I'm trying to work through a proof in Durrett's textbook of a martingale convergence theorem via an embedding of the martingale in Brownian motion, and am stuck verifying a detail as usual.
I'm looking at Theorem 8.8.2 in the above link (I'll leave out re-writing everything so it doesn't get cluttered). I can't seem to figure out the step where he conditions on the $\sigma$-field at time $T_{m-1}$ (top of p. 431):
$$ \sum_{m=1}^{\infty} E(t_m; M \geq m) = \sum_{m=1}^{\infty} E[E(t_M \mid \mathcal{B}_{T_{m-1}}); M \geq m] $$
I understand that $\{M \geq m\} \in \mathcal{F}_{m-1}$ so by conditioning each term of the sum on that set you are left with something involving the last term (Durrett's claim: $E(t_M \mid \mathcal{B}_{T_{m-1}})$), but my question is: why is it $t_M$ and not $T_M$?
In the LHS of the above, it looks like if you expand out the sum and cancel terms, then one ought to be left with $T_M-T_0$, where $T_0 = 0$ so you're just left with $T_M$ rather than $t_M = T_M - T_{M-1}$. I've gotta be missing something obvious here...