Proving the Kochen-Stone lemma using the Paley-Zygmund inequality I am trying to understand a proof to a lemma by Kochen and Stone which appears here, using the Paley-Zygmund inequality.
I'll repeat the proof in a detailed manner, and explain what bothers me about it.
Lemma (Kochen-Stone). $\ $ Let $A_n$ be a sequence of events with $\sum\mathbb{P}(A_n)=\infty$ and
\begin{equation*}
  \liminf_{k\to\infty}\frac{\sum_{1\le m,n \le k}\mathbb{P}(A_m\cap
  A_n)}{\left(\sum_{n=1}^k\mathbb{P}(A_n)\right)^2}<\infty
\end{equation*}
then, there is a positive probability that $A_n$ occur infinitely often.
Proof (partial). $\ $ Fix
$\ell<k$. Let $X=\sum_{n=\ell}^{k}1_{A_n}$; it follows that
\begin{equation*}
  \mathbb{E}(X)=\sum_{n=\ell}^{k}\mathbb(A_n)
\end{equation*}
and
\begin{equation*}
  \mathbb{E}(X^2)=\sum_{\ell\le m,n \le k}\mathbb{P}(A_n\cap A_m).
\end{equation*}
Using Paley-Zygmund inequality for $\theta=0$ (it's not mentioned in the wikipedia page, but the inequality holds for $\theta=0$ as well), we obtain
\begin{eqnarray*}
  \mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right)
  &=& \mathbb{P}(X>0)\\
  &\ge& \frac{\left(\sum_{n=\ell}^{k}\mathbb{P}(A_n)\right)^2}
           {\sum_{\ell\le m,n \le k}\mathbb{P}(A_n\cap A_m)}\\
  &\ge& \frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)
            -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2}
           {\sum_{1\le m,n \le k}\mathbb(A_n\cap A_m)
            -\sum_{1\le m,n < \ell}\mathbb(A_n\cap A_m)}
\end{eqnarray*}
Now, it holds that $\mathbb{P}(A_n\text{ occurs i.o.}) = \lim_{\ell\to\infty}\lim_{k\to\infty}\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right)$; however, I can't see how I can bound that probability away from 0. Am I missing some minor detail here?
 A: Here's a rough answer to your question. We have
$$
\frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)
            -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2}
           {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)
            -\sum_{1\le m,n < \ell}\mathbb{P}\mathbb(A_n\cap A_m)}\geq
\frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)
            -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2}
           {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)}
$$
Fix $l \in \mathbb{N}_1$. Since $\lim_{k \rightarrow \infty}\sum_{n = 1}^k \mathbb{P}(A_n) = \infty$, by assumption, if $k$ is sufficiently large,
$$
\left(\sum_{n=1}^{k}\mathbb{P}(A_n)
            -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2 \approx \left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2
$$
So if $k$ is sufficiently large,
$$
\frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)
            -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2}
           {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)} \approx
\frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2}
           {\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)} =
\frac{1}{\frac{\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)}{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2}}
$$
Suppose
$$
\liminf_{k\to\infty}\frac{\sum_{1\le m,n \le k}\mathbb{P}(A_m\cap
  A_n)}{\left(\sum_{n=1}^k\mathbb{P}(A_n)\right)^2} = c < \infty
$$
This means that no matter how large $k$ is, there is always some $k' \geq k$, such that
$$
\frac{\sum_{1\le m,n \le k'}\mathbb{P}(A_m\cap
  A_n)}{\left(\sum_{n=1}^{k'}\mathbb{P}(A_n)\right)^2} \approx c 
$$
So for infinitely many $k$'s
$$
\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right) \geq \frac{1}{\frac{\sum_{1\le m,n \le k}\mathbb{P}\mathbb(A_n\cap A_m)}{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)\right)^2}} \approx \frac{1}{c}
$$
Since $\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right)$ is decreasing in $k$, we have, approximately,
$$
\lim_{k \rightarrow \infty}\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right) \geq \frac{1}{c}
$$
Since $l$ was arbitrary,
$$
\lim_{l \rightarrow \infty}\lim_{k \rightarrow \infty}\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right) \geq \frac{1}{c} > 0
$$
A: Here is a proof from Rick Durrett's Probability.
Let $X_n = \sum_{k\le n} 1_{A_k}$, $Y_n = \frac{X_n}{E(X_n)}$, the Kochen-Stone lemma means $\limsup_{n\to\infty} \frac{1}{E(Y_n^2)}<\infty$.
Here is an inequality we will use: $P(Y_n > a) \ge \frac{(1-a)^2}{E(Y_n^2)}$, here $a$ is a constant. This comes from the Paley-Zygmund inequality, we can easily prove it with Cauchy-Schwarz inequality. By Borel-Cantelli lemma, we have $P(A_k, i.o.) = 1$. Since $P(\cup_{m\ge n} Y_m) \ge \max_{m\ge n} P(Y_m)$, it follows that $P(\limsup Y_n)\ge \limsup P(Y_n)$. Thus
$1 = P(A_n,i.o.) \ge P(\limsup Y_n > a) \ge \limsup P(Y_n > a) \ge (1-a)^2 \frac{1}{E(Y_n^2)}$,
Which means $\limsup_{n\to\infty} \frac{1}{E(Y_n^2)} < \infty$.
