Variant of the Strong Law of Large Numbers Let $X_1,X_2,\ldots$ be a i.i.d. sequence of random variables with uniform distribution on $[0,1]$, with $X_n: \Omega \to \mathbf{R}$ for each $n$. 

Question. Is it true that
  $$
\mathrm{Pr}\left(\left\{\omega \in \Omega: \lim_{n\to \infty}\frac{\sum_{1\le i\le j \le n}{\bf{1}}_{(-1/n,1/n)}{(X_i(\omega)-X_j(\omega))}}{n}=2\right\}\right)=1\,\,\,?
$$

Here ${\bf{1}}_A(z)$ is the characteristic function of $A$, that is, it is $1$ if $z \in A$ and $0$ otherwise.
 A: Yes. Here is a proof sketch: 
1) The statement is equivalent to the statement $\frac{Z_n}{n}\rightarrow 2$ with prob 1, where
$$ Z_n = \sum_{i=1}^n \sum_{j \in \{1, ..., n\}, j \neq i} 1_{\{|X_i-X_j|\leq 1/n\}}$$
2) We get $E[\frac{Z_n}{n}]\rightarrow 2$ and $Var(\frac{Z_n}{n}) = O(1/n)$.  Thus, $Var(\frac{Z_{n^2}}{n^2})=O(1/n^2)$ and hence
$$ \frac{Z_{n^2}}{n^2} \rightarrow 2 \quad \mbox{with prob 1} $$
3) For indices $k$ such that $n^2\leq k <(n+1)^2$, unfortunately we cannot "quite" say that $\frac{Z_{n^2}}{(n+1)^2} \leq \frac{Z_k}{k} \leq \frac{Z_{(n+1)^2}}{n^2}$ because the indicators $1_{\{|X_i-X_j|\leq 1/n\}}$ now have dependence on $n$.  So we have to go to step 4: 

Here is a fix: 
4)  Let $\{a_n\}_{n=1}^{\infty}$ be a deterministic sequence of (possibly negative) integers that satisfies  $a_n = O(\sqrt{n})$ and $n + a_n \in \{1, 2, 3, \ldots\}$ for all positive integers $n$.  Define: 
$$R_n(\{a_n\}) := \sum_{i=1}^n \sum_{j \in \{1, …, n\}, j \neq i} 1_{\{|X_i-X_j|\leq 1/(n+a_n)\}}$$
Now we can equally say $E[\frac{R_n(\{a_n\})}{n}]\rightarrow 2$ and $Var(\frac{R_n(\{a_n\})}{n})=O(1/n)$, and so $R_{n^2}(\{a_n\})/n^2\rightarrow 2$ with prob 1. Furthermore for any $k$ such that $n^2\leq k <(n+1)^2$ we get
$$ \frac{R_{n^2}(\{a_n\})}{(n+1)^2}\leq\frac{Z_k}{k} \leq \frac{R_{(n+1)^2}(\{b_n\})}{n^2}$$
for some sequences $\{a_n\}$ and $\{b_n\}$. 
A: One can use also tools related to $U$-statistics. 


*

*First step: for the indices $i=j$, the indicator is always one hence it suffices to prove the 
$$\frac 1n\sum_{1\leqslant i<j\leqslant n}\mathbf{1}_{(-1/n,1n)}\left(X_i-X_j\right)\to 1 \mbox{ a.s.}.$$

*Define $h_n(x,y):= \mathbf{1}_{(-1/n,1n)}\left(x-y\right)$ and $\mathcal F_i$ the $\sigma$-algebra generated by the random variables $X_1,\dots,X_i$. Then 
$$\mathbb E\left[h_n\left(X_i,X_j\right)\mid\mathcal F_{j-1}\right]=g_n\left(X_i\right),
$$
where $g_n(x)=2/n$ for $1/n\leqslant x\leqslant 1-1/n$, $g_n(1)=g_n(0)=1/n$ and $g_n$ is piecewise affine.

*We are thus reduces to show that 
$$
\frac 1n\sum_{1\leqslant i<j\leqslant n} \left(h_n\left(X_i,X_j\right)-\mathbb E\left[h_n\left(X_i,X_j\right)\mid\mathcal F_{j-1}\right]\right)\to 0 \mbox{ a.s.}
$$
$$
\frac 1n\sum_{i=1}^{n-1}\left(n-i\right)g_n\left(X_i\right)\to 1 \mbox{ a.s.}
$$

*The first part can be treated by looking at the fourth order moments and the fact that for a martingale differences sequence $\left(D_i\right)_{i\geqslant 1}$, 
$$\left\lVert \sum_{i=1}^nD_i\right\rVert_4^2\leqslant 3\sum_{i=1}^n\left\lVert D_i\right\rVert_4^2.
$$

*For the other part, control $g_n(X_i)-2/n$.

