Quantitative Borel-Cantelli

Question

I’ve been looking for a quantitative notion of the Borel-Cantelli lemma along the following lines:

Let $(X,\Omega,p)$ be a probability space, and let $(A_n)_n\subseteq \Omega$ be a sequence of measurable sets s.t $p(A_n)\geq 1-\delta$ for all $n$ (for some small $\delta>0$).

Then of course, the Borel-Cantelli lemma tells us that $p(\limsup A_n)>0$, and in fact a trivial observation from the proof of the lemma shows $p(\limsup A_n)\geq 1-\delta$.

However, I am interested in something further, relating to the lower density of sequences of indices to which elements belong: is there $\alpha>0$ s.t $$p\left(\left\{x\in X: \exists n_k\uparrow \infty\text{ with }\underline{d}((n_k)_k)\geq 1-\alpha\text{ s.t }x\in A_{n_k},\forall k\geq 0\right\}\right)\geq 1-\alpha?$$ where $\underline{d}(\cdot)$ denotes the lower-density of a sequence. In particular, can $\alpha>0$ be made small when $\delta>0$ is small?

Intuitively it seems to make sense, but so far no success in proving, and also searching online yielded nothing. Any interesting counter examples are also welcome.

Thanks.

Answer 1

Imagine that for all $k=1,2,\ldots$ all events $A_n$ for $n\in [k!,(k+1)!-1)$ are the same event $C_k$. Then if $x$ belongs to all $A_i$ along a subsequence of positive density yields that it belongs to all but finitely many $C_k$. It may well appear that there is no such $x$.