# Does martingale convergence hold for arbitrary time?

Let $\{\mathcal B_i:i\in I\}$ be a family of $\sigma$-algebras (over the same set $\Omega$) which are totally ordered by inclusion, in the sense that for any $i,j\in I$ either $\mathcal B_i\subset\mathcal B_j$ or $\mathcal B_j\subset\mathcal B_i$. I think this can be called a filtration with time $I$.

Let $\mathcal B=\sigma\left(\bigcup_{i\in I}\mathcal B_i\right)$ be the $\sigma$-algebra generated by all the $\mathcal B_i$, let $\mu$ be a probability on $\mathcal B$ and let $X\in L^2(\mathcal B,\mu)$ be a bounded random variable.

In what senses (if any) does the family $\{\mathbb{E}[X\mid\mathcal B_i];i\in I\}$ approximate $X$? More precisely:

Is it true that for every $\epsilon>0$ there is some $i\in I$ such that $\|X-\mathbb{E}[X\mid\mathcal B_i]\|_{L^2}<\epsilon$? ($L^2$ convergence)

Is it true that there exists a set $A\in\mathcal B$ with $\mu(A)=1$ and for every $\epsilon>0$ there is some $i\in I$ such that for each $\omega\in A$ we have $|X(\omega)-\mathbb{E}[X\mid\mathcal B_i](\omega)|<\epsilon$? (almost sure convergence)

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I think you'll want to look at Williams' Probability with Martingales: books.google.com/books?id=e9saZ0YSi-AC –  Steve Huntsman Nov 21 '12 at 17:49
That book, although quite nice for learning, doesn't seem to deal with uncountable time, which is what I was looking for. –  Joel Moreira Nov 29 '12 at 2:32

The sigma algebra generated by $X$ is countably generated. Thus $X$ is measurable for $$\sigma\left(\bigcup_{k=1}^\infty \mathcal{B}_{i_k}\right)$$ for some increasing sequence $i_1 \le i_2 \le \dots$ in $I$. So some of your desired results follow from the usual martingale convergence theory.

The almost sure convergence will fail, in general, for uncountable $I$. A reasonable replacement is called "essential convergence". That is what Sucheston and I used for such things in our book.

[plug: G. Edgar, L. Sucheston, Stopping Times and Directed Processes (Cambridge University Press 1992)]

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