Existence of Solution to SDE - Oksendal Theorem 5.2.1

Question

I am trying to understand a few lines in Oksendal's Theorem 5.2.1 (5th edition). This is the existence and uniqueness theorem for SDEs. The relevant lines are: Oksendal bottom of page 68 and Oksendal top of page 69.

I understand the application of the martingale inequality (the red line at the top of page 69) but I do not understand

1. The first inequality: $$P\left[\sup_{0\le t \le T} |Y_t^{(k+1)} - Y_t^{(k)}| > 2^{-k} \right] \le \\ P\left[\left(\int_0^T |b(s,Y_s^{(k)} - b(s,Y_s^{(k-1)}|ds\right)^2 > 2^{-2k-2} \right] + P\left [ \sup_{0\le t\le T} \left | \int_0^T (\sigma(s,Y_s^{k}) - \sigma(s,Y_s^{(k-1)}))dB_s\right| > 2^{-k-1}\right] $$ It is not clear to me how the probability is split in this way.
2. Then it looks to me that the author is applying the martingale inequality to the $\int (...) ds$ term on the next line as well as the $\int (...) dB_s$ term. The second of these makes sense to me, but the first intergal is not a margingale as I understand it so how is this step justified? The line in question is $$ \le 2^{2k+2} T \int_0^T E(|b(s,Y_s^{(k)}) - b(s,Y_s^{(k-1)})|^2)ds + 2^{2k+2}\int_0^TE[|\sigma(s,Y_s^{(k)}) - \sigma(s,Y_s^{(k-1)})|^2]ds $$ The second term is exactly the martingale inequality and the Ito isometry. The first looks like that as well, but I do not believe it can be applied.

Any help is greatly appreciated. Peter.

Answer 1

I understand now. I'll put the details here in case anyone comes across this problem. At the bottom of Page 68 the author establishes: $$ \sup_{0\le t \le T}|Y_t^{(k+1)}-Y_t^{k}| \le \int_0^T |b(s,Y_s^{k}) - b(s,Y_s^{(k-1)})|ds + \sup_{0\le t \le T} \left|\int_0^t(\sigma(s,Y_s^{(k)}) - \sigma(s,Y_s^{k-1}))dB_s\right | $$ and I'll write $A$ and $B$ for the two terms on the right. My first question was understanding why $$ P(A+B>\epsilon) \le P(A > \epsilon/2) + P(B > \epsilon/2) $$ is true. This follows because \begin{align} \{A + B > \epsilon\} & = \{A > \epsilon/2, B > \epsilon/2\} \cup \{A > \epsilon/2,\max(0,\epsilon-A) \le B \le \epsilon/2\} \cup \{B>\epsilon/2, \max(0,\epsilon-B)\le A \le \epsilon/2 \} \\ & \subset \{A > \epsilon/2\} \cup \{B > \epsilon/2\} \cup \{A > \epsilon/2 \} \cup \{B > \epsilon/2 \} \\ & = \{A>\epsilon/2\} \cup \{B>\epsilon/2\} \end{align} hence the probability inequality follows. I guess its quite trivial now I think about it.

The second question follows from using the Chebyshev inequality on the first integral.