Spectral averaging and the Krein spectral shift

We provide a new proof of a theorem of Birman and Solomyak that if $A(s) = A_{0} + sB$ with $B\geq 0$ trace class and $d\mu _{s} (\cdot ) = \text{Tr}(B^{1/2} E_{A(s)}(\cdot ) B^{1/2})$, then $\int ^{1}_{0} [d\mu _{s} (\lambda )]\, ds = \xi (\lambda )\, d\lambda$, where $\xi$ is the Krein spectral shift from $A(0)$ to $A(1)$. Our main point is that this is a simple consequence of the formula $\frac{d}{ds} \text{Tr}(f(A(s))=\text{Tr}(Bf'(A(s)))$.