In [this](http://www.ams.org/journals/bull/1965-71-01/S0002-9904-1965-11265-4/S0002-9904-1965-11265-4.pdf) classic paper, Sakai proves the following Radon-Nikodym theorem:



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> Let $M$ be a von Neumann algebra, and let $\phi$ and $\psi$ be two normal positive linear functionals on $M$. If $\psi \leq \phi$, then there is a positive operator $t\_0\in M$ such that $0 \leq t\_0 \leq 1$, and $\psi(x) = \phi(t\_0 x t\_0)$ for all $x \in M$.
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The paper provides no uniqueness result. One would naively expect that any two such operators $t\_0$ and $t\_1$ would satisfy $\phi((t\_1-t\_0)^2)=0$. I can find no such statement in the literature. Is this true?


Please note that $\phi$ is not assumed to be faithful.



A non-commutative Radon-Nikodym derivative.
Such t\_0 is unique if its support is at most p, where p is the support of ϕ.
Note that we can replace t\_0 by pt\_0p and the support of pt\_0p is at most p.


Without this additional condition t\_0 is highly non-unique, because
we can replace t\_0 by t\_0 + q,
where q is an arbitrary self-adjoint element with support at most 1-p such that t\_0 + q ≥ 0.
Using simple algebraic manipulations one can show that all solutions can be obtained in this way.


See Lemma 15.4 (page 104) in
Takesaki's book “Tomita's theory of modular Hilbert algebras and its applications”. [Note that Takesaki implicitly assumes that φ\_0 is faithful, hence you need to introduce
an additional condition on the support of h.]



> 
> One would naively expect that any two such operators t\_0 and t\_1 would satisfy ϕ((t\_1−t\_0)^2)=0. 
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> 
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This is a trivial corollary of the above statement characterizing all possible solutions.


