lim sup + lim inf property of Dirichlet density

Question

Let $A, A'$ be disjoint sets of primes whose union is the whole set of primes. We have the Dirichlet density $$ D(A) = \lim_{s \to 1^+} \frac{\sum_{p \in A} \frac{1}{p^s}}{\log(1/(s-1))}. $$ Provided both limits exist, then an easy formal property is $D(A) + D(A') = 1$. I'm wondering if you can say something stronger: namely, that the lim sup of the density with $A$ plus the lim inf of the density with $A'$ (or vice versa) is equal to $1$?

Answer 1

If we define

$$F_B(s) = \sum_{n \in B} \frac{1}{n^s}$$

for $B \subset \mathbb{N}$ and $s > 1$, then we have

$$F_{A'}(s) = F_{\mathbb{P}}(s) - F_A(s)$$

for all $s > 1$, where $\mathbb{P}$ is the set of all primes. Consequently

$$\frac{F_{A'}(s)}{F_{\mathbb{P}}(s)} = 1 - \frac{F_A(s)}{F_{\mathbb{P}}(s)}$$

for all $s > 1$ and hence

$$\liminf_{s \to 1^+} \frac{F_{A'}(s)}{F_{\mathbb{P}}(s)} = \liminf_{s \to 1^+}\biggl( 1 - \frac{F_A(s)}{F_{\mathbb{P}}(s)}\biggr) = 1 - \limsup_{s\to 1^+} \frac{F_A(s)}{F_{\mathbb{P}}(s)}.$$

Since

$$\lim_{s\to 1^+} \frac{F_{\mathbb{P}}(s)}{\log \frac{1}{s-1}} = 1,$$

the desired equality follows.