Let $(X_k)_{k\ge 1}$ be a sequence of i.i.d. positive random variables of exponential distribution $\mathcal E(1)$, i.e.
$$\mathbb P[X_k>x]=e^{-x},\quad \forall x\ge 0.$$
Fix some $s\ge 0$. For every $n\ge 1$, let $A^n_0:=\emptyset$ and define by recurrence the random sets
$$A^n_k:=\left\{X_k\le s+ \frac 1 n \sum_{i=0}^{k-1}I_{A^n_i}\right\},\quad \forall 1\le k\le n,$$
where $I_{A}$, for $A\subset\mathbb R$, denotes the indicator function on $A$, i.e. $I_{A}(x)=1$ for $x\in A$ and $I_{A}(x)=0$ for $x\notin A$. Define further
$$S_n:=\frac 1 n \sum_{i=1}^{n}I_{A^n_i}.$$
Can we prove or disprove that $\lim_{n\to\infty}S_n$ exists almost surely?
PS : For every $n\ge 1$, one has $I_{A_{0}^{n}}=0=I_{A_{0}^{n+1}}$,$I_{A_{1}^{n}}=I_{A_{1}^{n+1}}$, and thus
$$I_{A_{2}^{n}} \ge I_{A_{2}^{n+1}}.$$
By definition, it follows that
$$A_{3}^{n}=\left\{X_3\le s+ \frac 1 n (I_{A^n_1}+I_{A^n_2})\right\} \supset \left\{X_3\le s+ \frac 1 {n+1} (I_{A_{1}^{n+1}}+I_{A_{2}^{n+1}})\right\}=A_{3}^{n+1},$$
and thus $I_{A_{3}^{n}} \ge I_{A_{3}^{n+1}}$. By recurrence, one obtains $I_{A_{i}^{n}} \ge I_{A_{i}^{n+1}}$ for all $i=0,1,\ldots, n$. Therefore,
$$S_{n+1} \le \frac {I_{A_{n+1}^{n+1}}} {n+1} + \frac {n} {n+1}S_n.$$
However, I'm unable show that $S$ is, or is approximately, a super martingale.