In [the relevant Wikipedia entry](https://en.wikipedia.org/wiki/Direct_integral), I can read about how to define a direct integral on Hilbert spaces and Von-Neumann algebras.


Suppose that I want to define a direct integral on either Banach spaces or Banach algebras (in particular I am interested in the Schatten classes). Why can't I define a direct integral for those spaces?


**Disclaimer**: this question has been [crossposted from Math.SE](https://math.stackexchange.com/questions/4451664/why-is-it-difficult-to-define-a-direct-integral-of-banach-spaces-or-banach-algeb).



Why is it difficult to define a direct integral of Banach spaces or Banach algebras?
To see the problems one faces, start with the simplest example of a family of Banach spaces, the constant family $x\mapsto B$ with $x\in X$ and $X$ some measurable space. The direct integral will be some space of functions $X\to B$. If $B$ is a Hilbert space, then it has an orthonormal base $e\_i$, and we can consider the space of functions $f\colon X\to B$ such that
\begin{equation\*}
 x\mapsto \langle{e\_i, f(x)}\rangle
\end{equation\*}
is measurable for every $i$. There is an obvious way to stick an inner product in this space and assuming we have taken care that everything goes well, we ought to end up with a Hilbert space that I will denote by $X\otimes B$. This already fails for Banach spaces: there is no general concept of base for a Banach space and not every (separable) Banach space has a Schauder base. If we try to generalize from the coefficient functionals for some Schauder base to a family of projections, we will also fail, because some Banach spaces have very few projections. The naive way of considering the set of sections $s$ such that $x\mapsto \lVert{s(x)}\rVert$ is measurable does not work because this space is not even closed for the linear operations. But the way forward is clear enough: the direct integral will be some space of *measurable* functions $X\to B$ with the norm $\int\_{X}\lVert{f(x)}\rVert$ (or some $p$-version of it).


Now generalize to non-constant families $x\mapsto B\_x$. Since up to isometric isomorphism there is only one Hilbert space for each cardinal, whatever one can come up with it will boil down to a countable direct sum of the form
\begin{equation\*}
 \sum\_{n}X\_{n}\otimes B\_n
\end{equation\*}
where the $X\_n$ form a (countable) measurable partition of $X$ and $B\_n$ is the Hilbert space of dimension $n$. This of course fails for Banach spaces, because even if we restrict to familes of finite-dimensional Banach spaces, in each dimension the set of equivalence classes of isometric Banach spaces is indeed a set, but a very large one, so we must find a way to glue them all together such that there is a meaningful notion of section.


To proceed, it depends exactly on where you want to go. Here is one way: "section" leads to "sheaf", so one wants to consider sheaves of Banach spaces on the Boolean algebra of measurable sets of $X$. And at this point one needs again to exercise care, because the category of Banach spaces is not well-behaved so the original definition of sheaf must be strengthened.


