$\lim_{n} P(U_n)=0,$ $\sum_{n}P(U_n \cap U_{n+q_n}^c)<+\infty \implies P(\limsup_nU_n)=0$ Let $(\Omega,\mathcal{F},P)$ be a probability space, $(U_n)_n$ a sequence of elements in $\mathcal{F},(q_n)_n$ a sequence in $\mathbb{N}^*$ such that $\lim_{n \to+\infty} P(U_n)=0$ and $\sum_{n\in \mathbb{N}}P(U_n \cap U_{n+q_n}^c)<+\infty.$
Prove that $P(\limsup_n U_n)=0.$
From $\sum_nP(U_n \cap U_{n+q_n}^c)<+\infty$ we can deduce that $P(\limsup_n (U_n \cap U_{n+q_n}))=0,$ also $P(\liminf_n U_n) \leq \liminf_nP(U_n)=0.$
Notice that when $q_n=1,$ then we can conclude the result using that $(\limsup_nU_n)-(\liminf_nU_n)=\limsup_n(U_{n}-U_{n+1}).$
For the general case do we have this result? That is $(\limsup_n U_n)-(\liminf_n U_n)$ in term of $\limsup_n (U_n \cap U_{n+q_n}^c)$ ?
 A: For each $n\in\mathbb{N}$, define a sequence of positive integers
$(m_{n,k})_{k}$ recursively by $m_{n,1}=n$, $m_{n,k+1}=m_{n,k}+q(m_{n,k})$
for $k=1,2,\ldots$. Note that $m_{n,k+1}>m_{n,k}$, so $P\left(\cap_{k=1}^{\infty}U_{m_{n,k}}\right)=0$
because $P(U_{n})\rightarrow0$. Let $U=\cup_{n=1}^{\infty}\cap_{k=1}^{\infty}U_{m_{n,k}}$,
then $P(U)=0$.
We assert that for each $N\in\mathbb{N}$, $\cup_{n=N}^{\infty}U_{n}\subseteq\left[\cup_{n=N}^{\infty}\left(U_{n}\setminus U_{n+q(n)}\right)\right]\cup U$.
Let $\omega\in\cup_{n=N}^{\infty}U_{n}$, then $\omega\in U_{n}$
for some $n\geq N$. If there exists $k$ such that $\omega\notin U_{m_{n,k}}$,
we let $k$ be the smallest such integer. Note that $k>1$ and $\omega\in U_{m_{n,k-1}}\setminus U_{m_{n,k}}\subseteq\mbox{RHS}$.
If $\omega\in U_{m_{n,k}}$ for all $k$, then $\omega\in U\subseteq\mbox{RHS}$.
It follows that
\begin{eqnarray*}
P\left(\cup_{n=N}^{\infty}U_{n}\right) & \leq & P\left(\cup_{n=N}^{\infty}\left(U_{n}\setminus U_{n+q(n)}\right)\right)+P(U)\\
 & \leq & \sum_{n=N}^{\infty}P(U_{n}\setminus U_{n+q(n)})\\
 & \rightarrow & 0
\end{eqnarray*}
as $N\rightarrow\infty.$
Hence, $P\left(\limsup_{n}U_{n}\right)=\lim_{N\rightarrow\infty}P\left(\cup_{n=N}^{\infty}U_{n}\right)=0$.
