Comparison-like lemma Denote by $\precsim$ the order comes from "Murray-von Neumann" equivalence in the projection lattice of a von Numann algebra. Let $e$ and $f$ be two projection in a von Numann algebra $\mathfrak M$. Why is there a central projection $p$ such that 
$$ep\precsim‎ fp\quad \text{and}\quad (1-e)(1-p)\precsim‎(1-f)(1-p).$$
 A: This follows from the reduction theory for von Neumann algebras (alias direct integral decomposition).
Any von Neumann algebra is a direct integral of factors (i.e., von Neumann algebras with a trivial center).
In a factor any two projections are comparable: either e≾‎f or f≾‎e.
Given an arbitrary von Neumann algebra, decompose it as a direct integral
of factors over a measurable space X, and take p to be the central projection
corresponding to the set of points x∈X for which e_x≾‎f_x.
On the complement of this set we have f_x≾‎e_x.
(Reference: Theorem V.1.8 in Takesaki's Theory of Operator Algebras.)
The original post asks for (1−e)(1−p)≾‎(1−f)(1−p), a slightly different condition from f(1−p)≾‎e(1−p).
To deduce this variant from Takesaki's variant it suffices to show that in any factor f≾e and not e~f implies 1−e≾1−f. For simplicity, assume the factor to be σ-finite (the general case is treated similarly). If 1−f is infinite, the claim follows immediately. If 1−f (and hence 1−e) are finite and the factor itself is finite, the claim follows from tr(1−e)≤tr(1−f). Finally, if 1−f and 1−e are finite and the factor is infinite, then we must have e~f (because both are infinite), a contradiction.
A: This is proved in S.K. Berberian's book Baer $*$-Rings as Corollary 1 of section 14 on "Generalized Comparability," although you will need to skim some of the previous results in order to fill in all of the details of that proof. It holds more generally for AW*-algebras, which means there is a more algebraic proof than passing to direct integrals. 
