40% of zeros of Riemann's Zeta function - question. Conrey (1989) proved that at least 40% of zeros of the Riemann's zeta funvtion lie on critical lane. But what does exactly mean 40%? Cardinality of all RZF zeros is $\aleph_0$, and every infinite subset of a set of power $\aleph_0$ also has power $\aleph_0$.
Does that 40% mean that if we look at zeros with modulus $\leq M$ that $$\lim_{ M\rightarrow\infty} \frac{\text{zeros on critical lane with modulus} \leq M}{\text{all zeros with modulus} \leq M}\geq 0.4$$?
 A: Was a comment but it got too long so made it an answer:
It is highly unlikely (on known density theorems) that the limit above is not $1$ so $100$% of the zeroes are expected to lie on the critical line regardless of the truth of RH (remembering that for infinite sets there is a major difference between all and $100$%).
So far we know the following three facts:
the number of critical zeroes up to imaginary value $T$ is ~$T \log T$ (Riemann von Mangoldt)
the number of trivial zeroes up to absolute value $T$ is $O(T)$ (trivial)
the number of possible zeroes in the strip $\frac{1}{2} < \sigma < \Re s <1$ is $O_{\sigma}(T)$ (there are actually much more refined but of course more difficult estimates depending on $\sigma$) (classical, maybe Bohr and/or Landau?)
it follows that we already know that for any $\frac{1}{2} < \sigma<1$,  $100$% of zeroes regarded as the limit above with zeroes up to some form of magnitude $T$ (and we can include the trivial ones here if we so wish and use $|s| \le T$ or not include them and use the more usual $|\Im s| \le T$) lie in the strip $1-\sigma < \Re s < \sigma$
A: "For some $Y>0$, we count roots in the finite strip $\{x+\mathrm{i}y \mid 0 < x < 1 \text{ and } -Y < y < Y\}$.  Then look at what fraction of those are on the critical line or are far enough away from the critical line as we let $Y$ increase." from https://math.stackexchange.com/a/2680126/123905
The bit about "that far away or farther" is meaningful in the context of Bohr and Landau's result, which is also discussed there.
This is largely equivalent to your formulation.
