 Solution

The Bourgain bound for three-term-progression-free sets states that, if $N=|G|$ is odd and $A\subseteq G$ contains no nonconstant three-term arithmetic progression, then

Here is how it follows from the standard Bohr-set density increment lemma. Begin with $B_0=G$ and relative density ${\alpha }_0=\alpha =|A|/N$. Whenever the lemma gives its second alternative, replace the current set by the denser translate inside the smaller regular Bohr set. The density changes byso this can happen only $O({\alpha }^{-1})$ times. Throughout the iteration the rank is $O({\alpha }^{-1})$, the width is at least ${\alpha }^{O({\alpha }^{-1})}$, and the elementary lower bound for the size of a Bohr set givesAt the terminal stage the first alternative of the density-increment lemma holds. Combining it with the last display yieldsIf $\alpha <1/\log N$, Bourgain's bound is already true. Otherwise $\log (1/\alpha )\ll \log \log N$, and rearranging proves