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Let $_RM_R:={}_R(L^2R\otimes_{\mathbb C} L^2R)_R$, where the first $R$ acts on the first $L^2R$ and the second $R$ acts on the second $L^2R$. Its algebra of $R$-$R$-bimodule endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true. There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.

My argument above is not valid. Nevertheless, I still maintain that there is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Let $_RM_R:={}_R(L^2R\otimes_{\mathbb C} L^2R)_R$, where the first $R$ acts on the first $L^2R$ and the second $R$ acts on the second $L^2R$. Its algebra of $R$-$R$-bimodule endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true. There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.

My argument above is not valid. Nevertheless, I still maintain that there is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Let $R_1$ and $R_2$ be two factors such that one type II (or III) while the other is of type I.
Let $_{R_1}M_{R_2}:={}_{R_1}(L^2{R_1}\otimes_{\mathbb C} L^2{R_2})_{R_2}$, where the first $R_1$ acts on the first $L^2R_1$ and the second $R_2$ acts on the second $L^2R_2$. Its algebra of bimodule endomorphisms is $R_1^{\mathrm{op}}\,\bar\otimes\, R_2$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R_1^{\mathrm{op}}\bar\otimes R_2$ gives rise to a direct integral decomposition of $_{R_1}M_{R_2}$ into irreducible $R_1$-$R_2$-bimodules. But that is not true. There is no way of writing $M$ as a direct integral of irreducible bimodules as... there are no irreducible $R_1$-$R_2$-bimodules!

Let $_RM_R:={}_R(L^2R\otimes L^2R)_R$.
  Its algebra of endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true! There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.

My argument above is not valid. Nevertheless, I still maintain that there is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Let $_RM_R:={}_R(L^2R\otimes_{\mathbb C} L^2R)_R$, where the first $R$ acts on the first $L^2R$ and the second $R$ acts on the second $L^2R$. Its algebra of $R$-$R$-bimodule endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true. There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.

My argument above is not valid. Nevertheless, I still maintain that there is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Let $_RM_R:={}_R(L^2R\otimes L^2R)_R$.
Its algebra of endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true! There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.

Let $_RM_R:={}_R(L^2R\otimes L^2R)_R$.
Its algebra of endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.

Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true! There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.

Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.

My argument above is not valid. Nevertheless, I still maintain that there is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.