Dirichlet density vs natural density This is related to the edited form of this question. Suppose that I know that some set of primes has a certain Dirichlet density. What is the optimum statement one can make generally (for example, if the Dirichlet density is non-zero, is it true that the lower natural density is non-zero?
 A: Consider the set of integers $A$ which are in between $10^{n^2-n}$ and $10^{n^2}$ for some $n$. Then the upper natural density of $A$ is $1$, because among the $10^{n^2}$ first integers, at least $10^{n^2}-10^{n^2-n}$ are in $A$, so a proportion of $1-10^{-n}$ are in $A$. The lower density of $A$, on the other hand is $0$, for the number of integer up to $10^{n^2-n}$ of $A$ is at most $10^{(n-1)^2} = 10^{n^2-2n+1}$ so the proportion of those integers that are in $A$ is $10^{1-n}$. 
Now let's compute the logarithmic density of $A$. Each interval $[10^{n^2-n},10^{n^2}]$ in $A$ contributes $\log(10^{n^2}) - \log(10^{n^2-n}) + O(1) = n \log 10 + O(1)$. So if $m$ is any number between $10^{n^2}$ and $10^{(n+1)^2}$, the number of elements of $A$ up to $m$ is $n^2 \log(10)/2 + O(n)$, and when divided by $\log m \sim n^2 \log(10)$, one gets $1/2 + O(1/n)$. Hence the logarithmic density of $A$ is $1/2$.
Hence there is a set whose logarithmic density exists and is non-zero, hence whose Dirichlet density exists and is non-zero, but which have lower natural density 0 and upper natural density 1.
Of course, your question was about a set of primes, but then it suffices to replace $A$ by the set $A'$ of primes into $A$. But just applying the prime number theorem, we see that the computation of densities above are the same, and that proves that having a non-zero Dirichlet density says nothing about the lower or upper natural density.  
