Editing Direct integral (section)

== Direct integrals of von Neumann algebras ==

Let {''H''<sub>''x''</sub>}<sub>''x'' ∈ ''X''</sub> be a measurable family of Hilbert spaces. A family of [[von Neumann algebra]]s {''A''<sub>''x''</sub>}<sub>''x'' ∈ ''X''</sub>
with

:<math> A_x \subseteq \operatorname{L}(H_x) </math>

is measurable [[if and only if]] there is a countable set ''D'' of measurable operator families that pointwise generate {''A''<sub>''x''</sub>} <sub>''x'' ∈ ''X''</sub> as a von Neumann algebra in the following sense: For almost all ''x'' ∈ ''X'',

:<math> \operatorname{W^*}(\{S_x: S \in D\}) = A_x </math>

where W*(''S'') denotes the von Neumann algebra generated by the set ''S''. If {''A''<sub>''x''</sub>}<sub>''x'' ∈ ''X''</sub> is a measurable family of von Neumann algebras, the direct integral of von Neumann algebras

:<math> \int_X^\oplus A_x d\mu(x) </math>

consists of all operators of the form

:<math> \int_X^\oplus T_x d\mu(x)  </math>

for ''T''<sub>''x''</sub> ∈ ''A''<sub>''x''</sub>.

One of the main theorems of von Neumann and Murray in their original series of papers is a proof of the decomposition theorem: Any von Neumann algebra is a direct integral of factors. Precisely stated,

'''Theorem'''. If {''A''<sub>''x''</sub>}<sub>''x'' ∈ ''X''</sub> is a measurable family of von Neumann algebras and μ is standard, then the family of operator commutants is also measurable and

:<math> \bigg[\int_X^\oplus A_x d\mu(x)\bigg]' = \int_X^\oplus A'_x d\mu(x). </math>