Let $M$ be a von Neumann algebra and $e$ be a projection in $M$. Let $\{p_i\}_i$ be a maximal family of mutually orthogonal projections in $M$ such that $p_i \preceq e$ for every $i$. Then I want to show that $\displaystyle \sum_i p_i$ is the central support of $e$.
Let $z=\displaystyle \sum_i p_i$. Then in order to show that $z$ is the central support of $e$ it's enough to show $z \in M'\cap M$ and $z$ is the smallest projection such that $ze=e.$ Please help me to solve this. Thank you.
Let $z=\sum_jp_j$. By the comparison theorem, there exists $q$ central with $$\tag1 q (1-z)\preceq qe,\qquad\qquad (1-q)e\preceq(1-q)(1-z). $$ If $q(1-z)\ne0$ we contradict the maximally of $\{p_j\}$. So $q\leq z$. Then $$\tag2 (1-q)e\preceq (1-q)(1-z)=1-z. $$ This forces $(1-q)e=0$, for otherwise we again contradict the maximality. So $e≤q$. Then $$\tag3 e≤z.$$ Let $r≤z-q$. By comparison there exists central $q'$ with $$\tag4 q' (1-r)\preceq q'e. $$ As $q'(1-r)\perp z$, this once more contradicts the maximality if nonzero; thus $q'≤r$. We also have $$\tag5 (1-q')e\preceq (1-q')(1-r)=1-r, $$ and the maximality of $\{p_j\}$ gives us $(1-q')e=0$, that is $e≤q'$. But then $$\tag6 e≤q'≤r\perp q, $$ contradicting $e≤q$. This shows that $r=0$ and so $z=q$.
At this stage we have that $z$ is central and $e≤z$. Let $z'$ be the central support of $e$. So $e≤z'≤z$. Then $z''=z-z'$ is central and $z''e=0$. Then $e≤1-z''$; as $1-z''\perp z$, the maximality forces $z''=0$, that $z'=z$. So $z$ is the central support of $e$.