Yet another question from the depths of Durrett. Again in the proof of Theorem 8.8.3, the author notes that "by the orthogonality of martingale increments," $$ E \left( \sum_{m=1}^{[nt]} t_{n,m} - E(t_{n,m} \mid \mathcal{F}_{n,m-1}) \right)^2 = E \sum_{m=1}^{[nt]} \big[ t_{n,m}-E(t_{n,m}\mid\mathcal{F}_{n,m-1}) \big]^2 $$ I understand the orthogonality concept, but... where is the martingale? At first I tried making a martingale out of $Y_m = t_{n,m} - E(t_{n,m} \mid \mathcal{F}_{n,m-1})$, but it doesn't seem to work out.
I suspect this has to do with his observation that $$E(t_{n,m} \mid \mathcal{F}_{n,m-1}) = E(X_{n,m}^2 \mid \mathcal{F}_{n,m-1})$$ since $X_{n,m}$ is a martingale difference array, but it's not clear how this alone creates a martingale in terms of the $t_{n,m}$.
Define $Y_{n,m}=\sum_{k=1}^m\left[t_{n,k}-\mathbb{E}(t_{n,k}\mid\mathcal{F}_{n,k-1})\right]$. $Y_{n,m}$ is a MTG because
$$\mathbb{E}[Y_{n,m}\mid\mathcal{F}_{n,m-1}]=\sum_{k=1}^{m-1}\left[t_{n,k}-\mathbb{E}(t_{n,k}\mid\mathcal{F}_{n,k-1})\right]=Y_{n,m-1}$$
and
$$t_{n,m}-\mathbb{E}(t_{n,m}\mid\mathcal{F}_{n,m-1})=Y_{n,m}-Y_{n,m-1}$$
By Theorem 5.4.6 (Durrett) for $l\ne m$
$$\mathbb{E}[Y_{n,m}-Y_{n,m-1}][Y_{n,l}-Y_{n,l-1}]=0$$
so that
$$\mathbb{E}\left(\sum_{m=1}^{[nt]} t_{n,m}-E(t_{n,m} \mid \mathcal{F}_{n,m-1}) \right)^2=\mathbb{E}\left(\sum_{m=1}^{[nt]}[Y_{n,m}-Y_{n,m-1}] \right)^2$$ $$\mathbb{E}\sum_{m=1}^{[nt]}\left[Y_{n,m}-Y_{n,m-1} \right]^2+\sum_{m=1}^{[nt]}\sum_{l\ne m}\mathbb{E}\left[Y_{n,m}-Y_{n,m-1} \right]\left[Y_{n,l}-Y_{n,l-1} \right]$$ $$=\mathbb{E}\sum_{m=1}^{[nt]} \big[ t_{n,m}-E(t_{n,m}\mid\mathcal{F}_{n,m-1}) \big]^2$$