While studying some basic theory of Cartan subalgebras of von Neumann algebras I found the following fact that I couldn't prove:

- Let $H$ be a Hilbert space, $A$ and $B$ trace von Neumann subalgebras of a given von Neumann algebras $(M,\tau)$, $\tau$ is the trace of $M$ and the restrictions of $\tau$ to $A$ and $B$ give traces on them. Suppose that $H$ is a $(B,A)-$bimodule with $dim_AH_A<\infty$. Then there exists a nonzero projection $f\in B$ and an $(fBf,A)-$module $K\subset fH$ such that $K_A\subset L^2(A,\tau)_A$ as right $A-$module.

I recall that a subalgebra $A\subset M$ is called a Cartan subalgebra if it maximal abelian in $M$, that is $A'\cap M=A$ and if $N_M(A):=\{u\in U(M): uAu^*=A\}$ generates $M$.

If anyone has an idea and can help me understaning this point I would be grateful. I thank you all for the attention.

**Update**

I have just found out that this property is proved in the Book of Brown & Ozawa (C$^*$-algebras and finite-dimensional approximations), Proposition F.10, p. 482.