Let $Y = (Y_n)_{n\in\mathbb N}$ such that $\mathbb P(\{Y_0=2\})= 1/2=\mathbb P(\{Y_0=0\})$.
We consider the $Y$ adapted process $X = (X_n)_{n\in\mathbb N}$ given by $X_n=\prod_{i=1}^nY_i$ Show that there is $X\infty\in L^1(P)$ such that $X_n\to X_\infty$ almost everywhere, but $X$ does not converge to $X_\infty$ in L1(P) as $n\to\infty$.
I think I have to use the sub martingale theorem here. Am I right? But how can I use $\mathbb P(\{Y_0=2\})= 1/2=\mathbb P(\{Y_0=0\})$ here? Could anyone please help me? Thanks
Indeed, $X_n$ is be a martingale with respect to the filtration $(\mathcal F_n)$, where $\mathcal F_n$ is generated by $Y_1,\dots,Y_n$ provided $(Y_n)$ is independent (this is not assumed but it is crucial). It turns out that by independence, $$ \mathbb P(X_n\neq 0)=\mathbb P\left(\bigcap_{i=1}^n \{Y_i\neq 0\}\right)=\prod_{i=1}^n\mathbb P\left(Y_i\neq 0\right)=2^{-n} $$ hence $X_n\to 0=X_\infty$ almost surely. But $\mathbb E[X_n]=1$ hence the convergence in $\mathbb L^1$ cannot happen.