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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 $A \subseteq \mathbb R$.

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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 $a<b_{n,m} \rightarrow a_n$ 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.

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    $\begingroup$ 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 $a$. $\endgroup$ – nahp Sep 17 at 15:02
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    $\begingroup$ 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. $\endgroup$ – nahp Sep 17 at 15:12
  • $\begingroup$ 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. $\endgroup$ – Mindlack Sep 17 at 15:19

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