# $\int\frac{1}{\mu-\lambda}dP(\mu) (A-\lambda I)=I$ & Projection Valued Measures

Let $P_\Omega$ be a projection valued measure.

Let $d(\phi, P_\Omega \phi)$ denote integration with respect to the measure $(\phi, P_\Omega \phi)$.

We know that given a projection valued measure and a bounded Borel function on the support of $P_\Omega$ there is a unique operator $B$ s.t

$(\phi, B\phi) = \int f(\lambda)d(\phi,P_\Omega \phi)$.

$B$ is denoted by $\int f(\lambda)dP_\lambda$.

My question is, given a self adjoint operator $A$, and the fact that we know that

$A$ = $\int \lambda dP_\lambda$, how do we calculate $AB$ for $B$ a bounded operator defined by the p.v.m form?

Specifically, from this question: $\lambda\in\sigma(A)\Leftrightarrow P_{(\lambda-\epsilon,\lambda+\epsilon)}\neq 0\;\forall \epsilon>0$

where $P(\lambda-\epsilon,\lambda+\epsilon) = 0$

how can I show that for $R(\lambda) =$ $\int\frac{1}{\mu-\lambda}dP(\mu)=\int_{|\mu-\lambda| \ge \epsilon}\frac{1}{\mu-\lambda}dP(\mu)$,

$(A-\lambda I)R(\lambda)=R(\lambda)(A-\lambda I)=I$?