In his "Noncommutative Geometry" book Connes asserts (on p. 539) that for two standard probability spaces $(X,\mu_X)$, $(Y,\nu_Y)$ an $N$-$M$-bimodule for $M=L^\infty(X,\mu_X)$ and $N=L^\infty(Y,\mu_Y)$ is given by a measure class $\mu$ on $X\times Y$ with marginal projections $\mathrm{Pr_X}(\mu)$, $\mathrm{Pr_Y}(\mu)$ absolutely continuous w.r.t. $\mu_X$, $\mu_Y$ and by a $\mu$-measurable function $n: X\times Y \rightarrow \mathbb{Z}$. There is no proof for this fact in the book, and it is not clear for me, why does $\mu$ appear to be a countably-additive measure, not just a finitely-additive one.

It follows from the definition that $N$-$M$-bimodule is a representation $\pi$ of maximal $C^\star$-tensor product $N\otimes_{max} M^{o}$ such that restrictions of $\pi$ on $N$ and $M^o$ are both normal. Functionals of the form $$ \mu_\pi: z\in N\otimes_{max} M^{o} \rightarrow \langle \pi(z)\xi,\xi\rangle $$ where $\xi$ is a cyclic vector, are exactly binormal states, i.e. such normed positive functionals on $N\otimes_{max} M^{o}$ that the maps $(f,g)\rightarrow \mu_\pi(f\otimes g)$ are normal separately in both arguments.

In the commutative ("measure-theoretical") case binormal states can be associated (via the appropriate form of Riesz representation theorem) with finitely-additive probability measures on $X\times Y$ having countably-additive marginal projections, absolutely continuous w.r.t. reference measures $\mu_X$, $\mu_Y$ respectively. It is not a priori clear, why they are countably-additive themselves.

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    $\begingroup$ I'm not so sure about your last paragraph. Are you sure this is the association given by the Riesz representation theorem? $\endgroup$ Feb 12, 2015 at 18:43
  • $\begingroup$ I'm apologize for an inaccurate formulation. Of course, I mean that with any binormal state we can associate some set function of the specified type, but, as follows from the Connes' assertion, not every such set function corresponds to a binormal state. $\endgroup$ Feb 13, 2015 at 15:31
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    $\begingroup$ Yes, but that is what is unclear to me. How are you associating a finite-additive probability measure on $X \times Y$ to a bi-normal functional on $N \otimes_{max} M^o$? I think this is where the confusion lies. $\endgroup$ Feb 13, 2015 at 20:54
  • $\begingroup$ You are right, thank you. I got confused exactly at this point. $\endgroup$ Feb 14, 2015 at 16:36

1 Answer 1


The answer to my own question. Many thanks to Jesse Peterson for pointing out the confusing place.

Let $\mathcal A$, $\mathcal B$ be sigma-algebras of subsets of $X$ and $Y$ respectively. Define the following two algebras of subsets on $X\times Y$

  1. $\mathcal A \times \mathcal B$ be an algebra of all step-sets: finite unions of disjoint subsets of the form $A\times B$, $A\in \mathcal A$, $B\in \mathcal B$ ("measurable rectangles").
  2. $\mathcal A \otimes \mathcal B$ be a $\sigma$-algebra generated by the algebra of step-sets.

When we restrict a normalized state on $L^\infty(\mu_X) \otimes_{max} L^\infty(\mu_Y)$ to indicators of step-sets, we obtain a probability finitely-additive measure on the algebra $\mathcal A \times \mathcal B$ (due to positivity and the normalizing condition of a state). Its restrictions on $\mathcal A$, $\mathcal B$ appear to be countably additive absolutely continuous measures due to binormality: countable additivity corresponds normality, and absolute continuity follows from the Riesz representation of $(L^\infty(\mu_X))^*$ as the space of bounded absolutely continuous finitely-additive measures.

Then we should extend our finitely-additive measure from $\mathcal A \times \mathcal B$ to $\mathcal A \otimes \mathcal B$. It is not hard to prove, that countable additivity of marginals implies countable additivity of our measure on $\mathcal A \times \mathcal B$ (see "Lemma" below). By Hahn-Kolmogorov theorem, it has the unique countably-additive extension on $\mathcal A \otimes \mathcal B$.

It is straightforward to check that any measure of the specified type defines a binormal state on $L^\infty(\mu_X) \otimes_{max} L^\infty(\mu_Y)$ via integration.

Lemma. Any finitely additive measure $\gamma$ on $\mathcal A \times \mathcal B$ with countably additive marginals is countably additive on $\mathcal A \times \mathcal B$.

Proof. Since $\gamma$ is finitely additive, it follows that for a countable family $\{C_i\}$ of disjoint elements of $\mathcal A\times \mathcal B$ with $\bigcup_{i=1}^\infty C_i\in \mathcal A\times \mathcal B$ $$ \gamma\left(\bigcup_{i=1}^\infty C_i\right)\geq \sum_{i=1}^\infty\gamma(C_i) $$ Note that for a countable family $\{A_k\}$ of disjoint elements of $\mathcal A$ and any $B\in\mathcal B$ we have $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)=\gamma\left(\bigcup_{k=1}^\infty A_k \times Y\right)-\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right) $$ Then we can use countable additivity of the marginal $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)+\gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)=\gamma\left(\bigcup_{k=1}^\infty A_k \times Y\right)=\sum_{k=1}^\infty\gamma(A_k\times Y) $$ Since it is forbidden for $\gamma$ to have a strict inequality of the form $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times (Y\setminus B)\right)< \sum_{i=1}^\infty\gamma(A_k\times (Y\setminus B)) $$ we conclude that $$ \gamma\left(\bigcup_{k=1}^\infty A_k \times B\right)= \sum_{k=1}^\infty\gamma(A_k\times B) $$ and hence all measures of the form $\mu_B(A):=\gamma(A\times B)$, $B\in \mathcal B$ on $\mathcal A$ are countably additive. By analogous argument all measures $\nu_A(B):=\gamma(A\times B)$, $A\in \mathcal A$ on $\mathcal B$ are also countably additive.

Let $\{A_n\}$ be a countable family of disjoint elements of $\mathcal A$, $\{B_k\}$ be a countable family of disjoint elements of $\mathcal B$, $A=\bigcup_{n=1}^\infty A_n$, $B=\bigcup_{k=1}^\infty B_k$, $\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty (A_n\times B_k)\in \mathcal A \times \mathcal B$, then $$ \gamma\left(\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty A_n\times B_k\right)=\gamma\left(\bigcup_{k=1}^\infty A\times B_k\right)=\sum_{n=1}^\infty\gamma(A\times B_k)=\sum_{k=1}^\infty\sum_{n=1}^\infty\gamma(A_n\times B_k) $$ Since $$ \sum_{k=1}^\infty\sum_{n=1}^\infty\gamma(A_n\times B_k)=\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n=1}^\infty A_n\times B_k\right)=\gamma\left(\bigcup_{k=1}^\infty \bigcup_{n\neq k} A_n\times B_k\right)+\gamma\left(\bigcup_{n=1}^\infty A_n\times B_n\right) $$ and $$ \gamma\left(\bigcup_{k=1}^\infty \bigcup_{n\neq k} A_n\times B_k\right)\geq \sum_{k=1}^\infty\sum_{n\neq k}\gamma(A_n\times B_k) $$ we conclude that $$ \gamma\left(\bigcup_{n=1}^\infty A_n\times B_n\right)=\sum_{n=1}^\infty\gamma(A_n\times B_n) $$

Let $C=\bigcup_{n=1}^\infty D_n$, $C=\bigcup_{j=1}^N C_j$, $D_n=\bigcup_{i=1}^{M_n} D_{n,i}$, where the families $\{C_j\}$ and $\{D_{n,i}\}$ are families of disjoint measurable rectangles. It follows that $D_{n,i,j}=D_{n,i}\cap C_j$ is also a family of disjoint measurable rectangles. Since $C_j=\bigcup_{n=1}^\infty \bigcup_{i=1}^{M_n} D_{n,i,j}$, we can use the obtained result to show that $$ \gamma(C_j)=\sum_{n=1}^\infty \sum_{i=1}^{M_n} \gamma(D_{n,i,j}) $$ Since $D_{n,j}=\bigcup_{j=1}^{N} D_{n,i,j}$, it is obvious that $\gamma(D_{n,j})=\sum_{j=1}^N \gamma(D_{n,i,j})$. It is also true that $\gamma(C)=\sum_{j=1}^N \gamma(C_j)$, and it follows that $$ \gamma(C)=\sum_{i=1}^\infty \gamma(D_n) $$ which completes the proof of the lemma.

  • $\begingroup$ I don't follow the last part of the argument --- where does the decomposition $D_n = \bigcup_{i=1}^{M_n} D_{n,i}$ come from? But it seems to me that the usual proof of this lemma for product measures, where you integrate characteristic functions, should work. $\endgroup$
    – Nik Weaver
    Feb 14, 2015 at 19:29
  • $\begingroup$ $D_n$ is an element of the algebra of step-sets, hence it can be represented as a finite union of disjoint rectangles. I agree that a shorter proof for the lemma is definitely possible, and probably it already exists. $\endgroup$ Feb 15, 2015 at 9:06
  • $\begingroup$ I really don't think this proof is right. The result about $\gamma(\bigcup A_n \times B_n)$ is only proven when the sequences $\{A_n\}$ and $\{B_k\}$ are pairwise disjoint. $\endgroup$
    – Nik Weaver
    Feb 15, 2015 at 16:44
  • $\begingroup$ It is very likely that I am wrong, but I can't see the mistake. After we obtain the result for pairwise disjoint rectangles, we check that $\gamma(\bigcup D_n)=\sum \gamma(D_n)$ for arbitrary sequence of step-sets $\{D_n\}$. To show that, we split $\bigcup D_n$ into a union of disjoint rectangles $\{D_{n,i,j}\}$ and apply the result about pairwise disjoint sequences to it. $\endgroup$ Feb 16, 2015 at 15:07
  • $\begingroup$ The result is not obtained for pairwise disjoint rectangles, it is obtained for sequences $\{A_n\}$ and $\{B_k\}$ each of which is pairwise disjoint. $\endgroup$
    – Nik Weaver
    Feb 16, 2015 at 16:32

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