Equivalent definitions of $\mathrm{BMO}_p$ martingales

Question

I'm working through exercise 3.16 in Revuz and Yor. Assume $Y$ is a continuous UI martingale and $1\leq p<\infty$. Then these are equivalent

1. $\exists C\ \forall T$ stopping time $E[|Y_\infty-Y_T|^p| \mathscr{F}_T] \leq C^p$

2. $\exists C\ \forall T$ stopping time $E[|Y_\infty - Y_T|^p] \leq C^p P[T<\infty]$

3. $\exists C\ \forall T$ stopping time $\exists \alpha_T \in L^p(\mathscr{F}_T)$ with $E[|Y_\infty - \alpha_T|^p|\mathscr{F}_T] < C^p$

I've shown 1 and 2 are equivalent. As for 3, the book doesn't mention, but I think we need $Y_0 = 0$, so that 1,2 imply $Y_\infty \in L^p$. (Otherwise take $Y$ to be a constant martingale with $Y_0 \notin L^1$, then it satisfies 1,2 but not 3.) So let's just assume $Y_0 = 0$. The implication 1 implies 3 is trivial by taking $\alpha_T := Y_T$ (since we know $Y_T \in L^p$). However, I'm stuck on the reverse implication. I want to say that $\alpha_T$ has to be close to $Y_T$, but I don't see how to force it.

Answer 1

I figured it out. $|Y_T - \alpha_T|^p = |E[Y_\infty - \alpha_T|\mathscr{F}_T]|^p \leq E[|Y_\infty-\alpha_T|^p | \mathscr{F}_T] < C^p$. Thus $E[|Y_\infty-Y_T|^p|\mathscr{F}_T] \lesssim E[|Y_\infty - \alpha_T|^p|\mathscr{F}_T] + E[|\alpha_T-Y_T|^p|\mathscr{F}_T] \leq C^p + C^p$.