So the original question I wanted to ask was this one:


I'm currently a bit puzzled about the normalization for the Gelfand transform $U$:
So if we have a periodic Schrödinger operator $H$, then we can decompose it into a direct integral over the (what the physicists call) Brioullin zone $B$ (that is roughly the idea):
$$UHU^{-1} = \int\_{B}^{\oplus} H(k) dk$$
Now, I sometimes see references where a normalization appears in the measure of the direct-integral, although I don't understand why it is necessary!
As an example, let me attach the following 
[theorem from Reed Simon Analysis of Operators](https://i.stack.imgur.com/Gm86u.jpg)
I think I understand the proof, but I don't see any explanation why there is a normalization by $(2\pi)$ (of the direct integral's measure) in the the theorem. In my opinion this would mean that $(UHU^{-1}f)(\theta,x) = \frac{(Vf)(\theta,x)}{2\pi} = \frac{(Vf)(x)}{2\pi} $ which is obviously not what they calculated (in their calculation no $2\pi$ appears), but this the definition of $(\int\_{[0,2\pi)}^{\oplus} \frac{V(\theta)}{2\pi} d\theta f)(\theta, x)$ as far as I know. So something about this $2 \pi$ is weird here. Why is it there in the theorem although it does not show up in the proof?
If anything is unclear, please let me know.
**Now**, I noticed that somehow the definition of the direct integral of operators does not depend very much on the definition of the concrete measure, i.e.


if $d\lambda$ is one measure and the other one is equivalent to $d\lambda$, let us say $d\mu = 5d\lambda$ then the direct integrals


$$\int\_{M}^{\oplus} H(k) d \lambda(k) = \int\_{M}^{\oplus} H(k) d \mu(k)$$ are exactly the same. Is this really true or am I missing something? (This would of course also answer the previous question, if you consider the factor $\frac{1}{(2\pi)}$ as part of the measure. But somehow this looks extremely weird to me.



Question about normalization factors in the direct integral of operators
Well I think the point is that a "direct integral of operators" just means that you have a bounded measurable family of operators on the fibers of a bundle of Hilbert spaces, and you apply them all at once to get a single operator on the space of $L^2$ sections of the bundle. Of course the space of $L^2$ sections of the bundle depends on the choice of measure on the base space.


Once a measure is used to define the space of $L^2$ sections it wouldn't even make sense to talk about a direct integral of operators with respect to some other measure acting on that space. Maybe that is the confusion?


