0$ and let $00$, $B_{2R}(x_0)\subset\Omega$. Following the first part of the proof of Theorem \ref{Th1} step by step, we obtain the estimate \eqref{PTh1}. To estimate the last integral in \eqref{PTh1} we use the Young inequality \eqref{YIN} (here complementary functions are defined through \eqref{YF}) and for any $0<\varepsilon<\omega_{\infty}^{2}$ we obtain \begin{equation}\label{P2} \begin{aligned} &\int_{B_{R}}\omega^{2}(|Du-(Du)_{R}|) |Du-(Du)_{R}|^{2}\, dx \\ &\le\varepsilon\int_{B_{R}}|Du-(Du)_{R}|^{2} \ln_{+}^{\mu}\big(a\varepsilon|Du-(Du)_{R}|^{2}\big)\, dx +\int_{B_{R}}\overline{\Psi} (\frac{\omega_{R}^{2}}{\varepsilon})\, dx \\ &=\varepsilon I_1+I_2 \end{aligned} \end{equation} where $\omega_{R}^{2}(x)=\omega^{2}(|Du(x)-(Du)_{R}|)$. The term $I_1$ can be estimated by means of Proposition \ref{prop2} and we obtain \begin{equation} \label{I1} I_1\le C_{P}^{2}C_{Cacc}C_{q\mu}^{1-1/q} (2^na\varepsilon U_{2R})^{1-1/q}\phi(2R) =K(a\varepsilon U_{2R})^{1-1/q}\phi(2R) \end{equation} where $1 0 \end{equation*} and $m_{R}(t)=m(\{y\in B_{R}(x_0) :|Du-(Du)_{R}|>t\})$. Using the estimate $m_{R}(t)\le\kappa_{n}R^n$, $\kappa_{n}$ is the Lebesgue measure of the unit ball, we have (we use Lemma \ref{L6}) \begin{align} \widetilde{I}_2&\le\int_0^{t_0} \,\frac{d}{dt}\widetilde{\Psi} (\frac{\omega^{2}(t)}{\varepsilon})m_{R}(t)\, dt +\int_{t_0}^{\infty} \,\frac{d}{dt}\widetilde{\Psi} (\frac{\omega^{2}(t)}{\varepsilon})m_{R}(t)\, dt \nonumber \\ &\le\kappa_{n}R^n\int_0^{t_0} \,\frac{d}{dt}\widetilde{\Psi} (\frac{\omega^{2}(t)}{\varepsilon})\, dt +\sup_{t_00$ the average $U_{R}=0$ then it is clear that $x_0$ is the regular point. So in the next we can suppose $U_{R}$ is positive for all $R>0$. Inserting \eqref{P2}--\eqref{II2} into \eqref{PTh1} yields \begin{equation} \label{I3} \begin{aligned} \phi(\sigma) &\le 4L(\frac{\sigma}{R})^{n+2}\phi(R) +2n^{2}N^{2}(1+2L(\frac{\sigma}{R})^{n+2}) \\ &\times[\frac{\varepsilon\, K} {\nu^{2}}(2^na\varepsilon U_{2R})^{1-1/q} +\frac{1}{a\nu^{2}}(\frac{\widetilde{\Psi} \big(\frac{\omega^{2}(t_0)}{\varepsilon}\big)}{U_{2R}} +\frac{\mathcal{M}}{\sqrt{U_{2R}}})]\phi(2R). \end{aligned} \end{equation} In \eqref{I3} we can choose \begin{equation*} a=\frac{16\mathrm{e}n^{2}N^{2}}{\epsilon_0\nu^{2}c_0\, U_{2R}} \quad\text{for } U_{2R}>0 \end{equation*} where $0 0$, we obtain \begin{equation} \label{I5} \begin{aligned} \phi(\sigma) &\le 4L(\frac{\sigma}{R})^{n+2}\phi(R) +\frac{1}{2}\Big(1+2L(\frac{\sigma}{R})^{n+2}\Big)\\ &\quad\times\big[KK_1\epsilon_0^{\alpha+(\alpha-1)(1-1/q)} \nu^{(\beta-2)(2-1/q)} +\frac{\epsilon_0}{4\mathrm{e}^{2}} \big(\mathrm{e}+\mathcal{M}\,\sqrt{U_{2R}}\big)\big]\phi(2R) \\ &=4L\big(\frac{\sigma}{R}\big)^{n+2}\phi(R) +\frac{1}{2}\Big(1+2L(\frac{\sigma}{R})^{n+2}\Big)\\ &\quad\times \big[KK_1(\epsilon_0^{\alpha-1}\nu^{\beta-2})^{2-1/q} +\frac{c_0}{4}+\frac{\mathcal{M}}{4\mathrm{e}}c_0\,\sqrt{U_{2R}}\big] \epsilon_0\phi(2R) \end{aligned} \end{equation} where $K_1=4n^{2}N^{2}(2^{n+4} \mathrm{e}n^{2}N^{2}/c_0)^{1-1/q}$. The constants $\alpha$ and $\beta$ can be always chosen in such a way that \begin{equation*} KK_1(\epsilon_0^{\alpha-1}\nu^{\beta-2})^{2-1/q} \le\frac{1}{4} \end{equation*} and finally we have \begin{equation}\label{I7} \phi(\sigma)\le 4L(\frac{\sigma}{R})^{n+2}\phi(R) +\frac{1}{2}\big(1+2L(\frac{\sigma}{R})^{n+2}\big) \big(\frac{1}{2} +\frac{1}{10}\mathcal{M}c_0\,\sqrt{U_{2R}}\big)\epsilon_0\phi(2R). \end{equation} We can put \begin{equation*} B_1=\frac{1}{2}\epsilon_0, \quad B_2=\frac{1}{10}\mathcal{M}\epsilon_0 \end{equation*} and if we take into account assumption \eqref{T1} of Theorem \ref{Th2} we can use Lemma \ref{L3}. \end{proof} \begin{proof}[Proof of Theorem \ref{Th3}] Let $x_0\in\Omega_{\mathcal{R}}$ and $R_1>0$ be chosen in such a way that $B_{2R_1}(x_0)\subset\Omega$ and let $0 4. \end{cases} \end{equation*} If $m$ is chosen in a suitable way (with respect to $\lambda$) then our results can guarantee the interior regularity of the gradient of weak solution to the Dirichlet problem \eqref{R}. } \end{example} \begin{example}\label{Exam2} {\rm To illustrate some parameters from the proof of Theorem \ref{Th2} we can consider the following modulus of continuity \begin{equation*} \omega(t)= \begin{cases} \omega_0(t)=\frac{(1+s)^{s}\sqrt{\varepsilon}} {(1+\ln\frac{t_0\mathrm{e}^{s}}{t})^{s}} \quad &\text{for } 0 0,\\ \omega_1(t)=\sqrt{\varepsilon}\, kt^{\gamma}, &\text{for } t_0 0\\ \omega_{\infty}\ &\text{for } t>t_1 \end{cases} \end{equation*} where $\varepsilon>0$ is from \eqref{EPS}, $\omega_0(t_0)=\omega_1(t_0)=\sqrt{\varepsilon}<\omega_{\infty}$. For $\mathcal{M}$ from \eqref{Nat} (see \eqref{I2} and \eqref{I3} as well) where $\omega$ is the above function we obtain the estimate \begin{align*} \mathcal{M} &=\sup_{t_0 0, \\ \omega_1(t)=\sqrt{\varepsilon\ln(1+\theta(t))}, &\text{for}\quad t_0 t_1 \end{cases} \end{equation} where $\varepsilon>0$ is from \eqref{EPS}, $\omega_0(t_0)=\omega_1(t_0)=\sqrt{\varepsilon}<\omega_{\infty}$, $\theta(t)$ is a suitable increasing function such that $\lim_{t\to t_0^{+}}\theta(t)=\mathrm{e}-1$. For $\mathcal{M}$ defined by \eqref{Nat}, where $\omega$ is the above function, we obtain \begin{align*} \mathcal{M} &=\sup_{t_0 0$ is a constant), we can see that $\mathcal{M}\le 1$ for $t_0 0$ holds. Then, choosing $\Omega_0=B_{r}(0)$, $0