Is there a trivial construction of the trace on the Jones basic construction? Let $N$ be a type $II_{1}$-factor with trace $\tau$, and $B$ a von Neumann subalgebra. The existence of the semifinite trace on the Jones basic construction $\langle N, e_{B} \rangle$ is reasonably easy to establish if $B$ is a subfactor of $N$, but appears not to be so easy in general.

Question: What is the shortest known proof of the existence of the trace on the basic construction for a von Neumann subalgebra inclusion?

Two complete proofs of this appear in the excellent book Finite von Neumann algebras and Masas by Sinclair and Smith. I (and others far more adept than I) am curious if any other proofs of this exist in the literature.
EDIT: The previous question, as asked, is embarrassingly silly. What seems less than obvious is how to construct the trace and verify that what you have written down is actually a faithful, NORMAL, semifinite trace. If someone could indicate how to do this, I'd very much appreciate it.
 A: Perhaps I am missing some hypothesis, but I think the proof is just about the same whether or not $B$ is a factor.  Here is the proof from Jones' original paper, and I believe it does not use factoriality of B (and not even facoriality of $N$?)
Lemma. Let $J:L^2(N)\to L^2(N)$ be the modular conjugation.  Then $\langle N, e_B\rangle$ is $J B' J$. 
Indeed, clearly $N=JN'J \subset JB'J$ and $e_B=Je_BJ\subset JB'J$ since $L^2(B)$ is $B$-invariant; thus $\langle N,e_B\rangle \subset JN'J$.  On the other hand, $J\langle N, e_B\rangle J ' \subset JNJ ' \cap \{e_B\}' =\{e_B\}'\cap N$ since $JNJ'=N$ and $Je_BJ=e_B$.  If $x\in N$ commutes with $e_B$ then, denoting by $1\in L^2(N)$ the trace vector, $x e_B 1 = e_B x 1 \in L^2(B)$ so that $x\in L^2(B)$.  Thus $x=E_B(x)$ (where $E_B$ is the trace-preserving conditional expectation onto $B$) and so $x\in B$.  Thus $\{e_B'\}\cap N = B$ and as a result $J\langle N, e_B\rangle J' \subset B$.  By the bicommutant theorem you then get that $\langle N, e_B\rangle \subset JB'J$.
Now, given the Lemma, it is clear that $JB'J$ has a semi-finite trace, since $B'$ has a semi-finite trace (since $B$ has a finite trace).
