1.2. Roth’s theorem 41
Now if r is one of the periods in the above lemma, we have
En∈[N]|(fstr + fsml)(n + r) − (fstr + fsml)(n)| ε
and thus by shifting
En∈[N]|(fstr + fsml)(n + 2r) − (fstr + fsml)(n + r)| ε
and so by the triangle inequality
En∈[N]|(fstr + fsml)(n + 2r) − (fstr + fsml)(n)| ε.
Putting all this together using the triangle and H¨ older inequalities, we obtain
En∈[N](fstr + fsml)(n)(fstr + fsml)(n + r)(fstr + fsml)(n + 2r) ≥
Thus, if ε is suﬃciently small depending on δ, we have
En∈[N](fstr + fsml)(n)(fstr + fsml)(n + r)(fstr + fsml)(n + 2r)
N values of r, and thus
Λ(fstr + fsml,fstr + fsml,fstr + fsml)
if we then set F to be a suﬃciently rapidly growing function (depending
on δ), we obtain the claim from (1.15). This concludes the proof of Roth’s
Exercise 1.2.12. Use the energy increment method to establish a different
proof of Exercise 1.2.7. (Hint: For the multiple recurrence step, use a
pigeonhole principle argument rather than an appeal to equidistribution
We now briefly indicate how to translate the above arguments into the
ultralimit setting. We first need to construct an important measure on limit
sets, namely Loeb measure.
Exercise 1.2.13 (Construction of Loeb measure). Let N be an unbounded
natural number. Define the Loeb measure μ(A) of a limit subset A of [N]
to be the quantity st(|A|/N), thus for instance a set of cardinality o(N) will
have Loeb measure zero.
(i) Show that if a limit subset A of [N] is partitioned into countably
many disjoint limit subsets An, that all but finitely many of the An
are empty, and so μ(A) = μ(A1) + · · · + μ(An).
(ii) Define the outer measure μ∗(A) of a subset A of [N] (not necessarily
a limit subset) to be the infimum of
μ(An), where A1,A2,...
is a countable family of limit subsets of [N] that cover A, and call
a subset of [N] null if it has zero outer measure. Call a subset
Loeb measurable if it differs from a limit set by a null set. Show
that there is a unique extension of Loeb measure μ from limit sets
to Loeb measurable sets that is a countably additive probability