Statement of theorem: If $X_n$ is a uniformly integrable martingale, then $\lim_n X_n$ exists a.s. and in $L^1$, and
$$X_n=E(\lim_{n \to \infty} X_n \mid \mathcal F_n) \quad \text{a.s.}$$
I can't think of an example of a martingale that is not uniformly integrable for which $E(\lim_n X_n|\mathcal F_n)\neq X_n$ and $\lim_n X_n$ exists. For instance, let $X_1=1$, and for $n\ge2$, $X_n=X_{n-1}+\epsilon_n$, where $\epsilon_n=n$ with probability 1/2, $\epsilon_n=-n$ with probability 1/2, and $\epsilon_n\perp\epsilon_m$ for $n\neq m$. This is a martingale that is not u.i., but $\lim_n X_n$ does not exist.
Any help greatly appreciated!
Best Answer
A very elementary example is the following: consider the probability space $(0,1)$ with Lebesgue measure. Let $X_n=nI_{(0,\frac 1 n)}$. A straightforward argument shows that $\{X_n\}$ is a martingale. It converges almost surely to $0$. Obviously, $X_n=E(0|X_1,X_2,...,X_n)$ is not true. Hint for proving martingale property: $\sigma \{X_1,X_2,...,X_n\}=\sigma \{(0,1),(0,\frac 1 2),... ,(0,\frac 1 n)\}=\sigma\{(0,\frac 1 n),[\frac 1 n, \frac 1 {n-1}),...,[\frac 1 2 ,1)\}$ and this last sigma algebra consists precisely of unions of the intervals $(0,\frac 1 n),[\frac 1 n, \frac 1 {n-1}),...,[\frac 1 2 ,1)$. Hence it is enough to show that $EX_{n+1} I_A=EX_nI_A$ for each one of these intervals. This is easy.
PS: this martingale is very useful in providing counter-examples. Unfortunately it is not found in texts.