Let $b(n):\mathbb{N}\to S^1:=\{z\in\mathbb{C}:\,\lvert z\rvert = 1\}$ be a completely multiplicative sequence of integers, that is, $b(mn) = b(m)b(n)$ for all $m,n\in\mathbb{N}$. An example of such a sequence is $(\lambda(n))_{n=1}^\infty$, where $\lambda$ is Liouville's function. The values $b(n)$ are all determined by the values $b(p)$, where $p$ is a prime, since $b$ is completely multiplicative. We may also add the requirement that $b$ be finitely generated in the sense that $\{b(p):p\text{ is prime}\}$ is finite.
It's claimed in this paper, on page 8, footnote 6, that, supposing that we are given a sequence $b$ as above that has a mean value, changing the values of $b$ on a set of primes $P$ with $\sum_P 1/p<\infty$ does not affect whether or not $b$ has a mean value. That is, replacing $b$ with a sequence $b^*$ such that $b^*(p)\neq b(p)$ only when $p\in P$, we have that both $b$ and $b^*$ possess mean values.
I'm interested in what we can say about the mean values. Suppose that $b^*$ has a mean value $M^*$. Is, then, $M=M^*$?
I thought about the special case where $b=\lambda$ and changing the values on a finite set of primes, but not much has come to me yet. I've also thought about reasoning using the fact that $P$ as above must have (upper) density $0$ relative to any set of primes the sum of the reciprocal of whose elements diverges and so perhaps the change can be quantified as negligible in this regard, but if we suppose that $b$ is finitely generated, one would only have to change its value at finitely many primes to achieve a desired change in values over an arbitrary set of primes.
Any clues are much appreciated, thanks!
No. $b \equiv 1$ has mean value $1$ but changing $b(2) = -1$ yields mean value $1/3$.