Integration by parts involving empirical/counting measure

Question

Suppose that $\lambda_1(A) \geq \lambda_2(A) \geq \cdots \geq \lambda_n(A)$ are (real) eigenvalues of a Hermitian matrix $A$ and denote the empirical measure by $L_A:= \frac{1}{n}\sum_{i=1}^n \delta_{\lambda_i(A)}$. Let $F_A$ to be distribution function related to the (counting) measure $L_A$. It is stated in Example 6.4 of this book that we have the following integration by parts formula $$\int g\,\mathrm{d}L_A - \int g\,\mathrm{d}L_B = \int (F_A - F_B)\,\mathrm{d}g,$$ in which $g~ \colon \mathbb R \to \mathbb R$ is a bounded function with bounded total variation. I am curious whether the integration by parts in this case is connected to Abel's summation by parts, and if so, can anyone explicitly present the connection?

Answer 1

Sure, just take $g$ to be a distribution function of some discrete probability distribution $L_g$ (which is always bounded w/ bounded TV). Let $S$ be the union of all the supports of $L_g$, $L_A$, $L_B$, and let $I = [n]$ be the index for the set $S$, so that you can rewrite the LHS as:

$$\sum_{i \in [n]} g(s_i) (\delta_{s_i \in S_A} - \delta_{s_i \in S_B})$$

and the right hand side is

$$\sum_{i \in [n]} (g(s_i) - g(s_{i-1})) (F_A - F_B)(s_i)$$

This is basically summation by parts formula (where we note $\delta_{s_i \in S_A} = F_A(s_i) - F_A(s_{i-1}))$ , except the boundary terms, but that is fine, since $(F_A - F_B)(s_n) = 0$, and $g(s_0) = 0$, where $s_0$ is some other number smaller than the smallest number in $S$.

The wikipedia page on summation by parts explains it as well I think.