On Rudin's book of functional analysis he defines a resolution of the identity on a sigma algebra $\Sigma$ of a set $\Omega$ to be a mapping $P:\Sigma\rightarrow L(\mathcal{H})$ (the set of orthogonal projections on a Hilbert space $\mathcal{H}$) such that for all $E,F\in\Sigma$ and $x,y\in\mathcal{H}$ $P(E\cap F)=P(E)P(F)$, $P(E\cup F)=P(E)+P(F)$ and $P_{x,y}$ defined by $P_{x,y}(E)=\langle x, P(E) y\rangle$ is a complex valued measure.
Isn't $P(E\cup F)=P(E)+P(F)$ a consequence of $P_{x,y}$ being a complex valued measure?
Yes, you are right. Assume that for all $x,y\in \mathcal H$, $P_{x,y}$ is a complex measure. Suppose $E,F\in\Sigma$ with $E\cap F=\varnothing$. Then for all $x,y\in \mathcal H$, we have $$\langle x,P(E\cup F)y\rangle= P_{x,y}(E\cup F)=P_{x,y}(E)+P_{x,y}(F)=\langle x,(P(E)+P(F))y\rangle.$$ Since $x,y\in \mathcal H$ were arbitrary, it follows that $P(E\cup F)=P(E)+P(F)$.