# Direct integrals and fields of operators

Suppose we have a measure space $(X,\mu)$ and a measurable field of Hilbert spaces $H_x$ on it. We can form the direct integral ${\cal{H}} = \int H_x \ d \mu$, which is a Hilbert space.

Suppose now that I have a bounded operator $T$ on $\cal H$, about which I know that it is decomposable.

Do you know of any kind of a "formula" which will "compute" a measurable field of operators $T_x$, such that $\int T_x =T$?

-
Changed to a comment after noting your comment re: the measure not having an atom at $x$. In that case, however, and if $T$ is bounded and decomposable on $\mathcal{H} = \int^\oplus_X \mathcal{H}_x d\mu(x)$ with $\mathcal{H}_x = \mathcal{H}\_{x'}$ and $f \in \mathcal{H}$ then $Tf = \int^\oplus_X T_x f_x d\mu(x)$. See [here][1]. To get $T_x$, take $f$ with point support on $x$. [1]: books.google.com/books?id=yanslHJYvAIC&pg=PA120 –  Steve Huntsman Jan 15 '10 at 16:42
If $T$ is decomposable then $(Tf)_x := T_x f_x$. See appendices F2 and F3 (in particular, p. 417) of amazon.com/Crossed-Products-Algebras-Mathematical-Monographs/dp/… –  Steve Huntsman Jan 15 '10 at 17:12