[proofplan]
We use Roth's interval density-increment lemma as a black-box quantitative input, together with its fixed-density cutoff at density greater than $1/2$. Starting from a progression-free set of density $\alpha$, the lemma repeatedly passes to a long arithmetic progression on which the density increases by at least a fixed multiple of the square of the current density, as long as the current interval is long enough and the density is at most $1/2$. The fixed-density cutoff rules out the previously problematic stopping case in which the density first exceeds $1/2$ while the interval is still long. Thus the iteration must end by short length before $O(\alpha^{-1})$ successful increments, and the resulting length inequality implies the stated $N/(\log\log N)$ bound.
[/proofplan]
[step:State the external interval increment and cutoff inputs]
We use the following standard interval form of Roth's density-increment machinery. There exist absolute constants $c\in(0,1/4]$ and $M_*\in\mathbb N$ such that the following two assertions hold.
First, if $M\ge M_*$ and $B\subset\{1,\dots,M\}$ has density
\begin{align*}
\beta:=\frac{|B|}{M}\in(0,1/2]
\end{align*}
and contains no non-degenerate three-term arithmetic progression, then there is an arithmetic progression $P\subset\{1,\dots,M\}$ satisfying
\begin{align*}
|P|\ge c\beta^2M
\end{align*}
and
\begin{align*}
\frac{|B\cap P|}{|P|}\ge \beta+c\beta^2.
\end{align*}
Second, if $M\ge M_*$, then every subset $B\subset\{1,\dots,M\}$ with
\begin{align*}
\frac{|B|}{M}>\frac{1}{2}
\end{align*}
contains a non-degenerate three-term arithmetic progression.
The first assertion is the usual interval density-increment lemma obtained from Roth's large-Fourier-coefficient argument on intervals. The second assertion is the corresponding fixed-density Roth cutoff at the particular density $1/2$, with the threshold absorbed into the same constant $M_*$. In the sequel these two displayed statements are the only external inputs used.
[guided]
The proof needs two precise external facts. The first is quantitative: a progression-free set of density at most $1/2$ in a long interval has increased density on a long arithmetic progression. Formally, there are absolute constants $c\in(0,1/4]$ and $M_*\in\mathbb N$ such that whenever $M\ge M_*$, $B\subset\{1,\dots,M\}$, and
\begin{align*}
\beta:=\frac{|B|}{M}\in(0,1/2],
\end{align*}
with $B$ containing no non-degenerate three-term arithmetic progression, there is an arithmetic progression $P\subset\{1,\dots,M\}$ for which
\begin{align*}
|P|\ge c\beta^2M
\end{align*}
and
\begin{align*}
\frac{|B\cap P|}{|P|}\ge \beta+c\beta^2.
\end{align*}
The hypotheses that must be checked later are exactly: the ambient set is an interval of length at least $M_*$, the density lies in $(0,1/2]$, and the set is progression-free.
The second external fact is qualitative but essential for the stopping-time argument. With the same threshold $M_*$ enlarged if necessary, every subset $B\subset\{1,\dots,M\}$ with $M\ge M_*$ and density greater than $1/2$ contains a non-degenerate three-term arithmetic progression. This means that a progression-free set in a long interval can never be the reason the iteration stops by crossing density $1/2$. If the density crosses $1/2$, the current interval must already have length less than $M_*$. This is precisely the missing point in the failed proof.
[/guided]
[/step]
[step:Initialize the progression-free density iteration]
Let $N\ge M_*$, and let $A\subset\{1,\dots,N\}$ contain no non-degenerate three-term arithmetic progression. Define
\begin{align*}
\alpha:=\frac{|A|}{N}.
\end{align*}
If $\alpha=0$, then $|A|=0$, so the desired estimate is immediate. Assume henceforth that $\alpha>0$.
By the fixed-density cutoff from the previous step applied with $B=A$ and $M=N$, the inequality $\alpha>1/2$ is impossible. Hence
\begin{align*}
0<\alpha\le\frac{1}{2}.
\end{align*}
Set $I_0:=\{1,\dots,N\}$, $A_0:=A$, $N_0':=|I_0|=N$, and
\begin{align*}
\alpha_0:=\frac{|A_0|}{N_0'}=\alpha.
\end{align*}
Here the prime on $N_0'$ distinguishes this initial interval length from the final threshold constant in the theorem statement.
[/step]
[step:Perform one valid density-increment step]
Suppose that, for some integer $j\ge0$, we have constructed a finite arithmetic progression $I_j\subset\mathbb Z$, a subset $A_j\subset I_j$, its length $N_j':=|I_j|$, and its density
\begin{align*}
\alpha_j:=\frac{|A_j|}{N_j'}.
\end{align*}
Assume additionally that
\begin{align*}
N_j'\ge M_*
\end{align*}
and
\begin{align*}
0<\alpha_j\le\frac{1}{2}.
\end{align*}
Let
\begin{align*}
\phi_j:I_j\to\{1,\dots,N_j'\}
\end{align*}
be the unique increasing affine bijection. Define
\begin{align*}
B_j:=\phi_j(A_j)\subset\{1,\dots,N_j'\}.
\end{align*}
The map $\phi_j$ and its inverse preserve non-degenerate three-term arithmetic progressions, because affine maps preserve equations of the form $u+w=2v$ and preserve distinctness. Thus $B_j$ is progression-free and has density $\alpha_j$.
The interval density-increment input applies to $B_j$ with $M=N_j'$ and $\beta=\alpha_j$. Therefore there is an arithmetic progression $Q_j\subset\{1,\dots,N_j'\}$ such that
\begin{align*}
|Q_j|\ge c\alpha_j^2N_j'
\end{align*}
and
\begin{align*}
\frac{|B_j\cap Q_j|}{|Q_j|}\ge \alpha_j+c\alpha_j^2.
\end{align*}
Let $P_j:=\phi_j^{-1}(Q_j)\subset I_j$. Choose the increasing affine bijection
\begin{align*}
\psi_j:P_j\to\{1,\dots,|P_j|\}.
\end{align*}
Define
\begin{align*}
I_{j+1}:=\{1,\dots,|P_j|\}
\end{align*}
and
\begin{align*}
A_{j+1}:=\psi_j(A_j\cap P_j)\subset I_{j+1}.
\end{align*}
Then $A_{j+1}$ is progression-free, and with $N_{j+1}':=|I_{j+1}|$ we have
\begin{align*}
N_{j+1}'\ge c\alpha_j^2N_j'
\end{align*}
and
\begin{align*}
\alpha_{j+1}\ge \alpha_j+c\alpha_j^2.
\end{align*}
We repeat this construction exactly while both $N_j'\ge M_*$ and $0<\alpha_j\le1/2$ hold.
[/step]
[step:Bound the number of successful density increments]
Assume that the construction has made $r$ successful increments. For each $0\le j<r$, the increment step was applied only under the hypothesis $0<\alpha_j\le1/2$, and it produced
\begin{align*}
\alpha_{j+1}\ge \alpha_j+c\alpha_j^2.
\end{align*}
Since $0<c\alpha_j\le1$, the elementary inequality
\begin{align*}
\frac{1}{1+t}\le1-\frac{t}{2}\quad\text{for }0\le t\le1
\end{align*}
applied with $t=c\alpha_j$ gives
\begin{align*}
\frac{1}{\alpha_{j+1}}\le\frac{1}{\alpha_j(1+c\alpha_j)}\le\frac{1}{\alpha_j}-\frac{c}{2}.
\end{align*}
Summing this inequality for $j=0,\dots,r-1$ yields
\begin{align*}
\frac{1}{\alpha_r}\le\frac{1}{\alpha}-\frac{cr}{2}.
\end{align*}
Because $\alpha_r>0$, the right-hand side is positive. Therefore
\begin{align*}
r<\frac{2}{c\alpha}.
\end{align*}
Define
\begin{align*}
R:=\left\lceil\frac{2}{c\alpha}\right\rceil+1.
\end{align*}
Every possible number $r$ of successful increments satisfies $r<R$.
[/step]
[step:Force the iteration to stop only after the interval becomes short]
Let $s$ be the total number of successful increments before the iteration stops. The previous step gives
\begin{align*}
s<R.
\end{align*}
For every $0\le j<s$, the density sequence is nondecreasing, because
\begin{align*}
\alpha_{j+1}\ge\alpha_j+c\alpha_j^2\ge\alpha_j.
\end{align*}
Hence $\alpha_j\ge\alpha$ for every $0\le j<s$. The length estimate from each successful increment gives
\begin{align*}
N_{j+1}'\ge c\alpha_j^2N_j'\ge c\alpha^2N_j'.
\end{align*}
By induction on $j$, this implies
\begin{align*}
N_s'\ge N(c\alpha^2)^s.
\end{align*}
We now identify the stopping reason. The construction can stop only because $N_s'<M_*$ or because $\alpha_s>1/2$. If $\alpha_s>1/2$ and $N_s'\ge M_*$, then the fixed-density cutoff applied to $A_s\subset I_s$ gives a non-degenerate three-term arithmetic progression in $A_s$. Since $A_s$ is obtained from $A$ by a composition of affine bijections on arithmetic progressions, this would pull back to a non-degenerate three-term arithmetic progression in $A$, contradicting the hypothesis on $A$. Thus the alternative $\alpha_s>1/2$ is possible only when $N_s'<M_*$. Consequently, in all cases,
\begin{align*}
N_s'<M_*.
\end{align*}
Since $0<c\alpha^2\le1$ and $s<R$, we have
\begin{align*}
(c\alpha^2)^s\ge(c\alpha^2)^R.
\end{align*}
Combining this with the lower bound for $N_s'$ gives
\begin{align*}
M_*>N(c\alpha^2)^R.
\end{align*}
Taking logarithms, we obtain
\begin{align*}
\log N<\log M_*+R\log\left(\frac{1}{c\alpha^2}\right).
\end{align*}
[guided]
The delicate point is to prove that the iteration cannot stop merely because the density crossed $1/2$ while the current interval is still long. Let $s$ denote the number of successful increments before stopping. From the reciprocal estimate already proved, every such $s$ satisfies
\begin{align*}
s<R,
\end{align*}
where
\begin{align*}
R:=\left\lceil\frac{2}{c\alpha}\right\rceil+1.
\end{align*}
During successful increments the densities never decrease. Indeed, for each $0\le j<s$ we have
\begin{align*}
\alpha_{j+1}\ge\alpha_j+c\alpha_j^2\ge\alpha_j.
\end{align*}
Thus $\alpha_j\ge\alpha$ for all $0\le j<s$. The length part of the density-increment lemma says
\begin{align*}
N_{j+1}'\ge c\alpha_j^2N_j'.
\end{align*}
Substituting the monotonicity bound $\alpha_j\ge\alpha$ gives
\begin{align*}
N_{j+1}'\ge c\alpha^2N_j'.
\end{align*}
Iterating this inequality from $j=0$ to $j=s-1$ yields
\begin{align*}
N_s'\ge N(c\alpha^2)^s.
\end{align*}
Now examine why the process stopped. By construction, there are only two stopping conditions: either the interval has become short, meaning $N_s'<M_*$, or the density has crossed the permitted range for the density-increment lemma, meaning $\alpha_s>1/2$. Suppose the second condition occurs while $N_s'\ge M_*$. Then the fixed-density cutoff applies to the progression-free set $A_s\subset I_s$, because $I_s$ is an arithmetic progression of length at least $M_*$ and $A_s$ has density greater than $1/2$. The cutoff conclusion gives a non-degenerate three-term arithmetic progression in $A_s$.
This is impossible. Each passage from $A_j$ to $A_{j+1}$ was made by restricting to an arithmetic progression and then applying an affine bijection. Affine bijections preserve the equation $u+w=2v$ and preserve distinctness, so a non-degenerate three-term arithmetic progression in $A_s$ pulls back step by step to one in the original set $A$. That contradicts the hypothesis that $A$ is progression-free. Therefore the density-crossing stopping case can occur only after the interval is already short. In every case we have
\begin{align*}
N_s'<M_*.
\end{align*}
Combining shortness with the length lower bound gives
\begin{align*}
M_*>N(c\alpha^2)^s.
\end{align*}
Since $0<c\alpha^2\le1$ and $s<R$, replacing $s$ by the larger exponent $R$ only makes the right-hand side smaller, so
\begin{align*}
M_*>N(c\alpha^2)^R.
\end{align*}
Taking logarithms gives the quantitative length constraint
\begin{align*}
\log N<\log M_*+R\log\left(\frac{1}{c\alpha^2}\right).
\end{align*}
This is the rigorous replacement for the invalid dismissal of the $\alpha_s>1/2$ case.
[/guided]
[/step]
[step:Convert the length constraint into a logarithmic density bound]
Assume $0<\alpha\le1/2$. Since
\begin{align*}
R=\left\lceil\frac{2}{c\alpha}\right\rceil+1\le\frac{2}{c\alpha}+2,
\end{align*}
the previous step gives
\begin{align*}
\log N<\log M_*+\left(\frac{2}{c\alpha}+2\right)\log\left(\frac{1}{c\alpha^2}\right).
\end{align*}
For $0<\alpha\le1/2$, we have $\alpha^{-1}\ge2$ and
\begin{align*}
\log\left(\frac{1}{c\alpha^2}\right)=\log\left(\frac{1}{c}\right)+2\log\left(\frac{1}{\alpha}\right).
\end{align*}
Because $c\in(0,1/4]$, the fixed quantity $\log(1/c)$ is nonnegative. Define
\begin{align*}
K:=\max\left\{1,\log M_*,\log\left(\frac{1}{c}\right)\right\}.
\end{align*}
Then the last two displayed inequalities imply
\begin{align*}
\log N< K+\left(\frac{2}{c\alpha}+2\right)\left(K+2\log\left(\frac{1}{\alpha}\right)\right).
\end{align*}
Since $\alpha^{-1}\ge2$ and $\log(1/\alpha)\ge\log 2$, there is an absolute constant $C_1>0$, depending only on $c$ and $M_*$, such that
\begin{align*}
\log N\le C_1\alpha^{-1}\log\left(\frac{1}{\alpha}\right)
\end{align*}
whenever the iteration starts from a progression-free set of density $\alpha\in(0,1/2]$ in an interval of length $N\ge M_*$.
[/step]
[step:Deduce the stated weak Roth estimate]
Choose an absolute constant $C_2>4C_1$. We claim that, after increasing the final threshold, every admissible set satisfies
\begin{align*}
\alpha\le\frac{C_2}{\log\log N}.
\end{align*}
Suppose instead that
\begin{align*}
\alpha>\frac{C_2}{\log\log N}.
\end{align*}
For all sufficiently large $N$, the right-hand side is at most $1/2$ only in the nontrivial range; if $\alpha>1/2$, the fixed-density cutoff already contradicts progression-freeness. Thus we may apply the logarithmic density bound from the previous step. The assumed lower bound on $\alpha$ gives
\begin{align*}
\alpha^{-1}<\frac{\log\log N}{C_2}.
\end{align*}
Also, for all sufficiently large $N$,
\begin{align*}
\log\left(\frac{1}{\alpha}\right)\le2\log\log\log N.
\end{align*}
Substitution into
\begin{align*}
\log N\le C_1\alpha^{-1}\log\left(\frac{1}{\alpha}\right)
\end{align*}
gives
\begin{align*}
\log N\le\frac{2C_1}{C_2}(\log\log N)(\log\log\log N).
\end{align*}
Since $2C_1/C_2<1/2$ and
\begin{align*}
\frac{(\log\log N)(\log\log\log N)}{\log N}\to0
\end{align*}
as $N\to\infty$, this inequality is impossible for all sufficiently large $N$.
Therefore there exists an absolute threshold $N_0\in\mathbb N$, with $N_0\ge M_*$ and $\log\log N>0$ for every $N\ge N_0$, such that every progression-free $A\subset\{1,\dots,N\}$ with $N\ge N_0$ satisfies
\begin{align*}
\alpha\le\frac{C_2}{\log\log N}.
\end{align*}
Since $\alpha=|A|/N$, this is
\begin{align*}
|A|\le C_2\frac{N}{\log\log N}.
\end{align*}
Taking $C:=C_2$ completes the proof.
[/step]
