proof of convergence of a sequence with finite upcrossings

We show that a sequence $x_1,x_2,\mathrm{\dots }$ 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 .

Denoting the infimum limit and supremum limit by

$$l=\underset{n\to \mathrm{\infty }}{lim\; inf}{x}_n,u=\underset{n\to \mathrm{\infty }}{lim\; sup}{x}_n,$$

then $l\le u$ 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 . If $l>a$ then there is an $N$ such that $x_n>a$ for all $n\ge N$. So, all upcrossings of $[a,b]$ must start before time $N$, and we may conclude that $U[a,b]\le N$ is finite. On the other hand, if $l\le a$ then and we can infer that for all $n\ge N$ and some $N$. Again, this gives $U[a,b]\le N$.

Conversely, suppose that the sequence does not converge, so that $u>l$. Then choose 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 . This allows us to define infinite sequences $s_k,t_k$ by $t_0=0$ and

$t_k=inf\{m\ge s_k:X_m>b\},$

for $k\ge 1$. Clearly, and for all $k\ge 1$, so $U[a,b]=\mathrm{\infty }$.