Fix $a \in \mathbb{Z}$, $q \in \mathbb{N}$ and define an arithmetic function $f$ as follows: $f(n)= \log n$ if $n$ is prime, $0$ otherwise. Can we show that the following limit $\displaystyle \lim_{N\to \infty}\frac{1}{N}\sum _{n \leq N \\ n\equiv a \bmod q}f(n)$ exists and is equal to $1/{\phi(q)}$?
I tried using the Siegel-Walfisz theorem and obtained $$\frac{1}{N}\sum _{n \leq N \\ n\equiv a \bmod q}f(n)\leq \frac{1}{N}\sum _{n \leq N \\ n\equiv a \bmod q}\Lambda(n)=\frac{1}{\varphi(q)}+O(\exp(-C(\log N)^{\frac{1}{2}}))$$ for $\gcd(a,q)=1$. The problem with writing down this inequality is that it only gives us $\displaystyle \limsup_{N\to \infty}\frac{1}{N}\sum _{n \leq N \\ n\equiv a \bmod q}f(n)\leq 1/\varphi(q)$.
Also, for the case $\gcd(a,q)\not = 1$, the sum $\frac{1}{N}\sum _{n \leq N \\ n\equiv a \bmod q}f(n)$ becomes zero for $\gcd (a,q)$ composite ($n=q+kq$ and if $\gcd(a,q)=d$, then $d$ must divide $n$. In other words, for $d$ composite, each $n$ is composite and hence the sum is 0.) However, I am not able to say much for $\gcd(a,q)=d$ prime. Thanks for any tips!
Expanding on Daniel Fischer's hint, I have the following:
$\displaystyle \theta(x,q,a):=\sum _{p \leq x \\p \equiv a \bmod q} \log p$ and $\displaystyle \psi(x,q,a)=\sum_{n\leq x\\ n \equiv a\bmod q}\Lambda(n)$. We know that $\displaystyle \theta(x,q,a)-\psi(x,q,a)=O(\sqrt{x})$.
The sum in question, $\displaystyle \frac{1}{N}\sum _{n \leq N \\ n\equiv a \bmod q}f(n)=\frac{1}{N}\theta(N,q,a)=\frac{1}{N}\Big(\theta(N,q,a)-\psi(N,q,a)\Big)+\frac{1}{N}\psi(N,q,a)$ which is $O(\sqrt{N}/N)+\frac{1}{\varphi(q)}+O(\exp(-C(\log N)^{\frac{1}{2}}))$ for $\gcd(a,q)=1$ using Siegel-Walfisz. The desired limit is $\frac{1}{\varphi(q)}$.
Does this look right?