This is from a proof in Feller's Vol 2 (p.537). I am struggling to prove the following
and so it is clear that at some point $x_0$ either $|\Delta(x_0+)|=\eta$ or $|\Delta(x_0)|=\eta$.
I know that $F$ is increasing so all its one-sided limits exist and $G$ is continuous so all its limits alo exist. These functions are bounded so these limits are also bounded. Since $\Delta(x)$ is a difference of these two functions, its one-sided limits also exist everywhere. How to use the fact that $\Delta(x)$ vanishes at $\pm\infty$ to show that such a point $x_0$ exists? Obviously, $\Delta(x)=\sup_x\Delta(x)\text{ or }\inf_x\Delta(x)$.
This is a slightly different explanation of the same point in Chung's "A course in probability theory" (p.237). It says
Since $F-G$ vanishes at $\pm\infty$, there exists a sequence of numbers $\{x_n\}$ converging to a finite limit $b$ such that $F_n(x_n)-G_n(x_n)$ converges to $\eta$ or $-\eta$. Hence either $F(b)-G(b)=\eta$ or $F(b-)-G(b)=-\eta$.
I feel like I should use Bolzano–Weierstrass theorem here but unable to proceed. In particular, $f$ vanishing at $\pm\infty$ would mean
$$
\eta=\sup\{ \Delta(x) |x\in (-\infty,+\infty)\}=\sup\{ \Delta(x) |x\in [-K,K]\}
$$
for some a finite $K$. This would allow one to get a monotone converging subsequence of $x$. However, I still need to prove that $|\Delta()|$ either attains $\eta$ or that one of its jump points is equal to $\eta$.
Best Answer
Let’s assume that $\Delta$ is some function which vanishes at infinity and which has well defined left and right limits. Let $\eta=\sup_x |\Delta(x)|$.
If $|\Delta(x)|=\eta$ somewhere, then since $\Delta$ is continuous where defined, we’re done.
Otherwise, as the supremum, there’s some sequence of distinct points $x_n$ so that $|\Delta(x_n)|$ approaches $\eta$. Since $\Delta$ vanishes at infinity, $x_n$ is bounded. Splitting infinitely many points between whether $\Delta(x_n)$ are positive or negative, there must either be infinitely many of positive points or infinitely many negative.
Let’s assume there are infinitely many are positive (the negative case is identical) and remove the negative ones from the sequence. The remainder all must have $\Delta(x_n)$ approach $\eta$. This set of points $x_n$ is infinite and bounded, so by Bolzano Weistrauss, it has at least one limit point. Call a limit point $x_0$. There’s some subsequence which converges to $x_0$.
As a limit point there’s some subsequence of $x_n$, call it $y_n$, such that $y_n$ converges to $x_0$. Either infinitely many are less than $x_0$ or infinitely many are greater. Let’s handle the first case (the other is identical).
Taking the subsequence of points less than $x_0$, we get some sequence $y_n$ which converges from the left to $x_0$ and for which the limit of $\Delta(y_n)$ is $\eta$. Thus, since $\Delta$ has well-defined one sided limits, $\Delta(x_0^-)=\eta$.
Going back to full generality, we have that either the left limit or right limit is equal to either $-\eta$ or $\eta$ at some point $x_0$.