Let $M$ be a von Neumann algebra. Let $\Sigma$ be the set of $\sigma$-finite projections of $M$. In Takesaki's book "Theory of operator algebras II", chapter 7, p51, in the proof of theorem 1.11 the equivalence $(i)\iff (iii)$, it is claimed that if $\{p_i\}_{i \in I}$ is a maximal family of orthogonal projections in $\Sigma$, then
$$\sum_{i \in I} p_i = 1.$$
Of course, this begs the questions: does a von Neumann algebra even have $\sigma$-finite projections?
Recall that a projection $p\in M$ is called $\sigma$-finite iff $pMp$ is $\sigma$-finite, i.e. any family $\{p_j\}$ of orthogonal projections such that $p_j\le p$ for all $j$ is necessarily at most countable.
Best Answer
In any von Neumann algebra $M$, we can write $I=\sum_jp_j$ with each $p_j$ a $\sigma$-finite projection. The way to achieve this is by taking $\{p_j\}$ to be a maximal pairwise orthogonal family of cyclic projections.
First, it is clear that $\sum_jp_j=I$. For if $q\leq I-\sum_jp_j$ we take nonzero $x\in qH$ and take $q_0$ to be the projection onto $M'x$. Since $qM'x=M'qx=M'x$, we get $qq_0=q_0$, so $q_0\leq q$. The projection $q_0$ is cyclic and orthogonal to all $p_j$, contradicting the maximality of the family.
It remains to show that a cyclic projection $p$ is $\sigma$-finite. We have that $pH=\overline{M'x}$ for some $x\in pH$. Let $\{p_\alpha\}$ be a pairwise orthogonal family of subprojections of $p$. We have $$\overline{M'p_\alpha x}=\overline{p_\alpha \,M'x}.$$ As $M'x$ is dense in $pH$ and $p_\alpha$ is continuous, then $p_\alpha M'x$ is dense in $p_\alpha pH=p_\alpha H$. Thus we get that $\overline{M'p_\alpha x}=p_\alpha H$. So each $p_\alpha$ is cyclic with cyclic vector $p_\alpha x$. This guarantees that each $p_\alpha x\ne0$. Now, since $\sum_\alpha p_\alpha\leq p$, $$ \sum_\alpha \|p_\alpha x\|^2=\Big\|\sum_\alpha p_\alpha x\Big\|^2\leq\|px\|^2=\|x\|^2<\infty. $$ This shows that only countably many indices $\alpha$ can exist, as $\|p_\alpha x\|>0$ for all $\alpha$.