Doob upcrossing inequality

Put $Y_n=a+(X_n-a{)}^{+}$. The function $x\mapsto a+(x-a{)}^{+}$ is convex and $1$-Lipschitz up to an additive constant, so $Y$ is integrable and is a submartingale by F2. Moreover $X_s\le a$ iff $Y_s=a$, and $X_t\ge b$ iff $Y_t\ge b$; therefore $U_N[a,b](Y)=U_N[a,b](X)$.

Define $H_j\in \{0,1\}$ before the increment $Y_j-Y_{j-1}$: it switches from $0$ to $1$ after the first observation at level $a$, remains $1$ until an observation at least $b$, then repeats. This rule depends only on $Y_0,\dots ,Y_{j-1}$, so $H$ is predictable. Let $K_j=1-H_j$, also nonnegative and predictable.

Each completed holding interval contributes at least $b-a$ to $(H\cdot Y{)}_N={\sum }_{j=1}^N{H}_j(Y_j-Y_{j-1})$. Any final incomplete holding starts at $Y_s=a$ and contributes $Y_N-a\ge 0$. Hence, pathwise, $(b-a)U_N[a,b](Y)\le (H\cdot Y{)}_N\mathrm{.}$ No maximizing tuple from F1 was selected.

Since $H+K=1$, $Y_N-Y_0=(H\cdot Y{)}_N+(K\cdot Y{)}_N\mathrm{.}$ F3 gives $\mathbb{E}(K\cdot Y{)}_N\ge 0$, so step 2.1 yields $(b-a)\mathbb{E}{U}_N[a,b](X)\le \mathbb{E}(Y_N-Y_0)=\mathbb{E}(X_N-a{)}^{+}-\mathbb{E}(X_0-a{)}^{+}\mathrm{.}$ The last subtracted term is nonnegative, proving the second inequality. AC is used exactly as stated in F4.