The following question is about a well-known note posted here.
The note is nicely written and I am hoping that clarification of one line on page 35 will make things fall into place. The line is:
$$\sum_{n<M_k}\chi(\log p_n) = \sum_{n<N_k}\chi(\log p_n)\hspace{10mm}(1) $$
The article has assumed at this point (for contradiction) that $(\log p_n)$ is equidistributed mod 1.
$\chi(x)=1$ for $x\in [0,1/2)$ and $\chi(x)=0$ for $x\in [1/2,1).$
$N_k =\inf \{n: p_n>e^k \}$ and $M_k = \inf\{n:p_n>e^{k-1/2} \}$
My question is, why is this equation true (under the assumption or otherwise)? $N_k$ is a bigger number than $M_k.$
The way it is written, the equality above is part of the definition of $\chi(x).$ "Let $ \chi(x)$ be the function...defined by...[Eq.(1)]."
**I will edit the question shortly so there is no need to refer to the extrinsic linked material.
$\sum_{n < M_k} \chi(\log p_n) - \sum_{n < N_k} \chi(\log p_n) = \sum_{N_k \le n < M_k} \chi(\log p_n) $
If $N_k \le n < M_k$ then $e^{k-1/2} < p_n \le e^k$, and so $k-1/2 \le \log(p_n) < k$ and from there you get $\chi(\log p_n) = 0$ from the definition of $\chi$ (which I assume is extended with $1$-periodicity).
Hence $\sum_{N_k \le n < M_k} \chi(\log p_n) = \sum 0 = 0$, which proves the equality $\sum_{n < M_k} \chi(\log p_n) = \sum_{n < N_k} \chi(\log p_n) $