# Countably many discontinuities in a function of two variables

This question concerns an object from probability theory, but it should require only analysis to answer. The local time process $$(t,a) \mapsto L_t^a$$ of a continuous semimartingale is a real-valued function on $$[0,\infty) \times \mathbb R$$ with the following properties:

1. For every $$a$$, $$t \mapsto L_t^a$$ is continuous and (weakly) increasing.
2. For every $$t$$, $$a \mapsto L_t^a$$ is right-continuous and left-limited (cadlag).

Let $$\Delta L_t^a$$ denote $$L_t^a - L_t^{a-}$$, the size of the jump (if any) at $$(t,a)$$. Are the two properties above enough to verify the following claim (made in Revuz and Yor's Continuous Martingales and Brownian Motion, 3rd ed.)?

"[T]here are at most countably many $$x \in ]a,b[$$ such that $$\Delta L_s^x > 0$$ for some $$s \in [0,t]$$..." (Chapter VI.1, p. 230)

For fixed $$s$$, the cadlag function $$a \mapsto L_s^a$$ can only have countably many discontinuities (see this question, for example). However, $$[0,t]$$ is uncountable, so this observation does not give the claim automatically. It seems that continuity in the variable $$t$$ should yield the claim somehow, but I don't know how to show this. All I've found so far is that the left-limit function $$(t,a) \mapsto L_t^{a-}$$ need not be continuous is $$t$$. Based on a classical example that "a pointwise limit of continuous functions need not be continuous," notice that $$L_t^a = \begin{cases} (1-t)^{-1/a} 1_{[0,1]}(t), & a < 0 \\ 0, & a \geq 0 \end{cases}$$ has the following discontinuous left-limit function at zero: $$L_t^{0-} = 1_{\{0\}}(t)$$. Here, $$1_A$$ denotes the indicator function of $$A \subseteq \mathbb R$$.

Let $$L^t_a$$ be denoted $$f_t(a)$$, so that $$f_t$$ is a (pointwise) continuous weakly increasing family of cadlag functions.

Let’s show that for every sequence $$c_n$$ decreasing to $$a$$, every monotonous sequence (eg increasing but the decreasing case is similar) $$t_n$$ converging to $$t$$, then $$f_{t_n}(c_p)$$ converges uniformly in $$p$$ as $$n \rightarrow \infty$$ to $$f_t(c_p)$$. In particular, this implies $$f_{t_n}(c_n)$$ converges uniformly to $$f_t(a)$$.

Assume the opposite holds, then (up to extracting subsequences) there are $$\epsilon > 0$$ and $$p_n \rightarrow \infty$$ such that $$f_t(c{p_n}) > f_{t_n}(c_{p_n})+\epsilon$$ for each $$n$$. Up to re-extracting we assume $$p_n$$ increasing and we set $$c_{p_n}=c’_n$$.

Let $$m$$ be an integer, $$n \geq m$$, then $$f_t(c’_n) \geq f_{t_n}(c’_n)+\epsilon \geq f_{t_m}(c’_n)$$. As each $$f_s$$ is cadlag, as $$n$$ goes to infinity the inequality becomes $$f_t(a) \geq \epsilon+f_{t_m}(a)$$, ultimately contradicting the continuity of $$s \longmapsto f_s(a)$$.

Let $$S_{\epsilon}$$ be the set of pairs $$(a,t)$$ such that $$|f_t(a) - f_t(a^-)| > \epsilon$$, for each $$\epsilon>0$$.

Assume that there is a sequence $$(a_n,t_n)$$ in some $$S_{\epsilon}$$ with $$a_n$$ decreasing to some limit $$a > 0$$. Up to extracting a subsequence, we may assume that eg $$t_n$$ increases to $$s$$. For each $$n$$, we moreover have some increasing sequence $$a such that $$|f_{t_n}(b_{n,m})-f_{t_n}(a_n)| >\epsilon$$.

For each $$n$$, choose $$m_n$$ large enough so that $$b’_n=b_{n,m_n}$$ is decreasing (thus it converges to $$a$$). So $$|f_{t_n}(a_n)-f_{t_n}(b’_n)|>\epsilon$$. But by the first result both of these sequences converge to $$f_t(a)$$ and we get a contradiction.

So for any $$a \in S_{\epsilon}$$, there is no decreasing subsequence in $$S_{\epsilon}$$ converging to $$a$$. Thus there exists a rational $$q_a > a$$ such that $$S_{\epsilon} \cap [a,q_a]=\{a\}$$.

Thus $$a \in S_{\epsilon} \longmapsto q_a \in \mathbb{Q}$$ is injective, hence the countability of each $$S_{\epsilon}$$ – which is what we wanted to prove.

• I see, thank you for the help! It seems the two main results established in this argument are (1) that for any "diagonal" sequence $(c_n, t_n)$ with $c_n$ decreasing to $c$ and $t_n$ either increasing or decreasing to $t$, we have $f_{t_n}(c_n) \to f_t(c)$, and (2) if the set of values $a$ for which there is a jump at $a$ of size at least $\epsilon$ has a sequence $(a_n)$ decreasing to some limit $a$, then by using the "jump property" at each value $a_n$ and taking an appropriate subsequence, we may use (1) to show that there is no jump at $a$. – nahp Sep 17 at 15:02
• Actually, in (2) I should say that "we may use (1) to derive a contradiction," so that the points $a$ at which there are jumps of size at least $\epsilon$ do not "accumulate above" any real number; this establishes countability. – nahp Sep 17 at 15:12
• Note that (1) works because we can “make the diagonal sequence non-diagonal” thanks to the monotonicity. For (2), your second comment is correct indeed. – Mindlack Sep 17 at 15:19