# Augmented Filtrations and Martingales in the Martingale Representation Theorem

Note: This question is related to the following question about complete markets in continuous time. In the linked question, the answer mentions that complete markets in this setting is a result of the Martingale Representation theorem.

I'm trying to understand the statement of the theorem as given in its Wikipedia article:

Let $B_t$ be a Brownian motion on a standard filtered probability space $(\Omega, \mathcal F, \mathcal F_t, P)$, and let $\mathcal G_t$ be the augmentation of the filtration generated by $B$. If $X$ is a square integrable random variable measurable with respect to $\mathcal G_\infty$ then there exists a predictable process $C$ which is adapted with respect to $\mathcal G_t$, such that $$X = E[X] + \int_0^\infty C_s dB_s.$$ Consequently, $$E[X \mid G_t] = E[X] + \int_0^t C_s dB_s.$$

In this definition, where is the relationship to Martingales? I see that it is assumed that $X$ is square integrable with respect to $\mathcal G_\infty$ and I assume that it has something to do with the fact that $\mathcal G_t$ is an "augmentation of the filtration generated by $b$." Also, what is an "augmentation of a filtration?"

The augmentation of the filtration generated by $B$ is typically just called the augmented filtration and is defined using the below construction. It is more commonly referred to simply as the standard Brownian filtration.
• The collection $\mathcal{C}$ of all sets of probability $0$ in the sigma-field $\sigma\{B_{s} \: : \: s \leq t\}$
• The collection of null-sets, $\mathcal{N}$, of all $A$ such that $A \subset B$ for $B \in \mathcal{C}$. It is assumed $\mathbb{P}(A)=0$ for all such $A \in \mathcal{N}$.
Then the augmented filtration for Brownian motion to be the filtration given by $\{\mathcal{F}_{t}\}$, where for each $t$ we take $\mathcal{F}_{t}$ to be the smallest sigma-field containing $\sigma\{B_{s} \: : \: s \leq t\}$ and $\mathcal{N}$.
The relationship to martingales is simply that $X_{t}:=\mathbb{E}(X\:|\:\mathcal{G}_{t})$ is a martingale.
• Note that $X_t\equiv E[X|\mathcal{G}_t]$ being a martingale is equivalent to the law of iterated expectations: $E[X_{t+s}|\mathcal{G}_t]=E[E[X|\mathcal{G}_{t+s}]|\mathcal{G}_t]=E[X|\mathcal{G}_t]=X_t$ – nominally rigid Jan 1 '15 at 2:20