I read the following example in the book Counterexamples in Probability and Real Analysis by Gary L. Wise and Eric B. Hall:
Does anyone know simpler examples? I do have one! I would be glad to present it unless somebody gives an example more interesting than mine. (So this is a question to be eventually self answered.)
On
The simplest example I've found so far:
Let $f_1,f_2,...,f_n$ be random variables defined on $([0,1], \mathscr L,leb)$ such that
$f_n(x)=\begin {cases} n^2, \ \ if \ \ x\epsilon[0,\frac{1}{n})\\ 0, \ \ \ \ otherwise\\ \end {cases}$.
Then $\mathbb E[f_n|f_{n-1},f_{n-2} ...]=f_{n-1}$ and $\int_{0}^{1}f_n\ d\mathbb leb=n.$
Note that we don't really need $(\mathscr L,\mathbb {leb})$. However, explaining the simpler case would make the example more complicated.
$$EDITED:$$
My counter example is wrong. Here is the martingale I was having in mind:
$g_n(x)=\begin {cases} n, \ \ if \ \ x\epsilon[0,\frac{1}{n})\\ 0, \ \ \ \ otherwise\\ \end {cases}$.
In this case $\mathbb E[g_n|g_{n-1},g_{n-2} ...]=g_{n-1}$ is true. We did not have a martingale before. However, unfortunately $\int_{0}^{1}g_n\ d\mathbb leb=1 $ now. That is, my example does not work as a substitute for the more complicated examples. I apologize. What is true in the case of the $g-$martingale is that $\int_{0}^{1}|f_n|^2\ d\mathbb leb=n$. But this has nothing to do with my claim. I goofed this whole thing.
What about this: Let $N$ be geometric with success probability $1/2$ (so $Pr[N=k]=1/2^k$ for $k \in \{1, 2, 3, \ldots\}$). Take $X_1 =0$. For $1 \leq n < N$, independently define:
$$ X_{n+1} = \left\{ \begin{array}{ll} X_{n} + (|X_{n}|+8^{n+1}) &\mbox{ with prob $1/2$} \\ X_{n} - (|X_{n}|+8^{n+1}) & \mbox{ with prob $1/2$} \end{array} \right. $$
For $n> N$ define $X_n = X_N$. Then $|X_n|\geq 8^n$ whenever $1<n \leq N$ and so for $n>1$ we have $E[|X_n|] \geq 8^nPr[n\leq N] = \frac{8^n}{2^{n-1}}\rightarrow\infty$.