proof of convergence of a sequence with finite upcrossings

We show that a sequence $x_{1},x_{2},\ldots$ of real numbers converges to a limit in the extended real numbers if and only if the number of upcrossings $U[a,b]$ is finite for all $a<b$ .

Denoting the infimum limit and supremum limit by

l=\liminf_{n\rightarrow\infty}x_{n},\ u=\limsup_{n\rightarrow\infty}x_{n},

then $l\leq u$ and the sequence converges to a limit if and only if $l=u$ .

We first show that if the sequence converges then $U[a,b]$ is finite for $a<b$ . If $l>a$ then there is an $N$ such that $x_{n}>a$ for all $n\geq N$ . So, all upcrossings of $[a,b]$ must start before time $N$ , and we may conclude that $U[a,b]\leq N$ is finite. On the other hand, if $l\leq a$ then $u=l<b$ and we can infer that $x_{n}<b$ for all $n\geq N$ and some $N$ . Again, this gives $U[a,b]\leq N$ .

Conversely, suppose that the sequence does not converge, so that $u>l$ . Then choose $a<b$ in the interval $(l,u)$ . For any integer $n$ , there is then an $m>n$ such that $x_{m}>b$ and an $m>n$ with $x_{m}<a$ . This allows us to define infinite sequences $s_{k},t_{k}$ by $t_{0}=0$ and

	$\displaystyle s_{k}=\inf\left\{m\geq t_{k-1}\colon X_{m}<a\right\},$
	$\displaystyle t_{k}=\inf\left\{m\geq s_{k}\colon X_{m}>b\right\},$

for $k\geq 1$ . Clearly, $s_{1}<t_{1}<s_{2}<\cdots$ and $x_{s_{k}}<a<b<x_{t_{k}}$ for all $k\geq 1$ , so $U[a,b]=\infty$ .

Title	proof of convergence of a sequence with finite upcrossings
Canonical name	ProofOfConvergenceOfASequenceWithFiniteUpcrossings
Date of creation	2013-03-22 18:49:39
Last modified on	2013-03-22 18:49:39
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	4
Author	gel (22282)
Entry type	Proof
Classification	msc 40A05
Classification	msc 60G17