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Start with cash one, switch to $X/a$ when a nonnegative supermartingale first reaches at most $a$, and switch to cash $q=b/a$ when it next reaches at least $b$. Repeat with coefficients multiplied by $q$ after every completed upcrossing. At each switch the new value is at most the old one, so pasting supermartingales with downward jumps gives a supermartingale when stopped after any fixed number of switches. Its initial value is $\min (X_0/a,1)$, and its value at the $k$th completed crossing is exactly $q^k$. Applying its expected value bound proves the Dubins upcrossing inequality.

Interpret the comparison at $N$ on $\{N<\infty \}$; values at an infinite stopping time are not required. The stopping time events make $Y_n$ adapted, and $|Y_n|\le |X_n^1|+|X_n^2|$ gives integrability. On $\{N\le n\}$ the next value is $X_{n+1}^2$. On $\{N>n\}$ it is $X_{n+1}^1$ unless $N=n+1$, and on that latter event switching only decreases the value. Consequently the pointwise inequalityholds even though $\{N=n+1\}$ need not be known at time $n$. Both indicators displayed here are ${\mathcal{F}}_n$-measurable. Taking conditional expectations and using the two supermartingale inequalities therefore givesThus $\boxed{Y\text{ is a supermartingale}.}$ This proves pasting supermartingales with downward jumps without incorrectly treating the future switching event as predictable.