Equivalence Roth's Theorem I am trying to prove that the infinitary version of Roth's theorem (1) implies the finitary version of it (2):
(1) Any subset $A \subseteq \mathbb{N}$ of positive upper density contains a 3-term AP.
(2) $\forall \delta > 0$, $\exists N_0$ so that for all $N \ge N_0$ and all $A \subseteq [N]$ with $|A| \ge \delta N$, $A$ contains a 3-term AP.
I started by taking a bad $\delta > 0$ and positive integers $N_1 < N_2 < \dots$ with corresponding $A_j \subseteq [N_j]$ so that $|A_j| \ge \delta N_j$ and $A_j$ does not contain a 3-term AP. Then, each $(1_{A_j}(n))_{n=1}^\infty$ is an element of $\{0,1\}^\mathbb{N}$ which is compact (under product topology) and metrizable, so it has a convergent subsequence: say $(1_{A_{j_k}}(n))_{n=1}^\infty$ converges to some $(\omega_n)_{n=1}^\infty =: \omega$ as $k \to \infty$. It is easy to see that $\omega$ cannot have a 3-term AP. However, I am struggling to see why $\omega$ must have positive upper density. Couldn't it conceivably be the case that $A_j$ contains no elements of $[N_{j-1}]$ so that $\omega$ will just end up being the $0$ sequence? 
Is my whole proof attempt invalid or is my reasoning above incorrect? Thanks.
 A: You're right that that proof has a gap.
Here's one way to get a reduction. Given $A_j\subseteq [N_j]$ with $|A_j|\geq \delta N_j,$ we can find a single offset $a_j$ so that $A'_j=A_j-a_j\subset\mathbb Z$ satisfies $|A'_j\cap [-n,n]|\geq \delta n/8$ for each $n\leq N_j.$ This reduces $\delta$ slightly but it means that the limiting densities are bounded away from zero.
To construct $a_j,$ first include $A_j$ in $[2^k]$ with $k=\lceil \log_2 N_j\rceil,$ so $A_j$ has density at least $\delta/2.$ Then pick one of the halves $[1,2^{k-1}]$ or $[2^{k-1}+1,2^k]$ on which $A_j$ has density at least $\delta/2$. Then pick a half of this interval on which $A_j$ still has density at least $\delta/2,$ and so on. We end up with a single point $a_j$ such that for every $n\leq N_j$ there is an interval of length $2^{\lfloor \log_2 n\rfloor},$ including $a_j,$ where $A_j$ has density at least $\delta/2.$ Since that interval has length between $(n+1)/2$ and $n,$ the interval $[a_j-n,a_j+n]$ has density at least $\delta/8.$
