I was reading an article on Probabilistic Number Theory by M.Kac where I am not able to understand why a particular equation mentioned [here](https://projecteuclid.org/journalArticle/Download?urlId=bams%2F1183513946) in page $657$ equation $(7.7)$ is true?


I do understand that $\frac{(\nu(m)-\log\log n)^2}{\log\log n}<\omega^2$ and that it follows that $\frac{(\nu(m)-\log\log n)^2}{n\log\log n}<\frac{\omega^2}{n}$. But why after summing over $m$ we are getting a integral of $\omega^2$? Specifically, why did the inequality changed into an equality?


If somebody could explain this to me I would really appreciate it.



Why is $\sum_{m=1}^{n}\frac{(\nu(m)-\log\log n)^2}{n\log\log n}=\int_{-\infty}^{\infty}\omega^2\, \mathrm{d}\sigma_n(\omega)$?
Denote $f\_m=(\nu(m)-\log\log n)(\log\log n)^{-1/2}$ and define $\rho(\omega)=\sum\_{m=1}^n\delta(\omega-f\_m)$, with $\delta(x)$ the [Dirac delta function](https://en.wikipedia.org/wiki/Dirac_delta_function). Because of the identity $\int g(\omega)\delta(\omega-\omega\_0)\,d\omega=g(\omega\_0)$, one has
$$n^{-1}\sum\_{m=1}^n f\_m^2=n^{-1}\int \omega^2\rho(\omega)\,d\omega.$$


The function $\sigma(\omega)$ introduced by Kac is given by$^\ast$
$$\frac{d}{d\omega}\sigma(\omega)=n^{-1}\rho(\omega),$$
so one may equivalently write
$$n^{-1}\sum\_{m=1}^n f\_m^2=\int \omega^2\,d\sigma(\omega),$$
which is the formula desired by the OP.


$^\ast$ The relation between $\sigma(\omega)$ and $\rho(\omega)$ follows from the identity $d\theta(\omega)/d\omega=\delta(\omega)$, with $\theta(\omega)$ the unit step function.
