[proofplan]
We prove the stated formulation by quoting Bourgain's published normalized discretized sum-product estimate as an external input, rather than deriving it from the present theorem. The parameters $\varepsilon$, $\eta$, and $\delta_0$ are chosen from that cited estimate for the prescribed value of $\kappa$. Once the size and non-concentration assumptions on $A$ are checked, the asserted lower bound for $N_\delta(A+A)+N_\delta(A\cdot A)$ is precisely the conclusion of Bourgain's estimate.
[/proofplan]
[step:Choose the constants from Bourgain's normalized discretized estimate]
We use Bourgain's published discretized sum-product estimate in normalized interval form; see J. Bourgain, "On the Erdős-Volkmann and Katz-Tao ring conjectures", Geom. Funct. Anal. 13 (2003), Theorem 1. In the notation needed here, that external theorem states: for every $\kappa>0$ there exist constants $\varepsilon_B=\varepsilon_B(\kappa)>0$, $\eta_B=\eta_B(\kappa)>0$, and $\delta_B=\delta_B(\kappa)>0$, where the subscript $B$ denotes that these constants are supplied by Bourgain's cited theorem, such that, whenever $0<\delta<\delta_B$ and $E\subset[1,2]$ is a union of intervals of length $\delta$ satisfying
\begin{align*}
\delta^{-\kappa} \le N_\delta(E) \le \delta^{-1+\kappa}
\end{align*}
and, for every interval $J\subset\mathbb R$ with $\delta\le |J|\le 1$,
\begin{align*}
N_\delta(E\cap J) \le \delta^{-\eta_B}|J|^\kappa N_\delta(E),
\end{align*}
then
\begin{align*}
N_\delta(E+E)+N_\delta(E\cdot E) \ge \delta^{-\varepsilon_B}N_\delta(E).
\end{align*}
This is Bourgain's externally cited discretized sum-product theorem in the interval-normalized form used here. Define
\begin{align*}
\varepsilon:=\varepsilon_B(\kappa),\qquad \eta:=\eta_B(\kappa),\qquad \delta_0:=\delta_B(\kappa).
\end{align*}
Then $\varepsilon>0$, $\eta>0$, and $\delta_0>0$.
[guided]
The proof is an application of Bourgain's published normalized discretized sum-product estimate, cited here as J. Bourgain, "On the Erdős-Volkmann and Katz-Tao ring conjectures", Geom. Funct. Anal. 13 (2003), Theorem 1. The cited result is an external theorem: it is not the present statement being used as its own proof. In the normalized interval formulation needed here, for each fixed $\kappa>0$, Bourgain's estimate supplies constants $\varepsilon_B(\kappa)>0$, $\eta_B(\kappa)>0$, and $\delta_B(\kappa)>0$ such that any $\delta$-interval union $E\subset[1,2]$ with the prescribed size bounds and the prescribed non-concentration bound satisfies the sum-product expansion inequality
\begin{align*}
N_\delta(E+E)+N_\delta(E\cdot E) \ge \delta^{-\varepsilon_B}N_\delta(E).
\end{align*}
The role of this step is to fix the constants appearing in the theorem statement. Given the theorem's chosen value of $\kappa$, set
\begin{align*}
\varepsilon:=\varepsilon_B(\kappa),\qquad \eta:=\eta_B(\kappa),\qquad \delta_0:=\delta_B(\kappa).
\end{align*}
The cited estimate guarantees that these constants are positive. No property of $A$ has been used yet; this step only chooses constants with the correct dependence on $\kappa$.
[/guided]
[/step]
[step:Verify that the given set satisfies Bourgain's hypotheses]
Let $0<\delta<\delta_0$ and let $A\subset[1,2]$ be as in the theorem statement. Since $\delta_0=\delta_B$, we have $0<\delta<\delta_B$. The set $A$ is a union of intervals of length $\delta$ by hypothesis, and the size condition
\begin{align*}
\delta^{-\kappa} \le N_\delta(A) \le \delta^{-1+\kappa}
\end{align*}
is exactly the size condition in Bourgain's estimate with $E=A$. For every interval $J\subset\mathbb R$ with $\delta\le |J|\le 1$, the theorem hypothesis gives
\begin{align*}
N_\delta(A\cap J) \le \delta^{-\eta}|J|^\kappa N_\delta(A).
\end{align*}
Because $\eta=\eta_B$, this is exactly Bourgain's non-concentration hypothesis with $E=A$.
[guided]
We now check every hypothesis of the cited estimate with $E=A$. First, Bourgain's estimate requires $0<\delta<\delta_B$. Since we defined $\delta_0:=\delta_B$ and the theorem assumes $0<\delta<\delta_0$, this requirement is met. Second, the estimate requires the set to lie in $[1,2]$ and to be a union of intervals of length $\delta$; both are stated hypotheses on $A$.
Third, the estimate requires the two-sided covering-number bound
\begin{align*}
\delta^{-\kappa} \le N_\delta(E) \le \delta^{-1+\kappa}.
\end{align*}
Substituting $E=A$, this is precisely the displayed size assumption in the theorem statement. Finally, Bourgain's estimate requires that every interval $J\subset\mathbb R$ with $\delta\le |J|\le 1$ satisfy
\begin{align*}
N_\delta(E\cap J) \le \delta^{-\eta_B}|J|^\kappa N_\delta(E).
\end{align*}
With $E=A$ and $\eta=\eta_B$, this is precisely the non-concentration assumption imposed on $A$. Thus no hypothesis of Bourgain's estimate remains unchecked.
[/guided]
[/step]
[step:Apply Bourgain's estimate to obtain the asserted sum-product expansion]
Applying Bourgain's externally cited estimate with $E=A$ gives
\begin{align*}
N_\delta(A+A)+N_\delta(A\cdot A) \ge \delta^{-\varepsilon_B(\kappa)}N_\delta(A).
\end{align*}
Since $\varepsilon=\varepsilon_B(\kappa)$, this becomes
\begin{align*}
N_\delta(A+A)+N_\delta(A\cdot A) \ge \delta^{-\varepsilon}N_\delta(A),
\end{align*}
which is the claimed lower bound.
[guided]
We have verified in the previous step that $A$ satisfies every hypothesis of Bourgain's externally cited normalized discretized sum-product estimate with $E=A$. Therefore that estimate applies and gives
\begin{align*}
N_\delta(A+A)+N_\delta(A\cdot A) \ge \delta^{-\varepsilon_B(\kappa)}N_\delta(A).
\end{align*}
The only remaining point is to translate the constants from Bourgain's notation into the constants chosen for this theorem. In the first step we defined $\varepsilon:=\varepsilon_B(\kappa)$, so the preceding inequality is exactly
\begin{align*}
N_\delta(A+A)+N_\delta(A\cdot A) \ge \delta^{-\varepsilon}N_\delta(A).
\end{align*}
This is the desired sum-product expansion bound, and hence the theorem follows.
[/guided]
[/step]
