\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)=0.
\end{array}
\right.$$ By [**(UE)**]{} and the definitions, we have $$K^{-2}\cdot\mI\leq a^{\eps,R}_{t}(z)\leq K^2\cdot\mI,$$ and for all $|z-z'|\leq\eps$, $$\begin{aligned}
\|a^{\eps,R}_{t}(z)-a^{\eps,R}_{t}(z')\|&\leq\|a_{t}(z)-a_{t}(z')\|+\left\|\tfrac{\mE\big((a_t(Z_t)-a_t(z))\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}
{\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}\right\|\\
&+\left\|\tfrac{\mE\big((a_t(Z_t)-a_t(z'))\varrho_\eps(Z_t-z')1_{t<\tau_R}\big)}{\mE\big(\varrho_\eps(Z_t-z)1_{t<\tau_R}\big)}\right\|\leq 3\omega_a(\eps).\end{aligned}$$ Let $C_0$ be the same as in . By , one may choose $\delta_0$ small enough such that for all $\eps\in(0,\delta_0)$ and $R>0$, $$\begin{aligned}
\label{App}
\omega_{a^{\eps,R}}(\eps)\leq 3\omega_a(\delta_0)\leq\tfrac{1}{2(C_0+1)}.\end{aligned}$$

[**(b)**]{} In this step we show that for any $p>2(2d+1)$ and $f\in\mL^p(T)$, $$\begin{aligned}
\label{BB90}
\mE\left(\int^{T}_0 f_s(Z_s)\dif s\right)\leq C\|f\|_{\mL^p(T)}.\end{aligned}$$ By a standard density argument, we may assume $f\in C_c([0,T]\times\mR^{2d})$. Let $$u^{\eps,R}:=\cR^{a^{\eps,R},0}_{\lambda, T}(f),\ \ u^{\eps,R}_{\eps}:=u^{\eps,R}*\varrho_\eps,\ \ f_\eps:=f*\varrho_\eps.$$ By Itô’s formula, we have $$\begin{aligned}
\label{EY1}
\begin{split}
&\mE\Big(u^{\eps,R}_{\eps,T\wedge\tau_R}(Z_{T\wedge\tau_R})-u_{\eps,0}(Z_0)\Big)
=\mE\left(\int^{T\wedge\tau_R}_0\!\!\! (\p_s u^{\eps,R}_{\eps,s}+\sL^{a,b}_su^{\eps,R}_{\eps,s})(Z_s)\dif s\right).
\end{split}\end{aligned}$$ Noticing that by Definition \[Def31\], $$\p_su^{\eps,R}_{\eps, s}+(\sL^{a^{\eps,R},0}_su^{\eps,R}_s)*\varrho_\eps-\lambda u^{\eps,R}_{\eps, s}+f_\eps=0,$$ and by the definitions of $u^{\eps,R}_\eps$ and $a^{\eps,R}$, $$\begin{aligned}
&\mE\big(\tr(a_s\cdot\nabla^2_\v u^{\eps,R}_{\eps,t})(Z_s)1_{s<\tau_R}\big)\\
&=\int_{\mR^{2d}}\mE\big(\tr(a_s(Z_s)\cdot\nabla^2_\v u^{\eps,R}_s(z))\varrho_\eps(Z_s-z)1_{s<\tau_R}\big)\dif z\\
&=\int_{\mR^{2d}}\tr(a^{\eps,R}_s(z)\cdot\nabla^2_\v u^{\eps,R}_s(z))\mE\big(\varrho_\eps(Z_s-z)1_{s<\tau_R}\big)\dif z\\
&=\mE\Big((\tr(a^{\eps,R}_s\cdot\nabla^2_\v u^{\eps,R}_s)*\varrho_\eps)(Z_s)1_{s<\tau_R}\Big),\end{aligned}$$ by easy calculations, one sees that $$\begin{aligned}
&\mE\Big((\p_s u^{\eps,R}_{\eps,t}+\sL^{a,b}_su^{\eps,R}_{\eps,t})(Z_s)1_{s<\tau_R}\Big)\\
&=\int_{\mR^{2d}}\!\!\!\mE\Big((\dot X_s-\v)\cdot\nabla_x\varrho_\eps(Z_s-z)1_{s<\tau_R}\Big)u^{\eps,R}_s(z)\dif z\\
&\quad+\mE\Big((b_s\cdot\nabla_\v u^{\eps,R}_{\eps,t}+\lambda u^{\eps,R}_s-f_\eps)(Z_s)1_{s<\tau_R}\Big)\\
&\leq\|u^{\eps,R}_s\|_\infty\left(\lambda+\int_{\mR^{2d}}|\v|~|\nabla_x\varrho_\eps|(x,\v)\dif x\dif\v\right)\\
&\quad+\|\nabla_\v u^{\eps,R}_s\|_\infty\mE\big(|b_s|(Z_s)1_{s<\tau_R}\big)
-\mE\big( f_\eps(Z_s)1_{s<\tau_R}\big).\end{aligned}$$ Substituting this into , we obtain $$\begin{aligned}
&\mE\left(\int^{T\wedge\tau_R}_0f_\eps(Z_s)\dif s\right)\leq 
(\|\nabla_x\varrho\|_1+2+\lambda)\|u^{\eps,R}\|_{\mL^\infty(T)}\\
&\qquad+\|\nabla_\v u^{\eps,R}\|_{\mL^\infty(T)}\mE\left(\int^{T\wedge\tau_R}_0|b_s|(Z_s)\dif s\right).\end{aligned}$$ By , and with $b\equiv0$, there is a $C=C(d,p,K,T,\omega_a(\delta_0))>0$ such that for all $\eps\in(0,\delta_0)$ and $\lambda>1$, $$\begin{aligned}
&\mE\left(\int^{T\wedge\tau_R}_0f_{\eps,s}(Z_s)\dif s\right)\leq 
C(1+\lambda)T^{\frac{1}{2d+1}-\frac{1}{p}}\|f\|_{\mL^p(T)}\\
&\qquad+C\lambda^{\frac{1}{p}-\frac{1}{2(2d+1)}}\|f\|_{\mL^p(T)}\mE\left(\int^{T\wedge\tau_R}_0|b_s|(Z_s)\dif s\right),\end{aligned}$$ which implies, by letting $\eps\to 0$ and $\lambda$ large enough, that for any $\delta>0$, there is a $C_\delta>0$ 