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 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 $\omega^2$?
If somebody could explain this to me I would really appreciate it.
You want to look at $\displaystyle\frac{\nu(\cdot)-\log\log n}{\log \log n}$ as a discrete random variable $X_n$, whose distribution function is $\sigma_n$. Both the integral on the right and the sum on the left are then equal to $\mathbb{E}[{X_n}^2]$, after noting that the $\displaystyle\frac{1}{n}$ is simply $\mathbb{P}\left(X_n = \displaystyle\frac{\nu(m)-\log\log n}{\log \log n}\right)$, hence equation $(7.7)$.