# $L_1$ convergence for a product of indicator functions

Let $$X_1,X_2,\ldots$$ be a sequence of identically distributed random variables and let $$A\subset\mathbb{R}$$ be some set such that $$P(X_1\in A)<1$$. I have a product of indicator functions $$\lim_{N\rightarrow\infty}\prod_{n=1}^{N}1\{X_n\in A\}$$ and am interested in its limit behavior. Specifically, are there conditions on the dependence between the $$X_n$$, that are weaker than iid, and ensure convergence to zero in $$L_1$$ norm? That is $$\lim_{N\rightarrow\infty}E\left(\prod_{n=1}^{N}1\{X_n\in A\}\right) = \lim_{N\rightarrow\infty}P\left(\bigcap_{n=1}^{N}X_n\in A\right)=0.$$

I have already asked a question regarding this issue here. I know that ergodicity implies that the product converges to zero almost surely, but in the link Anthony Quas argues that this does not hold in $$L_1$$ norm.

Any hint or reference is appreciated, thank you in advance!

• I think perhaps you misunderstood Anthony's comment. When the product converges to zero almost surely, so that $P(\bigcap_{n=1}^\infty \{X_n \in A\}) = 0$, then it certainly converges in $L^1$ as well. This is just the "continuity from above" property of probability measures and follows immediately from countable additivity. (Or you could use the dominated convergence theorem.) I understood Anthony's comment to be saying that you can't always get exponentially fast convergence to zero. – Nate Eldredge Oct 19 '18 at 15:27
• You are right. I guess I stared at this problem for too long, I'll delete this question. Thank you for the reply! – Marc Oct 21 '18 at 11:57