I think that there is a typo in the paper referred to in the title of the question, available [here](http://www.ams.org/journals/bull/2018-55-01/S0273-0979-2017-01598-4/S0273-0979-2017-01598-4.pdf). While discussing Roth's proof of AP discrepancy, on page 2, the author states that



> 
> Roth [20] established the existence of such irregularities of distribution. Indeed, more generally he showed that if $A$ is a subset of the [positive] integers up to $N$ with $|A| = ρN$, then there exists an absolute positive constant $c$ such that
>  $$
> \sup\_{\stackrel{k \leq \sqrt{N}}{a\pmod k}} 
> \left| \sum\_{\stackrel{n \in A}{n \equiv a \pmod k}} 1-\frac{\rho N}{q} \right| \geq c \sqrt{\rho(1-\rho)}N^{1/4}
> $$
>  In other words, the only subsets of $[1, N ]$ that are evenly distributed in all arithmetic progressions with moduli below $\sqrt{N}$ are essentially the empty set (with $\rho=1$) or the whole set (with $\rho=1$).
> 
> 
> 


Shouldn't the $k$'s in this inequality should be replaced by $q$'s?



Typo in Soundararajan's *Bulletin of the AMS* article on Tao's resolution of The Erdos Discrepancy Problem?
I actually think there is a more significant flaw. A counterexample to the centered inequality is $\rho = \frac{1}{2}$ and $A = \{1,2,\dots,\frac{N}{2}\}$. Indeed, we always have $\sum\_{n \in A, n \equiv a \pmod{q}} 1 -\frac{N/2}{q} = O(1)$. 


What he actually meant to say is what he ended up saying in words: "evenly distributed in *all* arithmetic progressions with moduli below $\sqrt{N}$". The correct inequality would be something like $$\sup\_{\substack{q \le \sqrt{N} \\ a \pmod{q}} \\ 1 \le m\_1 < m\_2 \le N} \left| \sum\_{\substack{n \in A \\ n \equiv a \pmod{q} \\ m\_1 < n \le m\_2}} 1 - \frac{\rho (m\_2-m\_1)}{q}\right| \ge c\sqrt{\rho(1-\rho)}N^{1/4}.$$


And indeed, the [cited paper](http://matwbn.icm.edu.pl/ksiazki/aa/aa9/aa9125.pdf) says this same thing (note that, by the triangle inequality, we can remove the $m\_1$ in the above at the cost of changing $c$ to $\frac{c}{2}$).


