limit superior and telescoping sequence If we have ($x^k$ is a real-valued sequence and $\mu$ is a probability measure)
$\mu(\lim \sup_k \{\sup_{t\geq0}|x^{k+1}(t)-x^k(t)|\geq\frac{1}{2^k}\})=0$
then how can we prove that $x^k$ converges (uniformly w.r.t $t$) to some $x$ ?
From the Borel-Cantelli lemma, we know that 
$\sup_{t\geq0} |x^{k+1}-x^k| \to 0\thinspace \ \ \text{a.s.}$
but why do we have (uniform) convergence of $x^k$ ?
 A: From the Borel-Cantelli lemma, we know not only that $\sup_{t\ge 0}|x^{k+1}-x^k|\to0$, a.s., but also that
$$\mu\Bigl(\Bigl\{\sup_{t\ge0}|x^{k+1}-x^k|\ge 2^{-k}\Bigr\} \;\text{i.o.}\Bigr)=0,\quad \text{and}
\quad \sum_{k=1}^\infty\sup_{t\ge 0}|x^{k+1}-x^k|<\infty, \quad \text{a.s.}$$
From the convergence of last series, we could conclude the (a.s.) uniform convergence of $x^k$.
The proof of last conclusion: Firstly, letting $A_k=\{\omega:\sup_{t\ge0}|x^{k+1}(t,\omega)-x^k(t,\omega)|\ge 2^{-k}\}$,
$$\mu(A_k\;\text{i.o.})=\mu(\limsup_{k\to\infty}A_k)=0 \qquad\Longleftrightarrow\qquad \mu(\liminf_{k\to\infty}A_k^c)=1.$$
Now assume that $\omega\in\liminf_{k\to\infty}A_k^c$, then there exist a $K(\omega)$ such that 
$$\omega\in\bigcap_{k\ge K(\omega)}A_k^c,\quad\text{and}\quad
\sup_{t\ge 0}|x^{k+1}(t,\omega)-x^{k}(t,\omega)|<2^{-k},\quad\forall k\ge K(\omega).$$
Also for this $\omega$, $\sum_{k=1}^\infty\sup_{t\ge 0}|x^{k+1}(t,\omega)-x^k(t,\omega)|<\infty$.
This means that
\begin{gather*} \liminf_{k\to\infty}A_k^c\subset\Bigl\{\omega:\sum_{k=1}^\infty\sup_{t\ge 0}|x^{k+1}(t,\omega)-x^k(t,\omega)|<\infty\Bigr\}\\
\mu\Bigl(\Bigl\{\sup_{t\ge0}|x^{k+1}-x^k|\ge 2^{-k}\Bigr\} \;\text{i.o.}\Bigr)=0,\quad \Longrightarrow
\quad \sum_{k=1}^\infty\sup_{t\ge 0}|x^{k+1}-x^k|<\infty, \quad \text{a.s.}
\end{gather*}
