Cartan subalgebra and group measure space construction

Let $$N$$ be a $${\rm II}_1$$ factor. A maximal abelian self-adjoint subalgebra (MASA) is a $$*$$-subalgebra $$A \subset N$$ such that $$A' \cap N = A$$. It is called a Cartan subalgebra if moreover $$\mathcal{N}(A)''=N$$, with the normalizer $$\mathcal{N}(A) = \{u \in N \ | \ uAu^* = u^*Au = A, \ u \text{ unitary} \}.$$

Let $$\alpha$$ be a free ergodic action of a countable discrete group $$G$$ on a standard $$\sigma$$-finite measure space $$(X,\mu)$$, and let $$M$$ be the von Neumann algebra $$L^{\infty}(X,\mu) \rtimes_{\alpha} G$$. If $$M$$ is a $${\rm II}_1$$ factor, then $$A=L^{\infty}(X,\mu)$$ is a Cartan subalgebra. But assume that $$M$$ is a $${\rm II}_{\infty}$$ factor, then there is a $${\rm II}_1$$ factor $$N$$ such that $$M \simeq N \otimes B(H)$$.

Question: Should $$N$$ admit a Cartan subalgebra?
If no, could you provide a counter-example?
If yes, what is it?

First of all, there is a projection $$p \in M$$ of finite trace such that $$N \simeq pMp$$, and its equivalent class depends at most on $$tr(p)$$. Because $$G$$ is countable and $$M$$ a $${\rm II}_{\infty}$$ factor, the measure $$\mu$$ must be infinite, then we can choose a subspace $$Y \subset X$$ with $$\mu(Y)=tr(p)$$, and assume that $$N = 1_Y (A \rtimes_{\alpha} G) 1_Y,$$ with $$1_Y$$ the indicator function of $$Y$$.

Let's assume$$^1$$ that $$A$$ is a Cartan subalgebra of $$M$$.

By the work of Feldman and Moore Ergodic equivalence relations, cohomology, and von Neumann algebras, a von Neumann algebra has a Cartan subalgebra$$^2$$ iff it is of the form $$M(R,s)$$, i.e. generated by an equivalence relation $$R$$ (with all equivalent classes countable) on a standard Borel space $$X$$ (up to a cocycle twist $$s$$). The Cartan subalgebra is then the "diagonal", i.e. $$L^{\infty}(X)$$

Now let $$Y$$ be a subspace of $$X$$, then $$1_Y M(R,s) 1_Y = M(R_{|Y},s)$$, so has a Cartan subalgebra equals to the diagonal $$L^{\infty}(Y)$$.

Warning: The compression $$pMp$$ is not necessarily a group measure space construction, because even if the equivalent relation $$R_{|Y}$$ is always of the form $$R_H$$, the action of $$H$$ on $$Y$$ is not necessarily free, whereas freeness is used to show that $$M(R_H,1)$$ is a group measure space construction. Moreover, the restriction $$R_{|Y}$$ of an ergodic$$^3$$ equivalent relation $$R$$ is not necessarily ergodic. Now, $$pMp$$ is a factor, so if $$R_{|Y} = R_H$$ with $$H$$ acting freely then $$R_{|Y}$$ must be ergodic.

Acknowledgment: Thanks to Jesse Peterson for his help.

$$^1$$it is true in the $${\rm II}_{1}$$ case, but I did not check if it is always true in the $${\rm II}_{\infty}$$ case.
$$^2$$the general case requires the existence of a faithful normal conditional expectation $$E: M \to A$$.
$$^3$$every $$R$$-saturated Borel set has measure $$0$$ or $$1$$.