Given iid random variables, expected number of times that a random variable is greater than all previous variables. I'm given $X_1, X_2, \dots$ iid with continuous cdfs, and let $A_1 = \Omega, A_n=\{\omega:X_n
(\omega) > X_i(\omega) \textrm{ for } i < n\}$.
I'd like to show that $\lim_{n\rightarrow \infty}{\frac{1}{\log(n)}\sum_{j=1}^{n}{1_{A_j}}} = 1$ almost surely.
This is a pretty interesting result, but I'm pretty unsure of how I might go about showing this.
 A: Let me briefly sketch a proof. The first step should be to note that the $A_n$'s are independent and  $\mathsf{P}(A_n)=1/n$ for all $n$ (do you see why?). From Kolmogorov's inequality, for any $\epsilon>0$
$$\mathsf{P}\left(\max_{2^{n^2}\le k\le 2^{(n+1)^2}} \left|\sum_{j=1}^k \left(1_{A_j}-1/j\right)\right|> \epsilon n^2\right)\lesssim \frac{1}{\epsilon^2 n^2}.$$
Now we use Borel-Cantelli lemma to argue that almost surely, for $n$ large enough
$$\left|\sum_{j=1}^n \left(1_{A_j}-1/j\right)\right|\le \epsilon\log n.$$
What we just proved is
$$\forall \epsilon>0, \mathsf{P}\left(\exists n_0(\epsilon), \forall n>n_0(\epsilon), \left|\frac{1}{\log n}\sum_{j=1}^n 1_{A_j}-\frac{1}{\log n}\sum_{j=1}^n 1/j\right|\le \epsilon\right)=1.$$
As the intersection of a countable number of almost sure events is almost sure,
$$\mathsf{P}\left(\forall \epsilon\in\mathbb{Q}_+^*, \exists n_0(\epsilon), \forall n>n_0(\epsilon), \left|\frac{1}{\log n}\sum_{j=1}^n 1_{A_j}-\frac{1}{\log n}\sum_{j=1}^n 1/j\right|\le \epsilon\right)=1.$$
Now notice that if we take $n_0(\epsilon)$ large enough, from the asymptotics of the harmonic sum, it holds that for all $n>n_0(\epsilon)$, $\left|\frac{1}{\log n}\sum_{j=1}^n 1/j-1\right|\le \epsilon$. Hence:
$$\mathsf{P}\left(\forall \epsilon\in\mathbb{Q}_+^*, \exists n_0(\epsilon), \forall n>n_0(\epsilon), \left|\frac{1}{\log n}\sum_{j=1}^n 1_{A_j}-1\right|\le 2\epsilon\right)=1.$$
That's exactly the definition of almost sure convergence of $\sum_{j=1}^n 1_{A_j}/\log n$ to $1$.
