I have two doubts in the proof of the theorem below. If you want the detaIls can be found here.
Theorem A Let $q$ a given real number $q>p$. Let $u \in W^{1,p}$ be a solution to (E2). Assume that (H1) holds. Then there exists $\varepsilon_0>0, \varepsilon_0(q)$, such that if (H2) holds for some $0 < \varepsilon < \varepsilon_0$, then $u \in W^{1,q}$.
Given $q>p$ we study when $g\equiv M(| \nabla u|) \in L^{q/p}$ by standards arguments of measure theory, $g \in L^{q/p}$ if
\begin{equation} \sum_{k=1}^{\infty} A^{kq/p}w_g(A^{k}\lambda_0)< \infty, \tag{1} \end{equation} where $w_g$ is the distribution function of g.
Now take $A,\delta$ and the corresponding $\varepsilon>0$ given by lemma 1.3; by lemma 1.2 we obtain that $w_g(A\lambda_0) \le \delta w_g(\lambda_0) $ and by recurrence $w_g(A^k\lambda_0) \le \delta^k w_g( \lambda_0)$.
(My Ask)Then the candidate to $\varepsilon_0(q)$ is $\varepsilon=\varepsilon(\delta)$ given by lemma 1.3. But what fixed $0<\delta<1$? To apply the lemma 1.2 we need that i) $w_g(A \lambda_0)< \delta|Q´|$ where $w_g(t) = |\{x \in Q':|g(x)|>t\}|$ and $Q'$ is such that we hope obtain $g \in L^{q/p}$ in $Q'$ and ii) below, but by lemma 1.3 ii)holds only to dyadic cubes $Q_k \subset \overline{Q_k} \subset \dfrac{1}{4}Q$ from $Q$. To answer this, I think that is because we want to obtain local results, then we can consider $Q'$ such that the dyadic cubes $Q_k \subset \overline{Q_k} \subset \dfrac{1}{4}Q$ be also dyadic cubes from $Q'$. Am I right here?.
Then by (1) implies that $ \sum_{k=1}^{\infty} A^{kq/p}w_g(A^{k}\lambda_0) \le w_g(\lambda_0)\sum_{k=1}^{\infty} A^{kq/p} \delta^k.$ We need that $\delta A^{q/p}<1$. If $M(|\nabla u|^p)\in L^{q/p}$ a fortiori $\nabla u \in L^q$.
2.(Ask) We need $\delta <<1$ to obtain $\delta A^{q/p}<1$, but $\delta$ satisfies $ w_g(A\lambda_0) < \delta |Q´|$. Is not clear for me that we can choose $Q',\lambda_0$ that we can find $\delta$ such that $\delta A^{q/p}<1$ and $w_g(A\lambda_0) < \delta |Q´|$.
Lemma 1.2: Let $Q$ be a bounded cube in $\mathbb{R}^{N}$. Assume that $A$ and $B$ are measurable sets, $A \subset B \subset Q$, and that there exists a $\delta>0$ such that
i) $|A| < \delta|Q|$ and
ii) for each $Q_k$ dyadic cube obtained from $Q$ such that $|A \cap Q_k|> \delta_k$, its predecessor $\tilde{Q}_k \subset B$.
Then $|A|< \delta|B|$.
Lemma 1.3 Assume (H1) holds. Let $u \in W^{1,p}$ be a solutin to (E2) in $Q$. Denote by $A = \max(2^{N},2^{p+1}B\}$ with $B$ as in (H1). Then for $0<\delta<1$ fixed, there exist an $\varepsilon= \varepsilon(\delta)>0$ such that if hjypothesis (H2) holds for such $\varepsilon$, and $Q_k \subset \overline{Q_k} \subset \dfrac{1}{4}Q$ satisfies \begin{equation} |Q_k \cap \{x :M(\nabla u|^{p}) < A \lambda| > \delta |Q_k|, \end{equation} the predecessor $\overline{Q}_k$ satisfies $\overline{Q}_k \subset \{x:M(|\nabla u|^{p}) > A \lambda\}.$
Remark: $A$ does not depend on $Q$.