Identification between von Neumann algebras and measure spaces I could show that if $\pi:(X.\mathcal{M},\nu) \to (Y,\mathcal{N},\eta)$ is a measurable map with $\pi_*(\nu)<<\eta$, then the induced map $\tilde{\pi}: L^{\infty}(Y,\mathcal{N},\eta)\to L^{\infty}(X,\mathcal{M},\nu)$ with $\tilde{\pi}(f)=f\circ\pi$ is well defined. Moreover, I could show that $\|\tilde{\pi}(f)\|_{\infty} \le \|f\|_{\infty}$.  The link to the same can be found here. It is clear that $\tilde{\pi}$ defined above a $*$-homomorphism. Since $\tilde{\pi}\left(L^{\infty}(Y,\mathcal{N},\eta)\right)$is norm closed $*$-subalgebra of $L^{\infty}(X,\mathcal{M},\nu)$, it is also weak$^*$-closed, hence a von Neumann subalgebra of $L^{\infty}(X,\mathcal{M},\nu)$. 
I am trying to know if the converse to the above holds: 
Suppose that $\mathcal{A}$ is a von Neumann subalgebra of $L^{\infty}(X,\mathcal{M},\nu)$. Then there exists a measurable map $\pi:(X,\mathcal{M},\nu) \to (Y,\mathcal{N},\eta)$ with $\pi_*(\nu)<<\eta$ such that $\mathcal{A}$ is isomorphic to $L^{\infty}(Y,\mathcal{N},\eta)$ 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 $(X,\mathcal{M},\nu)$ be a standard Borel space and let $\mathcal{A}'\subset \mathcal{M}$ be a sub-sigma algebra which is $\nu$-complete. Then there exists a standard Borel space $(Y,\mathcal{N},\eta)$ and a measurable map $\pi: (X,\mathcal{M},\nu) \to (Y,\mathcal{N},\eta)$ with $\pi_*(\nu)=\eta$ and $\mathcal{A}'=\{\pi^{-1}(E): E \in \mathcal{N}\}$ 
I am trying to use this theorem to establish my claim. First, since $\mathcal{A}$ is an abelian von Neumann algebra, $\mathcal{A}$ is of the form $L^{\infty}(Z,\mathcal{N}',\eta')$. Let $\mathcal{A}'=\{\varphi(\chi_E): E \in \mathcal{N}'\}$, where $\varphi$ is the weak$^*$ isomorphism between $\mathcal{A}$ and $L^{\infty}(Z,\mathcal{N}',\eta')$. 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 $\mathcal{A}'$ is a sub-sigma algebra of $\mathcal{M}$. I want to show that $\mathcal{A}'$ is $\nu$-complete. Towards that end, let $S \subset\varphi(E)$ 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 $\varphi^{-1}(S) \in \mathcal{N}'$? 
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!!
 A: First, a   comment on your first paragraph. A norm-closed subset is not necessarily weak$^*$-closed. What happens here is that $\tilde\pi$ 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 $\mathcal A\subset L^\infty(X,\mathcal M,\nu)$. So you can take $Y$ as the support of $\mathcal A$ (i.e., the union of all $E$ such that $1_E\in\mathcal A$; if $\mathcal A$ has the same unit, you'll have $Y=X$). Then take $\mathcal N=\{E:\ 1_E\in\mathcal A\}$; 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 $L^\infty(Y,\mathcal N,\eta)\subset\mathcal A$. And, conversely, because $\mathcal N$ has all characteristic functions in $\mathcal A$ (i.e., all the projections), then $\mathcal A\subset L^\infty(Y,\mathcal N,\eta)$, as any von Neumann algebra is the norm-closure of its projections. 
