Dgdrxrt

I could show that if $pi:(X.mathcalM,nu) to (Y,mathcalN,eta)$ is a measurable map with $pi_*(nu)<<eta$, then the induced map $tildepi: L^infty(Y,mathcalN,eta)to L^infty(X,mathcalM,nu)$ with $tildepi(f)=fcircpi$ is well defined. Moreover, I could show that $|tildepi(f)|_infty le |f|_infty$. The link to the same can be found here. It is clear that $tildepi$ defined above a $*$-homomorphism. Since $tildepileft(L^infty(Y,mathcalN,eta)right)$is norm closed $*$-subalgebra of $L^infty(X,mathcalM,nu)$, it is also weak$^*$-closed, hence a von Neumann subalgebra of $L^infty(X,mathcalM,nu)$.

I am trying to know if the converse to the above holds:

Suppose that $mathcalA$ is a von Neumann subalgebra of $L^infty(X,mathcalM,nu)$. Then there exists a measurable map $pi:(X,mathcalM,nu) to (Y,mathcalN,eta)$ with $pi_*(nu)<<eta$ such that $mathcalA$ is isomorphic to $L^infty(Y,mathcalN,eta)$ as a von Neumann algebra.

I came across the theorem of Mackey which is known as Mackey's point realization which says the following: Let $(X,mathcalM,nu)$ be a standard Borel space and let $mathcalA'subset mathcalM$ be a sub-sigma algebra which is $nu$-complete. Then there exists a standard Borel space $(Y,mathcalN,eta)$ and a measurable map $pi: (X,mathcalM,nu) to (Y,mathcalN,eta)$ with $pi_*(nu)=eta$ and $mathcalA'=pi^-1(E): E in mathcalN$

I am trying to use this theorem to establish my claim. First, since $mathcalA$ is an abelian von Neumann algebra, $mathcalA$ is of the form $L^infty(Z,mathcalN',eta')$. Let $mathcalA'=varphi(chi_E): E in mathcalN'$, where $varphi$ is the weak$^*$ isomorphism between $mathcalA$ and $L^infty(Z,mathcalN',eta')$. I think that $varphi$ must take characteristic functions to characteristic functions and that $varphi(chi_E)=chi_varphi(E)$. If this is the case, then $mathcalA'$ is a sub-sigma algebra of $mathcalM$. I want to show that $mathcalA'$ is $nu$-complete. Towards that end, let $S subsetvarphi(E)$ such that $nu(varphi(E))=0$. The completeness of $nu$ is not a big deal as I can always complete the sub-sigma algebra by adding the subsets of the measure zero sets but I am not sure if there is any other way of proving this. The only other possibility is that $S=varphi(varphi^-1(S))$ but why should $varphi^-1(S) in mathcalN'$?

Assuming all these, I am a little confused as to what I have shown in all these and how to go forward. A little hint would suffice.

Thanks for the help!!

First, a comment on your first paragraph. A norm-closed subset is not necessarily weak$^*$-closed. What happens here is that $tildepi$ is weak$^*$-continuous.

As for your question, I think you don't gain anything from already having $mathcal A$ represented as $L^infty$. You already have $mathcal Asubset L^infty(X,mathcal M,nu)$. So you can take $Y$ as the support of $mathcal A$ (i.e., the union of all $E$ such that $1_Einmathcal A$; if $mathcal A$ has the same unit, you'll have $Y=X$). Then take $mathcal N=E: 1_Einmathcal A$; it is not hard to check that this is a $sigma$-algebra, using that $mathcal A$ is a von Neumann algebra. And let $eta=nu|_mathcal N$. You have by definition that all simple functions and their norm limits are in $mathcal A$, so $L^infty(Y,mathcal N,eta)subsetmathcal A$. And, conversely, because $mathcal N$ has all characteristic functions in $mathcal A$ (i.e., all the projections), then $mathcal Asubset L^infty(Y,mathcal N,eta)$, as any von Neumann algebra is the norm-closure of its projections.