 Solution

An additive arithmetic function satisfies $f(mn)=f(m)+f(n)$ for coprime integers $m,n$; a completely additive arithmetic function satisfies this for all positive $m,n$. In either case $f(1)=0$. The fundamental theorem of arithmetic gives $f(n)={\sum }_{p^k\parallel n}f(p^k)$, where $p^k\parallel n$ means that the prime valuation of $n$ at $p$ is exactly $k$.

Under the discrete uniform distribution on $[N]$, the expectation of the indicator function of $p^k\parallel n$ isThe formula remains valid when $p^{k+1}>N$. Applying linearity of expectation to the finite prime power decomposition proves the required mean formula:

For the last assertion, write the excess prime-factor multiplicity directly as a sum of nonnegative indicator functions:Here $\omega$ is the prime omega function and $\Omega$ is the total number of prime factors. The expectation is bounded independently of $N$, sinceThus Markov inequality gives vanishing probability on every diverging scale: