Let $I\subset \mathbb{R}$ be bounded and $f:I\rightarrow \mathbb{R}$ be a real function. The number of upcrossings of $f$ across the interval $[a,b]$, denoted by $U_f[a,b]$ is defined as the supremum of the nonnegative integers such that there exists $s_k,t_k\in I$ satisfying $s_1<t_1<s_2<t_2<\cdots<s_n<t_n$, and for which $f(s_k)\le a<b\le f(t_k)$, i.e.
$$ \begin{aligned} U_f[a,b]:= \sup\big\{n\in\mathbb{N}\ |\ &(\exists s_1,...,s_n\in I)(\exists t_1,...,t_n\in I)(s_1<t_1<...<s_n<t_n \\ &\& \ (\forall k\leq n)(f(s_k)\le a<b\le f(t_k)))\} \end{aligned}\tag{1}\label{uc}$$
The following lemma is taken from La Gall's Brownian Motion, Martingales ad Stochastic Calculus(Lemma 3.16). No proof is given in the text and the domain of the function is taken to be a countable dense set $D\subset\mathbb{R}_+$, as in stochastic analysis a common trope is to prove results in the discrete case first then use a limiting argument to go to the continuous case, however in this particular case I don't think we need the domain to be a countable set, if I am wrong please let me know why? In the text the function is assumed to be bounded and the upcrossings finite on the sets $D\cap [0,T]$ for every $T\in D$. So I am not sure what is more general, by considering the entire domain, but only bounded sets or the countable subsets?
Lemma. If $f:I\rightarrow \mathbb{R}$ is bounded and for every $a<b$, $U_f[a,b]<\infty$ then $\lim\limits_{s\downarrow t}f(s)$ and $\lim\limits_{s\uparrow t}f(s)$ exist for all $t\in I$.
We can just consider the right limits as the left case dually follows, my argument is a proof by contradiction and is as follows:
Suppose that the right limit does not exist for a particular $x\in I$. Then for a decreasing sequence $\{x_n\}\subset I$ converging to $x\in I$ we have $$l:=\liminf\limits_{n\rightarrow\infty}f(x_n)<\limsup\limits_{n\rightarrow\infty}f(x_n):=u.\tag{2}$$ Now take $a,b\in(l,u)$, with $a<b$, then for any $n\in\mathbb{N}$ there must be a $n'>n$ such that $f(x_{n'})>b$ (otherwise $u\le b$ an impossibility); and similarly there is a $n^*>n$ such that $f(x_{n^*})<a$. We can then define the following sequence: $$ \begin{aligned} t_0 &=\inf I \\ s_k &=\inf\{m\geq t_{k-1} \ | \ f(x_m)\le a\} &\\ t_k &=\inf\{m\geq s_{k} \ | \ f(x_m)\ge b\} \ &\text{for}\ k\in\mathbb{N}. \end{aligned}\tag{3}\label{ucs} $$
Where we see that $s_1<t_1<s_2<t_2\cdots$ and $(\forall k\in\mathbb{N})(f(s_k)\le a<b\le f(t_k)$, thus $U_f[a,b]=\infty$, and the supposition was wrong and the right limit exists for every $x\in I$.
Would appreciate any feedback and input in particular please let me know:
- If the assumptions and the statement of the lemma is correct, and what is the most general case of the lemma?
- If the proof of the lemma is correct; if there are errors can they be fixed or if the whole argument has the be scrapped? If the latter please let me know how to proceed?
- If there is a better argument, a neater one than the one above please let me know?
- Is the formula for $U_f[a,b]$ in (1) correct? Should the conjunction (&) be replaced with an implication, $(\Rightarrow)$? Is there a better way to write a formula for $U_f[a,b]$?
Thanks in advance, any feedback will be appreciated.
Generally, a bounded sequence $(x_n)_n$ has a subsequence $(x_{n(k)})_k$ that decreases to $\limsup_{n \to \infty}x_n$. Similar result holds for $\liminf_{n \to \infty}x_n$.
Let $(t_n)_n$ be a sequence in $D$ decreasing to $t >0.$ If $l<u$, then we have two subsequences of $(f(t_n))_n$ that decreases to $l$ and increases to $u$, respectively. This implies $M^{f}_{ab}([0,t_0]\cap D)=\infty$.