Let $N\subseteq M$ be an inclusion of semi-finite factors with normal faithful semi-finite traces $\operatorname{Tr}_N$ and $\operatorname{Tr}_M$ respectively. Let $T: M^+\to \widehat{N^+}$ be the unique trace-preserving normal faithful semi-finite operator valued weight.

As for normal weights, we define \begin{align*} \mathfrak{n}_T &= \left\{x\in M \,\middle|\, T(x^*x)\in N^+\right\} \\ \mathfrak{m}_T &= \mathfrak{n}_T^*\mathfrak{n}_T \\ \mathfrak{p}_T &= \left\{x\in M^+\,\middle|\,T(x)\in N^+\right\} \end{align*} It is straightforward to show that $\mathfrak{n}_T$ is a left ideal, $\mathfrak{m}_T\subset \mathfrak{n}_T\cap \mathfrak{n}_T^*$ is a hereditary $*$-algebra, both $\mathfrak{n}_T$ and $\mathfrak{m}_T$ are $N-N$ bimodules, and $\mathfrak{m}_T$ is spanned by its positive part $\mathfrak{m}_T^+$, which in turn is equal to $\mathfrak{p}_T$. Moreover, $T$ has a canonical extension to a map $\mathfrak{m}_T\to N$.

Thus if $x\in \mathfrak{m}_T$ is self-adjoint, we can write $x=x_1-x_2$ with $x_1,x_2\in \mathfrak{p}_T$. But it is not clear to me if $\mathfrak{m}_T$ is closed under taking the Hahn-Jordan decomposition of $x$. That is:

Suppose $x\in \mathfrak{m}_T$ is self-adjoint and $x=x_+-x_-$ is the Hahn-Jordan decomposition of $x$, where $x_+$ and $x_-$ are positive and $x_+x_-=0$. Does it follow that $x_\pm \in \mathfrak{p}_T$?