\mathcal {U}} \to (B \mathbin {\overline {\otimes }} N)^{\mathcal {U}}$ . Since M is a factor, $Z(A \mathbin {\overline {\otimes }} M) = A$ , so by Proposition 2.8, the center of $(A \mathbin {\overline {\otimes }} M)^{\mathcal {U}}$ is $A^{\mathcal {U}}$ . Similarly, $Z((B \mathbin {\overline {\otimes }} N)^{\mathcal {U}}) = B^{\mathcal {U}}$ . Hence, $\pi $ must restrict to a trace-preserving isomorphism between $A^{\mathcal {U}}$ and $B^{\mathcal {U}}$ . We may thus identify these algebras via $\pi $ .

Let $\mathcal {V}$ be any character on $A^{\mathcal {U}} = B^{\mathcal {U}}$ . Let E denote the center-valued trace of $(A \mathbin {\overline {\otimes }} M)^{\mathcal {U}}$ as well as the center-valued trace of $(B \mathbin {\overline {\otimes }} N)^{\mathcal {U}}$ . We again identify the two algebras via $\pi $ and note that the $\pi $ must also preserve the center-valued trace. Hence, we must have

$$ \begin{align*} [(A \mathbin{\overline{\otimes}} M)^{\mathcal{U}}]^{/E, \mathcal{V}} = [(B \mathbin{\overline{\otimes}} N)^{\mathcal{U}}]^{/E, \mathcal{V}}. \end{align*} $$

Note that $(A \mathbin {\overline {\otimes }} M)^{\mathcal {U}}$ is, by definition, a quotient of $C \mathbin {\overline {\otimes }} A \mathbin {\overline {\otimes }} M$ . Let the natural quotient map be $q: C \mathbin {\overline {\otimes }} A \mathbin {\overline {\otimes }} M \to (A \mathbin {\overline {\otimes }} M)^{\mathcal {U}}$ . Then,

$$ \begin{align*} [(A \mathbin{\overline{\otimes}} M)^{\mathcal{U}}]^{/E, \mathcal{V}} = (C \mathbin{\overline{\otimes}} A \mathbin{\overline{\otimes}} M)/q^{-1}(I_{E, \mathcal{V}}). \end{align*} $$

We also have

$$ \begin{align*} q^{-1}(I_{E, \mathcal{V}}) = \{x \in C \mathbin{\overline{\otimes}} A \mathbin{\overline{\otimes}} M: \mathcal{V}(E(q(x^\ast x))) = 0\}. \end{align*} $$

It follows from Dixmier averaging as in the proof of Proposition 2.8 that $E \circ q = q \circ (\mathrm {Id} \otimes \mathrm {Id} \otimes \tau )$ . Since q restricts to a *-homomorphism $C \mathbin {\overline {\otimes }} A \to A^{\mathcal {U}}$ , we see that $\mathcal {V} \circ q|_{C \mathbin {\overline {\otimes }} A}$ is a character on $C \mathbin {\overline {\otimes }} A$ . Hence,

$$ \begin{align*} q^{-1}(I_{E, \mathcal{V}}) = I_{\mathrm{Id} \otimes \mathrm{Id} \otimes \tau, \mathcal{V} \circ q}. \end{align*} $$

But this means $[(A \mathbin {\overline {\otimes }} M)^{\mathcal {U}}]^{/E, \mathcal {V}} = M^{\mathcal {V} \circ q}$ . Similarly, $[(B \mathbin {\overline {\otimes }} N)^{\mathcal {U}}]^{/E, \mathcal {V}}$ is a generalized ultrapower of N as well. Hence, M and N admit generalized ultrapowers that are isomorphic, and so $M \equiv N$ by Proposition C and the transitivity of elementary equivalence.

Proof. Consider rational polynomial formulas defined in a similar way to formulas, except that

• • The basic formulas are $\operatorname {\mathrm {Re}} \operatorname {\mathrm {tr}}(p(x_1,\dots ,x_n))$ and $\operatorname {\mathrm {Im}} \operatorname {\mathrm {tr}}(p(x_1,\dots ,x_n))$ where p is a non-commutative $*$ -polynomial with coefficients in $\mathbb {Q}[i]$ .

• • The connectives are polynomial functions $\mathbb {R}^k \to \mathbb {R}$ with rational coefficients.

The set of rational polynomial formulas is countable if the underlying set of variables is countable. The rational polynomials in variables $X \sqcup I$ are dense in the space of all formulas with respect to ${\lVert {\cdot }\rVert }_u$ . To show this, one proceeds by induction on the complexity of the formula. For the base case, it suffices to approximate the coefficients of each non-commutative $*$ -polynomial by elements of $\mathbb {Q}[i]$ and estimate the uniform norm of the difference. For applying a connective f to formulas $\phi _1$ , …, $\phi _k$ , recall by the Stone-Weierstrass theorem that f can be uniformly approximated by a polynomial on $[-{\lVert {\varphi _1}\rVert }_u,{\lVert {\varphi _1}\rVert }_u] \times \dots \times [-{\lVert {\varphi _k}\rVert }_u, {\lVert {\varphi _k}\rVert }_u]$ . The case of applying a quantifier is immediate since

$$\begin{align*}{\lVert{\sup_y \varphi(x_1,\dots,x_n,y) - \sup_y \psi(x_1,\dots,x_n,y)} \rVert}_u \leq {\lVert {\phi(x_1,\dots,x_n) - \psi(x_1,\dots,x_n)} \rVert}_u. \end{align*}$$

Hence, rational polynomial formulas are dense in the space of formulas. The same holds when we restrict to formulas where the free variables are in X and the bound variables are in I. This shows separability of $\mathcal {P}_X$ since this is the separation-completion of the space of such formulas. Since $\mathcal {P}_X$ is separable, the weak- $*$ topology on the unit ball in the dual space is metrizable, and hence also the space of states and the space of characters are metrizable.

Proof. The case where $\varphi (\overline {x}) = \operatorname {\mathrm {Re}} \operatorname {\mathrm {tr}}(p(\overline {x}))$ or $\operatorname {\mathrm {Im}} \operatorname {\mathrm {tr}}(p(\overline {x}))$ follows by expanding the polynomial p termwise and using the triangle inequality and submultiplicativity of the operator norm. Next, suppose that $\varphi = f(\varphi _1,\dots ,\varphi _k)$ for some continuous $f: \mathbb {R}^k \to \mathbb {R}$ . Since each of the $\varphi _j$ ’s is bounded by some constant, the output $f(\varphi _1,\dots ,\varphi _k)$ will also be bounded, by the spectral mapping property for continuous functional calculus. Finally, suppose that $\varphi (\overline {x}) = \sup _y \psi (\overline {x},y)$ . Clearly, if ${\lVert {\psi _E^M(\overline {x},y)}\rVert } \leq C$ for all y, then ${\lVert {\varphi _E^M(x)}\rVert }\leq C$ as well.

Proof. We proceed by induction on complexity of the formula. First, suppose $\varphi (\overline {x}) = \operatorname {\mathrm {Re}} \operatorname {\mathrm {tr}}(p(\overline {x}))$ , and hence, $\varphi _E^M(x_{1,j},\dots ,x_{n,j}) = \operatorname {\mathrm {Re}} E[p(x_{1,j},\dots ,x_{n,j})]$ . Write $x_i = \sum _{j \in J} e_j x_{i,j}$ . Since $e_j$ is central,

$$ \begin{align*} e_j \operatorname{\mathrm{Re}} E[p(x_{1,j},\dots,x_{n,j})] &= e_j \operatorname{\mathrm{Re}} E[p(e_j x_{1,j},\dots,e_j x_{n,j})] \\ &= e_j \operatorname{\mathrm{Re}} E[p(e_j x_1,\dots,e_j x_n)] \\ &= e_j \operatorname{\mathrm{Re}} E[p(x_1,\dots,x_n)]. \end{align*} $$

Summing over j completes the proof for this case, and the imaginary part is similar.

Next, suppose that $\varphi = f(\varphi _1,\dots ,\varphi _k)$ for some continuous $f: \mathbb {R}^k \to \mathbb {R}$ , such that $\varphi _j$ satisfies the induction hypothesis. Let $\overline {x}_j = (x_{1,j},\dots ,x_{n,j})$ . Then we have

$$\begin{align*}e_j f((\varphi_1)_E^M(\overline{x}_j),\dots,(\varphi_k)_E^M(\overline{x})) = e_j f(e_j (\varphi_1)_E^M(\overline{x}), \dots, e_j (\varphi_k)_E^M(\overline{x})) \end{align*}$$

because the mapping $N \to e_jN$ , $y \mapsto e_j y$ is a $*$ -homomorphism of commutative $\mathrm {C}^*$ -algebras and therefore respects functional calculus. From here, we apply the induction hypothesis to $\varphi _i$ and then argue similarly as in the first case.

Now suppose that $\varphi (\overline {x}) = \sup _y \psi (\overline {x},y)$ where $\psi $ satisfies the induction hypothesis. First, note that if $(z_i)_{i \in I}$ is a family of self-adjoint operators in N and e is a projection in N, then

$$\begin{align*}e \sup_i z_i = \sup_i e z_i. \end{align*}$$

Indeed, clearly $e \sup _i z_i$ is an upper bound for $ez_i$ for each i, and hence, $e \sup _i z_i \geq \sup _i e z_i$ . However, let $C = \sup _i {\lVert {z_i}\rVert }$ . Then clearly, $(1 - e)C + \sup _i ez_i$ is an upper bound for each $z_i$ , so that $\sup _i z_i \leq (1 - e)C + \sup _i ez_i$ , which implies

$$\begin{align*}e \sup_i z_i \leq e \sup_i ez_i \leq \sup_i ez_i \leq e \sup_i z_i, \end{align*}$$

which proves the claim. Again, let $x_i = \sum _j e_j x_{i,j}$ . Note that

$$ \begin{align*} e_j \varphi_E^M(x_{1,j},\dots,x_{n,j}) &= e_j \sup_{y \in (M)_1} \psi_E^M(x_{1,j},\dots,x_{n,j},y) \\&= \sup_{y \in (M)_1} e_j \psi_E^M(x_{1,j},\dots,x_{n,j},y) \\&= \sup_{y \in (M)_1} e_j \psi_E^M(e_j x_{1,j},\dots,e_jx_{n,j}, e_jy) \\&= \sup_{y \in (M)_1} e_j \psi_E^M(e_j x_1,\dots,e_j x_n, e_j y), \end{align*} $$

and by the reverse manipulations as before, this equals $e_j \varphi _E^M(x_1,\dots ,x_n)$ . Hence,

$$\begin{align*}\sum_{j \in J} e_j \varphi_E^M(x_{1,j},\dots,x_{n,j}) = \sum_{j \in J} e_j \varphi_E^M(x_1,\dots,x_n) = \varphi_E^M(x_1,\dots,x_n) \end{align*}$$

as desired. The case of an infimum is symmetrical; in fact, one can use the identity

$$\begin{align*}\inf_{y \in (M)_1} \varphi_E^M(\overline{x},y) = -\sup_{y \in (M)_1} [-\varphi_E^M(\overline{x},y)] \end{align*}$$

to reduce it to the case of suprema and connectives.

Proof. Let $\Omega $ be the Gelfand spectrum of N, so that $N \cong C(\Omega )$ . We shall regard all $z_i$ as well as $z = \sup _{i \in I} z_i$ as continuous functions on $\Omega $ . Then by [Reference TakesakiTak79, Corollary III.1.16], there exists an open dense set $O \subseteq \Omega $ s.t. for any $t \in O$ , $z(t) = \sup _{i \in I} z_i(t)$ . We now consider a maximal collection $\{K_j\}_{j \in J}$ of nonempty pairwise disjoint clopen subsets s.t. for each j there exists $i_j \in I$ with $|z(t) - z_{i_j}(t)| < \epsilon $ for all $t \in K_j$ . We claim that $\cup K_j$ is dense. Indeed, assume otherwise, then $U = \Omega \setminus \overline {\cup K_j}$ is a nonempty open set, whence so is $O \cap U$ . Fix any $t_0 \in O \cap U$ , as $z(t_0) = \sup _{i \in I} z_i(t_0)$ , there exists some $z_i$ s.t. $|z(t_0) - z_i(t_0)| < \epsilon $ . Continuity then implies $|z(t) - z_i(t)| < \epsilon $ in some neighborhood K of $t_0$ . As clopen sets form a basis of topology for $\Omega $ (see [Reference TakesakiTak79, Definition III.1.6]), such a neighborhood can be chosen to be clopen and a subset of U. But then adding K to $\{K_j\}$ shows the latter collection is not maximal, a contradiction. This shows that $\cup K_j$ is indeed dense. Let $e_j = 1_{K_j}$ . Then $|\sum _j e_j z_{i_j}(t) - z(t)| \leq \epsilon $ on a dense subset of $\Omega $ , and hence by continuity, ${\lVert {\sum _j e_j z_{i_j}(t) - z(t)} \rVert } \leq \epsilon $ .

Proof of Proposition 3.13.

As $\varphi ^M_E(\overline {x}) = \sup _{y \in (M)_1} \psi ^M_E(\overline {x}, y)$ , by Lemma 3.15, there exists a PVM $\{e_j\}_{j \in J}$ over N and elements $y_j \in (M)_1$ such that

$$\begin{align*}\sum_j e_j\psi^M_E(\overline{x}, y_j) \geq \varphi^M_E(\overline{x}) - \epsilon. \end{align*}$$

Let $y = \sum _j e_j y_j$ . By Lemma 3.14, to the decompositions $x_i = \sum _j e_j x_i$ and $y = \sum _j e_j y_j$ , we have

$$\begin{align*}\psi^M_E(\overline{x}, y) = \sum_j e_j \psi_E^M(x_1,\dots,x_n, y_j) \geq \varphi_E^M(\overline{x}) - \epsilon, \end{align*}$$

which concludes the proof.

Proof. We proceed by induction on the complexity of formulas. First, consider a basic formula $\operatorname {\mathrm {Re}} \operatorname {\mathrm {tr}}(p(x_1,\dots ,x_n))$ . By decomposing p into monomials, it suffices to show that $E[x_{i_1} \dots x_{i_k}]$ is uniformly continuous on $(M)_1^n$ . This follows easily from the inequalities ${\lVert {xy}\rVert }_2 \leq \min ({\lVert {x}\rVert } {\lVert {y} \rVert }_2, {\lVert {x}\rVert }_2 {\lVert {y}\rVert })$ and ${\lVert {E(x)}\rVert }_2 \leq {\lVert {x}\rVert }_2$ .

Next, suppose that $\varphi = f(\varphi _1,\dots ,\varphi _k)$ for some $f: \mathbb {R}^k \to \mathbb {R}$ continuous, and formulas $\varphi _1$ , …, $\varphi _k$ satisfying the conclusion of the lemma. By Lemma 3.12, each formula $\varphi _j$ is bounded by some constant $C_j$ . To obtain the conclusion for $\varphi $ , it suffices to show that the application of f by functional calculus defines a ${\lVert {\cdot }\rVert }_2$ -uniformly continuous function

$$\begin{align*}((N)_{C_1} \cap N_{\operatorname{\mathrm{sa}}}) \times \dots \times ((N)_{C_k} \cap N_{\operatorname{\mathrm{sa}}}) \to N_{\operatorname{\mathrm{sa}}}. \end{align*}$$

Note that in the case where f is a polynomial, this follows from the same reasoning as in the first step. Now fix a sequence of polynomials $f_j$ such that $f_j \to f$ uniformly on $[-C_1,C_1] \times \dots \times [-C_k,C_k]$ . The spectral mapping theorem implies uniform convergence of $f_j \to f$ on N (and with bounds independent of the particular choice of N). Moreover, since uniform continuity is preserved under uniform limits, we see that f is ${\lVert {\cdot }\rVert }_2$ -uniformly continuous.

Next, consider the case where $\varphi (\overline {x}) = \sup _y \psi (\overline {x},y)$ . Let $\delta : [0,\infty ) \to [0,\infty )$ be the modulus of continuity associated to $\psi $ , and suppose that $\max _j {\lVert {x_j - x_j'}\rVert }_2 \leq \delta (\epsilon / 4)$ . By Proposition 3.13, there exists y such that

$$\begin{align*}\psi_E^M(\overline{x},y) \geq \varphi_E^M(\overline{x}) - \epsilon/4. \end{align*}$$

Furthermore, note that

$$\begin{align*}\varphi_E^M(\overline{x}') \geq \psi_E^M(\overline{x}',y). \end{align*}$$

Letting $[\cdot ]_+$ denote the positive part of a self-adjoint operator, we have

$$\begin{align*}[\varphi_E^M(\overline{x}) - \varphi_E^M(\overline{x}')]_+ \leq [\varphi_E^M(\overline{x}) - \psi_E^M(\overline{x},y)]_+ + [\psi_E^M(\overline{x},y) - \psi_E^M(\overline{x}',y)]_+ + [\psi_E^M(\overline{x}',y) - \varphi_E^M(\overline{x}',y)]_+. \end{align*}$$

On the right-hand side, the first term is bounded by $\epsilon /4$ by our choice of y, and the last term is zero by definition of $\varphi $ . Hence,

$$\begin{align*}{\lVert { [\varphi_E^M(\overline{x}) - \varphi_E^M(\overline{x}')]_+ }\rVert}_2 \leq \frac{\epsilon}{4} + {\lVert {\psi_E^M(\overline{x},y) - \psi_E^M(\overline{x}',y)}\rVert}_2 \leq \frac{\epsilon}{2}, \end{align*}$$

where we have applied the uniform continuity condition on $\psi $ . A symmetrical argument shows that

$$\begin{align*}{\lVert { [\varphi_E^M(\overline{x}') - \varphi_E^M(\overline{x})]_+ }\rVert}_2 \leq \frac{\epsilon}{2}. \end{align*}$$

Overall,

$$\begin{align*}{\lVert { \varphi_E^M(\overline{x}) - \varphi_E^M(\overline{x}') }\rVert}_2 = {\lVert{ [\varphi_E^M(\overline{x}) - \varphi_E^M(\overline{x}')]_+ } \rVert}_2 + {\lVert { [\varphi_E^M(\overline{x}') - \varphi_E^M(\overline{x})]_+ }\rVert}_2 \leq \epsilon, \end{align*}$$

provided that $\max _j {\lVert {x_j - x_j'}\rVert }_2 \leq \delta (\epsilon /4)$ . As before, the case of an infimum is symmetrical to the case of a supremum.

Proof. We proceed by induction on the complexity of formulas. The cases for basic formulas and adding connectives are straightforward and left to the reader. Now consider $\varphi = \sup _y \psi (\overline {x}, y)$ where the claim is already proved for $\psi $ .

Let $z(\omega ) = \varphi ^{M_\omega }(\overline {x}(\omega ))$ . First, let us verify that z is measurable (as in [Reference Farah and GhasemiFG24, Lemma 1.2]). Let $(e_n)_{n> 0}$ be as in Definition 1.1, so that $(e_n(\omega ))_{n> 0}$ is ${\lVert {\cdot }\rVert }_2$ -dense in $(M_\omega )_1$ for every $\omega $ . Note that $\varphi ^{M_\omega }$ is uniformly continuous on $(M_\omega )_1$ with respect to ${\lVert {\cdot }\rVert }_2$ by Lemma 3.16 in the case $N = \mathbb {C}$ . Therefore, for almost every $\omega $ ,

$$ \begin{align*} \varphi^{M_\omega}(\overline{x}(\omega)) = \sup_{y_0 \in (M_\omega)_1} \psi^{M_\omega}(\overline{x}(\omega), y_0) = \sup_{n> 0} \psi^{M_\omega}(\overline{x}(\omega), e_n(\omega)). \end{align*} $$

Thus, $z(\omega ) = \varphi ^{M_\omega }(\overline {x}(\omega ))$ is the pointwise supremum of a countable collection of measurable functions, and hence, it is measurable.

We show that this agrees with the operator-valued supremum by proving two directions. Given $y \in (M)_1$ , writing y as a function $\omega \mapsto y(\omega ) \in (M_\omega )_1$ defined almost everywhere, we have by induction hypothesis

$$ \begin{align*} \begin{split} \psi^M_E(\overline{x}, y)(\omega) &= \psi^{M_\omega}(\overline{x}(\omega), y(\omega))\\ &\leq \sup_{y_0 \in (M_\omega)_1} \psi^{M_\omega}(\overline{x}(\omega), y_0)\\ &= \varphi^{M_\omega}(\overline{x}(\omega)). \end{split} \end{align*} $$

Hence, z is an upper bound for $\psi _E^M(\overline {x},y)$ in $N_{\operatorname {\mathrm {sa}}}$ , and

$$\begin{align*}\varphi_E^M(\overline{x}) = \sup_{y \in (M)_1} \psi_E^M(\overline{x},y) \leq z. \end{align*}$$

However, since the operator supremum in $L^\infty (\Omega )$ agrees with the pointwise supremum almost everywhere,

$$ \begin{align*} \begin{split} z(\omega) = \varphi^{M_\omega}(\overline{x}(\omega)) &= \sup_{n> 0} \psi^{M_\omega}(\overline{x}(\omega), e_n(\omega))\\ &= \left[ \sup_{n > 0} \psi^M_E(\overline{x}, e_n) \right](\omega)\\ &\leq \left[\sup_{y \in (M)_1} \psi^M_E(\overline{x}, y) \right](\omega)\\ &= \left[\varphi^M_E(\overline{x})\right](\omega). \end{split} \end{align*} $$

Thus, $z \leq \varphi _E^M(\overline {x})$ . Thus, we have shown $z = \varphi _E^M(\overline {x})$ , as desired.

Proof. We proceed by induction on the complexity of $\varphi $ . For atomic formulas, this holds because for a non-commutative $*$ -polynomial p,

$$\begin{align*}\tau_{E,\mathcal{U}}(p(\pi_{E,\mathcal{U}}(x)) = \tau_{E,\mathcal{U}}(\pi_{E,\mathcal{U}}(p(x)) = \mathcal{U} \circ E[p(x)]. \end{align*}$$

The case of adding connectives can also be proved easily from the fact that $\mathcal {U}$ is a $*$ -homomorphism from N to $\mathbb {C}$ and therefore commutes with continuous functional calculus.

Next, suppose that $\varphi (\overline {x}) = \sup _y \psi (\overline {x},y)$ ; the case of an infimum, of course, is symmetrical. We first show that $\varphi ^{M^{/E, \mathcal {U}}}(\pi _{E,\mathcal {U}}(\overline {x})) \leq \mathcal {U}({\varphi ^M_E}(\overline {x}))$ . As noted in Proposition 2.2, any element $\hat {y}$ of norm at most $1$ in $M^{/E, \mathcal {U}}$ can be lifted to an element y of norm at most $1$ in M. From the induction hypothesis

$$\begin{align*}\psi^{M^{/E, \mathcal{U}}}(\pi_{E,\mathcal{U}}(\overline{x}),\hat{y}) = \mathcal{U}(\psi_E^M(\overline{x},y)) \leq \mathcal{U}(\varphi_E^M(\overline{x})), \end{align*}$$

and since $\hat {y}$ was arbitrary, $\varphi ^{M^{/E,\mathcal {U}}}(\pi _{E,\mathcal {U}}(\overline {x})) \leq \mathcal {U}(\varphi _E^M(\overline {x}))$ . For the other direction, fix $\epsilon> 0$ . Then by Proposition 3.13, there exists $y \in (M)_1$ s.t. $\psi ^M_E(\overline {x}, y) \geq \varphi ^M_E(\overline {x}) - \epsilon $ . Then by induction hypothesis,

$$ \begin{align*} \begin{split} \mathcal{U}({\varphi^M_E}(\overline{x})) &\leq \mathcal{U}({\psi^M_E}(\overline{x}, y)) + \epsilon\\ &= \psi^{M^{/E, \mathcal{U}}}(\pi_{E,\mathcal{U}}(\overline{x}), \pi_{E,\mathcal{U}}(y)) + \epsilon\\ &\leq \varphi^{M^{/E, \mathcal{U}}}(\pi_{E,\mathcal{U}}(\overline{x})) + \epsilon. \end{split} \end{align*} $$

Since $\epsilon $ was arbitrary, $\mathcal {U}({\varphi ^M_E}(\overline {x})) \leq \varphi ^{M^{/E, \mathcal {U}}}(\pi _{E,\mathcal {U}}(\overline {x}))$ , so the proof is complete.

Proof. We proceed by induction on the complexity of the formula. As in the proof of Lemma 3.14, the cases for atomic formulae and adding connectives follows from the fact that $\tilde {M} \ni x \mapsto (e_j \otimes 1)x \in (e_j \otimes 1)\tilde {M}$ is a *-homomorphism. As before, the $\sup $ and $\inf $ cases are symmetrical, and so suppose that $\varphi (\overline {x}) = \sup _y \psi (\overline {x},y)$ where $\psi $ satisfies the induction hypothesis.

Let $e_j$ , $m_{i,j}$ and $x_i$ be as in the theorem statement. Let $\epsilon> 0$ . For each j, by Proposition 3.13, there exists $n_j$ such that

$$\begin{align*}\psi_E^M(\overline{m}_j,n_j) \geq \varphi_E^M(\overline{m}_j) - \epsilon. \end{align*}$$

Let $y = \sum _{j \in J} e_j \otimes n_j$ . Applying the induction hypothesis to $\psi $ , we obtain

$$\begin{align*}\varphi_{\tilde{E}}^{\tilde{M}}(\overline{x}) \geq \psi_{\tilde{E}}^{\tilde{M}}(\overline{x},y) = \sum_{j \in J} e_j \otimes \psi_E^M(\overline{m}_j, n_j) \geq \sum_{j \in J} e_j \otimes (\varphi_E^M(\overline{m}_j) - \epsilon) \geq \sum_{j \in J} e_j \otimes \varphi_E^M(\overline{m}_j) - \epsilon. \end{align*}$$

Therefore, $\varphi _{\tilde {E}}^{\tilde {M}}(\overline {x}) \geq \sum _{j \in J} e_j \otimes \varphi _E^M(\overline {m}_j)$ .

In order to prove the reverse inequality, we start by considering y in certain a dense subset of $(\tilde {M})_1$ with respect to the strong- $*$ operator topology (SOT- $*$ ). Let

$$\begin{align*}S = \left\{\sum_{k \in K} p_k \otimes n_k, \{p_k\}_{k \in K}\textrm{ is a PVM over }A, n_k \in (M)_1 \right\}. \end{align*}$$

By the construction of tensor products and by the Kaplansky density theorem, S is dense in $(\tilde {M})_1$ . Suppose that $y = \sum _{k \in K} p_k \otimes n_k$ is in S. Then by the induction hypothesis applied to the partition $(e_j p_k)_{j \in J, k \in K}$ , we have

$$\begin{align*}\psi_{\tilde{E}}^{\tilde{M}}(\overline{x},y) = \sum_{j \in J, k \in K} e_j p_k \otimes \psi_E^M(\overline{m}_j,n_k) \leq \sum_{j \in J, k \in K} e_j p_k \otimes \varphi_E^M(\overline{m}_j) = \sum_{j \in J} e_j \otimes \varphi_E^M(\overline{m}_j). \end{align*}$$

Next, suppose that y is a general element in $(\tilde {M})_1$ and write y as the limit of a net $y_i \in S$ . Since $y_i \to y$ in SOT- $*$ , we have that for every normal state $\tau $ on $A \overline {\otimes } N$ , we have ${\lVert {y_\ell - y}\rVert }_{2,\tau \circ \tilde {E}} \to 0$ . By Lemma 3.16, this implies that

$$\begin{align*}{\lVert {\psi_{\tilde{E}}^{\tilde{M}}(\overline{x},y_\ell) - \psi_{\tilde{E}}^{\tilde{M}}(\overline{x},y) }\rVert}_{2, \tau \circ \tilde{E}} \to 0. \end{align*}$$

Therefore, since this holds for every normal state, we have $\psi _{\tilde {E}}^{\tilde {M}}(\overline {x},y_\ell ) \to \psi _{\tilde {E}}^{\tilde {M}}(\overline {x},y)$ . The preceding argument showed that

$$\begin{align*}\psi_{\tilde{E}}^{\tilde{M}}(\overline{x},y_\ell) \leq \sum_{j \in J} e_j \otimes \varphi_E^M(\overline{m}_j), \end{align*}$$

and therefore,

$$\begin{align*}\psi_{\tilde{E}}^{\tilde{M}}(\overline{x},y) \leq \sum_{j \in J} e_j \otimes \varphi_E^M(\overline{m}_j). \end{align*}$$

Since $y \in (\tilde {M})_1$ was arbitrary, $\varphi _{\tilde {E}}^{\tilde {M}}(\overline {x}) \leq \sum _{j \in J} e_j \otimes \varphi _E^M(\overline {m}_j)$ . We have now shown both inequalities so the proof is complete.

Proof. (1) $\implies $ (2). If $(M,\tau _M) \equiv (N,\tau _N)$ , then the Keisler-Shelah theorem implies that $M^{\mathcal {U}} \cong N^{\mathcal {U}}$ for some ultrafilter $\mathcal {U}$ on a discrete index set (i.e., a pure state on $\ell ^\infty (I)$ for some index set).

(2) $\implies $ (1). This follows from Proposition 3.20 and the transitivity of elementary equivalence.

Proof. First, to see why $\operatorname {\mathrm {Th}}(M_\omega ,\tau _\omega )$ is a measurable function, it suffices to show that each open set is measurable. Recall that $\mathbb {T}_{\operatorname {\mathrm {tr}}}$ is separable and metrizable, and the topology is generated by the continuous functions given by elements of $\mathcal {P}_{\varnothing }$ , so it suffices to check that $\omega \mapsto \operatorname {\mathrm {Th}}(M_\omega ,\tau _\omega )[\varphi ]$ is measurable for each $\varphi \in \mathcal {P}_{\varnothing }$ . Since formulas give a dense subset of $\mathcal {P}_{\varnothing }$ , it suffices to check measurability for formulas, which follows Lemma 3.17.

Next, to show that the distribution of $\omega \mapsto \operatorname {\mathrm {Th}}(M_\omega ,\tau _\omega )$ agrees with the abstract distribution of theories, it suffices to check that both measures agree when integrated against any continuous function on $\mathbb {T}_{\operatorname {\mathrm {tr}}}$ . By Lemma 3.17, for a sentence $\varphi $ , we have

$$\begin{align*}\tau_N(\varphi_E^M) = \int_{\Omega} \varphi^{M_\omega}\,d\mu(\omega) = \int_{\Omega} \operatorname{\mathrm{Th}}(M_\omega,\tau_\omega)[\varphi]\,d\mu(\omega). \end{align*}$$

By density, this extends to all functions in $\mathcal {P}_{\varnothing } = C(\mathbb {T}_{\operatorname {\mathrm {tr}}};\mathbb {R})$ , so the claim is proved.

Proof. Let q denote the natural quotient map $q: A \overline {\otimes } M \to M^{\mathcal {U}}$ . Note that the trace $\mathcal {U} \circ (\operatorname {\mathrm {id}}_A \otimes \tau )$ on $A \overline {\otimes } M$ restricts on $A \overline {\otimes } N$ to the trace $\mathcal {U} \circ (\operatorname {\mathrm {id}}_A \otimes \tau |_N)$ , and hence, we have a natural trace-preserving embedding $N^{\mathcal {U}} \to M^{\mathcal {U}}$ . Furthermore, letting $\tilde {E}: M^{\mathcal {U}} \to N^{\mathcal {U}}$ be the unique trace-preserving conditional expectation, we have

$$\begin{align*}\tilde{E} \circ q = q \circ (\operatorname{\mathrm{id}}_A \otimes E). \end{align*}$$

Indeed, for each $x \in A \overline {\otimes } M$ , $(\operatorname {\mathrm {id}}_A \otimes E)(x)$ is an element in $A \overline {\otimes } N$ satisfying

$$\begin{align*}(\operatorname{\mathrm{id}}_A \otimes \tau)((\operatorname{\mathrm{id}}_A \otimes E)(x) y) = (\operatorname{\mathrm{id}}_A \otimes \tau)(xy) \text{ for } y \in A \overline{\otimes} N. \end{align*}$$

By applying $\mathcal {U}$ to both sides, we see that $q((\operatorname {\mathrm {id}}_A \otimes E)(x))$ is an element in $N^{\mathcal {U}}$ with the same inner product as $q(x)$ with every element of $N^{\mathcal {U}}$ .

Let $\mathcal {V}$ be a character on $N^{\mathcal {U}}$ . We claim that $(M^{\mathcal {U}})^{/\tilde {E}, \mathcal {V}} \cong (A \mathbin {\overline {\otimes }} M)^{/\operatorname {\mathrm {id}}_A \otimes E, \mathcal {V} \circ q}$ . Indeed, $(A \overline {\otimes } M)^{/\operatorname {\mathrm {id}}_A \otimes E,\mathcal {V} \circ q}$ is obtained by quotienting $A \overline {\otimes } M$ by the ideal $\{x: \mathcal {V} \circ q \circ (\operatorname {\mathrm {id}}_A \otimes E)(x^*x) = 0\}$ . Meanwhile, $(M^{\mathcal {U}})^{/\tilde {E}, \mathcal {V}}$ is obtained by quotienting out the ideal $\{x: \mathcal {U} \circ (\operatorname {\mathrm {id}}_A \otimes \tau )(x^*x) = 0\}$ , and then further quotienting by $\{y \in M^{\mathcal {U}}: \mathcal {V} \circ \tilde {E}(y^*y) = 0\}$ . But by the first paragraph, $\mathcal {V} \circ \tilde {E} \circ q = \mathcal {V} \circ q \circ (\operatorname {\mathrm {id}}_A \otimes E)$ , and hence, these two quotients are the same.

By Theorem 3.18, for every sentence $\varphi $ ,

$$\begin{align*}\mathcal{V}\left[\varphi_{\tilde{E}}^{M^{\mathcal{U}}} \right] = \varphi^{(M^{\mathcal{U}})^{/\tilde{E}, \mathcal{V}}} = \varphi^{(A \mathbin{\overline{\otimes}} M)^{/\operatorname{\mathrm{id}}_A \otimes E, \mathcal{V} \circ q}} = \mathcal{V} \circ q \left[ \varphi_{\operatorname{\mathrm{id}}_A \otimes E}^{A \overline{\otimes} M} \right]. \end{align*}$$

Since $\mathcal {V}$ was an arbitrary pure state,

$$\begin{align*}\varphi_{\tilde{E}}^{M^{\mathcal{U}}} = q\left[ \varphi_{\operatorname{\mathrm{id}}_A \otimes E}^{A \overline{\otimes} M} \right]. \end{align*}$$

By Proposition 3.21,

$$\begin{align*}\varphi_{\operatorname{\mathrm{id}}_A \otimes E}^{A \overline{\otimes} M} = 1_A \otimes \varphi_E^M. \end{align*}$$

Therefore,

$$\begin{align*}\varphi_{\tilde{E}}^{M^{\mathcal{U}}} = q \left[ 1_A \otimes \varphi_E^M \right], \end{align*}$$

or in other words, $\varphi _{\tilde {E}}^{M^{\mathcal {U}}}$ equals the image of $\varphi _E^M$ under the diagonal embedding $N \to N^{\mathcal {U}}$ . Since the diagonal embedding is trace-preserving, we have

$$\begin{align*}\tau_{N^{\mathcal{U}}}[\varphi_{\tilde{E}}^{M^{\mathcal{U}}}] = \tau_N[\varphi_E^M]. \end{align*}$$

Since this holds for all sentences $\varphi $ , we have proved that the distributions of theories agree.

Next, consider the second claim regarding the case where $N = Z(M)$ . By Proposition 2.8, $Z(M^{\mathcal {U}}) = Z(M)^{\mathcal {U}}$ , so this follows from the first claim.

Proof. By the Keisler-Shelah theorem (Theorem 3.23), there exists some ultrafilter such that $M^{\mathcal {U}} \cong N^{\mathcal {U}}$ , which of course implies that $M^{\mathcal {U}}$ and $N^{\mathcal {U}}$ have the same distribution of theories. By Proposition 4.3, M and $M^{\mathcal {U}}$ have the same distribution of theories, and N and $N^{\mathcal {U}}$ have the same distribution of theories. Hence, M and N have the same distribution of theories.

Proof. (1) $\implies $ (2). Note that $([0,1],\operatorname {\mathrm {Leb}})$ is isomorphic to $([0,1] \times [0,1], \mu _{X,Y})$ as a measure space since $(X,Y)$ generate the entire Borel $\sigma $ -algebra on $[0,1]$ . Hence, it suffices to produce an isomorphism $([0,1] \times [0,1], \mu _{X,Y}) \to ([0,1],\mu _X) \times ([0,1],\operatorname {\mathrm {Leb}})$ that preserves the first coordinate.

Let $F_{\nu _x}: [0,1] \to [0,1]$ be the cumulative distribution function of $\nu _x$ – that is, $F_{\nu _x}(y) = \nu _x([0,y])$ . Since $\nu _x$ has no atoms, $F_{\nu _x}$ is continuous. Since $F_{\nu _x}$ is an increasing surjective continuous function $[0,1] \to [0,1]$ , there is a unique strictly increasing, right-continuous function $G_x: [0,1] \to [0,1]$ such that $F_{\nu _x} \circ G_x = \operatorname {\mathrm {id}}$ . Let $F(x,y) = F_{\nu _x}(y)$ and let $G(x,y) = G_x(y)$ . We claim that F and G are Borel-measurable. Indeed, note that

$$\begin{align*}\{(x,y): F(x,y) \leq t\} = \{(x,y): \nu_x([0,y]) \leq t\}, \end{align*}$$

which is Borel-measurable by the properties of the disintegration. However,

$$\begin{align*}\{(x,y): G(x,y) \leq t \} = \{(x,y): y \leq F(x,t) \}, \end{align*}$$

which is Borel-measurable since F is Borel-measurable. Note that $(F_{\nu _x})_* \nu _x = \operatorname {\mathrm {Leb}}$ since $(F_{\nu _x})_* \nu _x([0,t]) = \nu _x(F_{\nu _x}^{-1}([0,t])) = \nu _x([0,G_x(t)]) = F_{\nu _x}(G_x(t)) = t$ .

Now let $\tilde {F}(x,y) = (x, F(x,y))$ and $\tilde {G}(x,y) = (x,G(x,y))$ . Note that $F(x,G(x,y)) = y$ by construction, and $\mu _{X,Y}$ -almost everywhere we have $G(x,F(x,y)) = y$ since the set of jump discontinuities of $G_x$ has measure zero, and hence, $\tilde {F}$ and $\tilde {G}$ are inverses of each other almost everywhere. Moreover, $\tilde {F}_* \mu _{X,Y} = \mu _X \times \operatorname {\mathrm {Leb}}$ because using the disintegration theorem, for $h \in C([0,1] \times [0,1])$ ,

$$ \begin{align*} \int_{[0,1]^2} h \circ \tilde{F}\,d\mu_{X,Y} &= \int_{[0,1]} \int_{[0,1]} h(x,F(x,y))\,d\nu_x(y)\,d\mu_X(x) \\ &= \int_{[0,1]} \int_{[0,1]} h(x,y) \,d(F_{\nu_x})_*\nu_x(y)\,d\mu_X(x) \\ &= \int_{[0,1]} \int_{[0,1]} h(x,y)\,dy\,d\mu_X(x) \\ &= \int_{[0,1] \times [0,1]} h\,d(\mu_X \times \operatorname{\mathrm{Leb}}). \end{align*} $$

(2) $\implies $ (3). Let $\Phi $ be as in (2). Let $\Gamma : [0,1] \to [0,1] \times [0,1]$ be any isomorphism of measure spaces (for instance, one can take $\Gamma = \Phi $ ). Then let $\Psi = (\Phi ^{-1} \times \operatorname {\mathrm {id}}_{[0,1]}) \circ (\operatorname {\mathrm {id}}_{[0,1]} \times \Gamma ) \circ \Phi $ . We leave the verification to the reader.

(3) $\implies $ (4). This is immediate.

(4) $\implies $ (5). Let $\Psi $ be as in (4), and let $\tilde {X} = X \circ \pi _1 \circ \Psi $ . We can use the maps $\Psi $ and $\Psi ^{-1}$ to transform between random variables on $[0,1] \times [0,1]$ and $[0,1]$ ; hence, it suffices to show that there are random variables $Z_n: [0,1] \times [0,1] \to \{-1,1\}$ , such that for all random variables $W: [0,1] \times [0,1] \to [0,1]$ , we have ${\lVert {E[Z_n W \mid \tilde {X}]} \rVert }_{L^2} \to 0$ . Let $f_n: [0,1] \to \{-1,1\}$ be a function that alternates between $\pm 1$ on intervals of length $2^{-n}$ . Let $Z_n = f_n \circ \pi _2$ . Given Borel $W_1$ , $W_2: [0,1] \to \mathbb {R}$ , consider the tensor product $W_1 \otimes W_2$ on $[0,1] \times [0,1]$ . Then

$$ \begin{align*} {\lVert{E[Z_n (W_1 \otimes W_2) \mid \tilde{X}]} \rVert}_{L^2} &\leq {\lVert{E[Z_n(W_1 \otimes W_2) \mid \pi_1]} \rVert}_{L^2} \\ &\qquad\qquad\qquad\qquad = {\lVert {(W_1 \otimes 1) E[Z_n W_2]}\rVert}_{L^2} = {\lVert{W_1} \rVert}_{L^2} | E[Z_n W_2]| \to 0. \end{align*} $$

Since simple tensors span a dense subset of the probability space, the claim follows.

(5) $\implies $ (1). Use the same notation from the first step. Let $S_\epsilon $ be the set of $(x,y)$ such that $\nu _x$ has an atom of size $\geq \epsilon $ at y. We can write

$$\begin{align*}S_\epsilon = \{(x,y): \lim_{z \to y^-} F_{\nu_x}(z) \leq F_{\nu_x}(y) - \epsilon \}, \end{align*}$$

and hence, $S_\epsilon $ is a Borel set since $(x,y) \mapsto F_{\nu _x}(y)$ is Borel-measurable. Let $(S_\epsilon )_x$ be the slice $\{y: (x,y) \in S_\epsilon \}$ . From this, it is not hard to see that

$$\begin{align*}f_\epsilon(x) = \begin{cases} \inf (S_\epsilon)_x, & (S_\epsilon)_x \neq \varnothing, \\ -1, \text{ else.} \end{cases} \end{align*}$$

is a Borel function. Let $(Z_n)_{n \in \mathbb {N}}$ be a sequence of random variables as in (5). Let $W_\epsilon = \mathbf {1}_{Y = f_\epsilon (X)}$ . By passing to a subsequence, assume that $E[W_\epsilon Z_n \mid X] \to 0$ almost surely (that is, $\mu _X$ -almost everywhere). But note that when $(S_\epsilon )_x \neq \varnothing $ , we have

$$\begin{align*}|E[W_\epsilon Z_n \mid X](x)| = \left| \int_{[0,1]} Z_n(x,y) \mathbf{1}_{y = f_\epsilon(x)}\,d\nu_x(y) \right| = |Z_n(x,f_\epsilon(x)) \nu_x(\{f_\epsilon(x)))| \geq \epsilon. \end{align*}$$

Therefore, whenever $E[W_\epsilon Z_n \mid X](x) \to 0$ , we have $(S_\epsilon )_x = \varnothing $ . Hence, for $\mu _X$ -almost every x, we have $(S_\epsilon )_x = \varnothing $ . Since this holds for every $\epsilon $ , we conclude that for $\mu _X$ -almost every x, the measure $\nu _x$ has no atoms.

Proof of Theorem A.

The measure space $(\Omega _1,\mu _1)$ can be decomposed into countably many atoms and an atomless space, which we in turn can assume is $([0,1],\operatorname {\mathrm {Leb}})$ . Thus, suppose that $\Omega _1 = I \sqcup [0,1]$ and

$$\begin{align*}(M,\tau) = \bigoplus_{i \in I} \alpha_i (M_i,\tau_i) \oplus \alpha_0 \int_{[0,1]}^{\oplus} (M_\omega,\tau_\omega)\,d\omega, \end{align*}$$

where $(\alpha _i)_{i \in I}$ are the weights of the atoms and $\alpha _0 = 1 - \sum _{i \in I} \alpha _i$ . Let $(M_0,\tau _0) = \int _{[0,1]}^{\oplus } (M_\omega ,\tau _\omega )\,d\omega $ . Similarly, suppose that

$$\begin{align*}(N,\sigma) = \bigoplus_{j \in J} \beta_i (M_j,\sigma_j) \oplus \beta_0 \int_{[0,1]}^{\oplus} (N_\omega,\sigma_\omega)\,d\omega, \end{align*}$$

and let $(N_0,\sigma _0) = \int _{[0,1]}^{\oplus } (N_\omega ,\sigma _\omega )\,d\omega $ .

Since we assumed $(M,\tau ) \equiv (N,\sigma )$ , the Keisler-Shelah theorem yields some ultrafilter $\mathcal {U}$ on some index set S such that $(M,\tau )^{\mathcal {U}} \cong (N,\sigma )^{\mathcal {U}}$ . Note that there is an isomorphism

$$\begin{align*}(M,\tau)^{\mathcal{U}} \cong \bigoplus_{i \in \{0\} \cup I} \alpha_i (M_i,\tau_i)^{\mathcal{U}} \end{align*}$$

in the obvious way. Indeed, any element in the ultrapower $(M,\tau )^{\mathcal {U}}$ can be lifted to some tuple $(x_s)_{s \in S}$ over the index set S, which in turn can be decomposed as a direct sum of elements $x_{s,i} \in M_i$ , $i \in I \cup \{0\}$ . Moreover, ${\lVert {x_s}\rVert }_2 \to 0$ along the ultrafilter if and only if ${\lVert {x_{s,i}}\rVert }_2 \to 0$ along the ultrafilter for each i. The elements $(M_i,\tau _i)^{\mathcal {U}}$ are factors for each $i \in I$ (by Proposition 2.8 for instance), while $(M_0,\tau _0)^{\mathcal {U}}$ has diffuse center. Thus, the elements $1_{M_i}$ are exactly the minimal central projections in $(M,\tau )^{\mathcal {U}}$ for $i \in I$ .

The analogous claims hold for $(N,\sigma )$ as well. Hence, $(M,\tau )^{\mathcal {U}} \cong (N,\sigma )^{\mathcal {U}}$ becomes

$$\begin{align*}\bigoplus_{i \in I} \alpha_i (M_i,\tau_i)^{\mathcal{U}} \oplus \alpha_0 (M_0,\tau_0)^{\mathcal{U}} \cong \bigoplus_{j \in J} \beta_j (N_j,\sigma_j)^{\mathcal{U}} \oplus \beta_0 (N,\sigma)^{\mathcal{U}}. \end{align*}$$

The isomorphism must map minimal central projections on the left-hand side to minimal central projections on the right-hand side. Therefore, there is a bijection $f_1: I \to J$ such that the identity in $(M_i,\tau _i)^{\mathcal {U}}$ is mapped to the identity in $(N_{f_1(i)},\sigma _{f_1(i)})^{\mathcal {U}}$ under the isomorphism. It easily follows that $\alpha _i = \beta _{f_1(i)}$ and $(M_i,\tau _i)^{\mathcal {U}} \cong (N_{f_1(i)},\sigma _{f_1(i)})^{\mathcal {U}}$ , which means that $(M_i,\tau _i) \equiv (N_{f(i)},\sigma _{f(i)})$ . Moreover, the remaining summands with diffuse center in the direct sum decomposition must also correspond under the isomorphism, so that $(M_0,\tau _0)^{\mathcal {U}} \cong (N_0,\sigma _0)^{\mathcal {U}}$ , which means that $(M_0,\tau _0) \equiv (N_0,\sigma _0)$ .

By Corollary 4.4, $(M_0,\tau _0)$ and $(N_0,\sigma _0)$ have the same distribution of theories. By Lemma 4.2, this means that the random variables $X: \omega \mapsto \operatorname {\mathrm {Th}}(M_\omega ,\tau _\omega )$ and $Y: \omega \mapsto \operatorname {\mathrm {Th}}(N_\omega ,\sigma _\omega )$ have the same distribution, or $\mu _X = \mu _Y$ as probability measures on $\mathbb {T}_{\operatorname {\mathrm {tr}}}$ . Let $g: \mathbb {T}_{\operatorname {\mathrm {tr}}} \to [0,1]$ be an isomorphism of standard Borel spaces, and let $\tilde {X} = g \circ X$ . Since we assumed that the diffuse part of the direct integral decomposition satisfies stability under tensorization, it is easy to see that the random variable $\tilde {X}$ satisfies condition (3) of Lemma 4.6. Hence, by condition (2), there exists a measurable isomorphism $\tilde {\Phi }: ([0,1],\operatorname {\mathrm {Leb}}) \to ([0,1],\mu _{\tilde {X}}) \times ([0,1],\operatorname {\mathrm {Leb}})$ such that $\tilde {X} = \pi _1 \circ \tilde {\Phi }$ . Let $\Phi = (g^{-1} \times \operatorname {\mathrm {id}}) \circ \tilde {\Phi }: ([0,1],\operatorname {\mathrm {Leb}}) \to (\mathbb {T}_{\operatorname {\mathrm {tr}}},\mu _{X}) \times ([0,1],\operatorname {\mathrm {Leb}})$ ; then we have $X = \pi _1 \circ \Phi $ . Similarly, obtain an isomorphism $\Psi : ([0,1],\operatorname {\mathrm {Leb}}) \to (\mathbb {T}_{\operatorname {\mathrm {tr}}},\mu _{Y}) \times ([0,1],\operatorname {\mathrm {Leb}})$ such that $Y = \pi _1 \circ \Psi $ .

Then $f_0 = \Psi ^{-1} \circ \Phi $ gives an isomorphism $([0,1],\operatorname {\mathrm {Leb}}) \to ([0,1],\operatorname {\mathrm {Leb}})$ such that $Y \circ f_0 = X$ , meaning that $\operatorname {\mathrm {Th}}(N_{f_0(\omega )},\sigma _{f_0(\omega )}) = \operatorname {\mathrm {Th}}(M_\omega ,\tau _\omega )$ . Let $f: I \sqcup [0,1] \to J \sqcup [0,1]$ be the function given by $f|_I = f_1$ and $f|_{[0,1]} = f_0$ . Then f is a bijection such that $(N_{f(\omega )},\sigma _{f(\omega )}) \equiv (M_\omega , \tau _\omega )$ .