# Sticking martingales together

Let $$(\Omega, \mathcal F, P)$$ be a probability space with sigma algebras $$\mathcal F_{n, t}$$ for $$n \in \mathbb N$$, $$t \in [0, 1]$$, where for all $$n$$, $$\mathcal F_{n, t} \subset \mathcal F_{n, h}\$$ whenever $$h > t$$. Assume also for each $$n$$, the filtration $$\mathcal \{ \mathcal F_{n, t} \}$$ is right continuous in $$t$$. Finally, assume $$\mathcal F_{0, 0}\$$ contains all $$P$$-null sets.

Let $$\mathcal G_t := \sigma( \{ \mathcal F_{n, t} \}_{n \in \mathbb N} \ )$$ be the sigma algebra generated by $$\mathcal F_{n, t}$$ for fixed $$t$$.

Given a sequence $$M_n$$ of $$\{ \mathcal F_{n, t} \}$$-martingales such that $$\sup_{n \in \mathbb N} \int |M_n (t, \omega)| dt dP < \infty$$,

define for each $$k \in \mathbb Z^+$$, the process $$S_k$$ on $$(0, 1]$$ given by

$$S_k (t) := \frac{1}{2^k t} \sum_{h = 0}^{2^k - 1} 1_{[h/2^k, (h+1)/2^k]} (t) \sum_{j = 0}^{h-1} M_j (t).$$

As the $$S_k$$ are equibounded in $$L^1 (\Omega \times (0, 1])$$ , it follows from a result of Komlos that a subsequence of the $$S_k$$ Cesaro converges in $$L^1 (\Omega \times (0, 1])$$ to an “almost” $$L_1$$ process, say $$X$$ - in the sense that $$X_t$$ is in $$L^1 (\Omega)$$ for (Lebesgue) almost every $$t$$. In what follows we restrict the process $$X$$, and the filtration $$\mathcal G_t$$ to those times $$t$$.

Question: Is $$X$$ a $$\mathcal G_t$$-martingale?

• Your $S_k(t)$ does not depend on $t$. – Iosif Pinelis Apr 7 at 13:39
• So, it’s left implicit in the indicator function, but it’s not obvious as is so I’ll change it. – Nate River Apr 7 at 15:09
• If $M_j = M$ is independent of $j$, then $X(t) = t M(1)$ so no. It's not even adapted to ${\mathcal G}$... – Martin Hairer Apr 7 at 15:17
• Sorry, why is $M_j$ independent of $j$? – Nate River Apr 7 at 23:14
• Oh I understand, I guess one would also need that the $\mathcal F_{j, t}$ are independent of $\mathcal F_{h, t}$ for $h > j$? – Nate River Apr 7 at 23:29