I am studying the notes "Advanced stochastic analysis" by Martin Hairer for a seminar. In the sixth section, Hairer proves Norris' lemma (Lemma 6.6) giving an explicit exponent in the proof, using the concept of almost true events.
I do not understand the last step in the picture below
I do understand the first implication for $B$, but not the second part: The exponential martingale inequality gives a bound on the martingale term. How does this impact the bounded variation part? Can the exponential martingale inequality be extended to (continuous) semimartingales?
In the following the norm $\|\cdot\|_\infty$ refers to the sup-norm taken over the time interval $[0,1]$. Remember the exponential martingale inequality: $$ P(\|M\|_\infty\geq x,\langle M\rangle_1\leq y)\leq 2e^{-x^2/(2y)}. $$ for a continuous local martingale $M$ starting at $0$. Apply this with $M_t=\int_0^t B_s\,dW_s$ to get $$ P(\|M\|_\infty\geq\varepsilon^{\frac{1}{36}},\|B\|_\infty\leq\varepsilon^{\frac{1}{17}})\leq 2\exp\left(-\frac{1}{2\varepsilon^{\frac{1}{17}-\frac{1}{18}}}\right), $$ whence $\{\|B\|_\infty\leq \varepsilon^{\frac{1}{17}}\}\Rightarrow_\varepsilon\{\|M\|_\infty\leq\varepsilon^{\frac{1}{36}}\}$ (note the typo in the notes). On the other hand on the event $\{\|Z\|_\infty<\varepsilon,\|M\|_\infty\leq\varepsilon^{\frac{1}{36}}\}$, $$ \varepsilon>\left\|\int_0^\cdot A_s\,ds\right\|_\infty-\varepsilon^{\frac{1}{36}}, $$ whence $\left\|\int_0^\cdot A_s\,ds\right\|_\infty<2\varepsilon^{\frac{1}{36}}$. Finally use \begin{align} P\left(\left\|\int_0^\cdot A_s\,ds\right\|_\infty\geq 2\varepsilon^{\frac{1}{36}},\|Z\|_\infty<\varepsilon\right)\leq& P\left(\left\|\int_0^\cdot A_s\,ds\right\|_\infty\geq 2\varepsilon^{\frac{1}{36}},\|Z\|_\infty<\varepsilon,\|M\|_\infty\leq\varepsilon^{\frac{1}{36}}\right)\\ &+P(\|Z\|_\infty<\varepsilon,\|M\|_\infty>\varepsilon^{\frac{1}{36}})=0+O(\varepsilon^p) \end{align} for all $p>0$. Hence, $$ \{\|Z\|_\infty<\varepsilon\}\Rightarrow_\varepsilon\left\{\left\|\int_0^\cdot A_s\,ds\right\|_\infty<2\varepsilon^{\frac{1}{36}}\right\}, $$ as required.