I'm trying to solve the question (Ex. 4.2.10) from Durret's textbook.
For every positive supermartingale $X_n$, $n\geq0$, the number of upcrossing $U$ of $[a,b]$ satisfies $$P(U\geq k)\leq(\frac{a}{b})^kE\min(X_0/a,1).$$
To prove this, we let $N_0=-1$ and for $j \geq 1$ let
$$N_{2j-1}=\inf\{m>N_{2j-2}:X_m\leq a\},\\N_{2j}=\inf\{m>N_{2j-1}:X_m\geq b\}.$$
Let $Y_n=1$ for $0\leq n \lt N_1$ and for $j\geq 1$
$$Y_n= \begin{cases}(b/a)^{j-1}(X_n/a) & \mbox{for}&N_{2j-1}\leq n \lt N_{2j}\\ (b/a)^{j}&\mbox{for}&N_{2j}\leq n \lt N_{2j+1}\end{cases}$$
(i) Use the switching principle in the previous exercise and induction to show that $Z_n^j=Y_{n\bigwedge N_j}$is a supermartingale. (ii) Use $EY_{n\bigwedge N_{2k}} \leq E Y_{0}$ and let $n\rightarrow\infty$ to get Dubins' inequality.
The following is my try following the hints given by Durret.
(i) The first one is simple, since
$$Z_n^1 =Y_{n\bigwedge N_1}=1*\mathbf{1}_{(n<N_1)}+\frac{X_{N_1}}{a}*\mathbf{1}_{(n\geq N_1)} \ \& \ at \ n=N_1, \ 1\geq \frac{X_n}{a}\\
Z_n^2 = [1*\mathbf{1}_{(n<N_1)}+\frac{X_n}{a}*\mathbf{1}_{(n\geq N_1)}]*\mathbf{1}_{(n<N_2)}+\frac{b}{a}*\mathbf{1}_{(n\geq N_2)} \ \& \ at \ n=N_2,\ \frac{X_{N_2}}{a}\geq\frac{b}{a}$$
and so on, we can prove (i) easily.
(ii) I am stuck at the second one. I cannot figure out the final step.
Since $Y_{n\wedge N_{2k}}$ is a martingale, $EY_{n\wedge N_{2k}}\leq EY_0 = 1$. Then $$ 1\geq\lim_{n\to\infty}EY_{n\wedge N_{2k}}\\=E\lim_{n\to\infty}Y_{n\wedge N_{2k}}\\=E\left[\lim_{n\to\infty}Y_n\mathbf{1}_{(U<k)}+(\frac{b}{a})^k\mathbf{1}_{U\geq k}\right],$$where the first equality is by bounded convergence theorem and the second inequality is by usual expansion. Then we can get $$P(U\geq k)\left(\frac{b}{a}\right)^k\leq1-E\lim_{n\to\infty}Y_n\mathbf{1}_{(U>k)}.$$But then what should I do to get the final answer? I will be extremely grateful if you can provide any suggestion.
Hint: show that $E(Y_{n\wedge N_{2k}}) \le E(\min(\frac{X_0}a,1))$ instead of $E(Y_{n\wedge N_{2k}}) \le 1$