$\E (|M_t - M_S|^p \| F_S) 
\leq c_p b (t-S)^{p/2} \leq c_p b t^{p/2}$. Hence by Markov’s inequality, $\P (M_t - M_S \geq y \| \F_S) \leq c_p b t^{p/2} / y^p$. Therefore, $$\begin{aligned}
\E ((M_t^+)^r \one_{\tau > t} \| \F_S) & \leq & \E ((M_t^+)^r \one_{M_t > t} 
   \| \F_S) \\
& = & t^r \P (M_t \geq t \| \F_S) + \int_t^\infty r y^{r-1} 
   \P (M_t \geq y \| \F_S) \, dy \\
& \leq & t^r \P (M_t - M_S \geq t/4 \| \F_S) + \int_{t/4}^\infty 
   r y^{r-1} \P (M_t - M_S \geq y \| \F_S) \, dy \\
& \leq & c_p b 4^p t^{r - p/2} + \int_{t/4}^\infty 
   r y^{r-1} c_b p t^{p/2} y^{-p} \, dy \\
& \leq & c_p b (4^p + {r \over p - r} 4^{p-r}) t^{r-p/2} \\
& \leq & c_2 (b , p , r) t^{r - p/2} ,\end{aligned}$$ where $c_2 (b , p , r) := c_p b (4^p + 4^{p-r} r / (p-r))$. Combining with (\[eq:S\]) gives $$\label{eq:term1}
\E \left( (M_t^+)^r \one_G \right) \leq 2^{p/2} c_2 C' t^{r - p} .$$

We will bound the second term by decomposing according to the value of $T$. A preliminary 