# Why is the supremum of the sequence of running maximum of cadlag processes also cadlag?

I have a question while reading the proof of Theorem 6 from the following post: https://almostsuremath.com/2009/12/24/local-martingales/

So here we assume $$X^n$$ to be a sequence of local submartingales, i.e. processes that are locally a cadlag adapted submartingale. Also, we know from the assumption that (passing to a subsequence), $$X^n$$ converges to $$X$$ uniformly on compacts. So we define $$M_t := \sup_n \sup_{s \le t} |X_t^n|.$$ My questions are, why is this process cadlag, and the jumps $$|\Delta M| \le \sup_n |\Delta X^n|$$?

I know that the running maximum $$\sup_{s \le t} |X_t|$$ of a cadlag process $$X$$ is right continuous since it is increasing and it can only have jumps where $$X$$ has a jump which can be approximated from the right. However, I cannot see why the supremum over $$n$$ of these sequences will also be right-continuous and have left hand limits.

Moreover, how do the jumps of this process $$M$$ look like so that they are bounded above by the supremum of the jumps $$|\Delta X^n|$$?

I would greatly appreciate some help with these details.

Theorem 6 Let $$\{X^n\}_{n=1,2,\ldots}$$ be a sequence of local martingales (resp. local submartingales, local supermartingales) converging ucp to a limit $$_X$$. If $$\sup_n \sup_{s \leq t} |\Delta X_s^n|$$ is locally integrable then $$_X$$ is a local martingale (resp. local submartingale, local supermartingale).

Proof: It is enough to prove the submartingale case, as the martingale and supermartingale cases follow from applying this to $$_{-X}$$.

First, as it is a ucp limit of cadlag adapted processes, $$_X$$ will be cadlag and adapted. Passing to a subsequence if necessary, we may suppose that $$_{X^n}$$ converges to $$_X$$ uniformly on compacts. Then, $$M_t = \sup_n \sup_{s \leq t}|X_t^n|$$ is cadlag, adapted, and increasing. It has jumps $$|\Delta M| \leq \sup_n |\Delta X^n|$$ which, by the condition of the theorem, is locally integrable. Therefore, $$_M$$ is locally integrable. Let $$\tau_k$$ be a localizing sequence, so that $$1_{\{\tau_k > 0\}}M^{\tau_k}$$ is integrable. Then, $$1_{\{\tau_k > 0\}}(X^n)^{\tau_k}$$ are local submartingales bounded by $$1_{\{\tau_k\}}M^{\tau_k}$$ and, in particular, are of class (DL). So, they are proper submartingales converging to $$1_{\{\tau_k > 0\}}X^{\tau_k}$$ and, applying bounded convergence to this limit, $$1_{\{\tau_k > 0\}}X^{\tau_k}$$ is a submartingale. Therefore, $$\tau_k$$ is a localizing sequence for $$_X$$, showing that it is a local submartingale. $$\square$$

• I think I can write an answer. I'll be doing that shortly. Jan 28, 2022 at 19:33

We have $$M_t = \sup_{n \geq 1} \sup_{s \leq t} |X_s^n|$$ We consider the processes $$M^n_t =\sup_{m \leq n} \sup_{s \leq t} |X^m_s|$$

For $$n \geq 1$$. We claim that $$M^n_t$$ are increasing adapted processes.

Indeed, for each $$1 \leq m \leq n$$, the processes $$Y^m_t = \sup_{s \leq t} |X^m_s|$$ are increasing (obviously) and adapted. Once you observe this, $$M^n_t = \max_{m \leq n} Y^m_t$$ is a finite maximum of adapted processes, hence increasing and adapted.

Now, for any (bounded) sequence of real numbers $$a_n$$, we have $$\sup_{n \geq m} a_m \to \sup_{n \geq 1} a_n$$. Therefore,$$M_t^n \to M_t$$ pointwise, which shows that $$M_t$$ is adapted as well.

Claim : Let $$f:\mathbb R^+ \to \mathbb R$$ be a right continuous function. Then $$g(t) = \sup_{s \leq t} f(s)$$ is a cadlag function.

Proof : Note that $$g(t)$$ is an increasing function, and therefore admits left and right limits at every point. It is therefore sufficient to prove that $$g(t)$$ is right-continuous.

Let $$t_n \downarrow t$$, we must prove that $$g(t_n) \to g(t)$$. Suppose not. Then, for some $$\epsilon>0$$, $$g(t_n) > g(t)+\epsilon$$ for all $$n$$. By the definition of the supremum, there exist $$s_n \leq t_n$$ so that $$f(s_n) > g(t)+\epsilon \geq f(t) +\epsilon$$ for all $$n$$.

We claim that $$s_n > t$$. Indeed, if $$s_n \leq t$$ then $$f(s_n) \leq g(s_n) \leq g(t)$$ which contradicts the choice $$f(s_n)>g(t)+\epsilon$$ made above. Consequently, by the squeeze theorem, $$s_n \to t$$ and therefore $$f(s_n)\to f(t)$$ by right-continuity, but we also know that $$f(s_n)-f(t) > \epsilon$$ for all $$n$$. This is a contradiction : consequently, $$g$$ is right continuous. $$\blacksquare$$

As $$X^m_t$$ is right-continuous, it follows from the above that $$Y^m_t$$ is cadlag. Finally, $$M^n_t$$, for any $$n$$, is a finite maximum over some of these cadlag functions, and is therefore cadlag itself (to see this, you can use induction, along with the formula $$\max\{x,y\} = \frac{x+y+|x-y|}{2}$$).

To show that $$M_t$$ is cadlag, we first note that with respect to $$t$$, it is an increasing process, because if $$s \leq t$$ then $$M^n_s \leq M^n_t$$ for each $$n$$ and you can take the limit. Therefore, the left and right limit functions $$M_{s-}, M_{s+}$$ are well-defined for sure, since every increasing function admits right and left limits at each point.

To show right-continuity, we must prove that $$M_{s+} = M_s$$ almost surely. Suppose not. In that case, on a set of non-zero probability, $$M_{s+} \neq M_s$$. In particular, for some $$\epsilon>0$$, the set $$\{M_{s+} - M_s > \epsilon\}$$ has non-zero probability. We note that $$M_{s+} - M_s > \epsilon$$ if and only if there is a sequence $$s_n \downarrow s$$ such that $$M_{s_n} > M_s + \epsilon$$ for every $$n$$.

However, by the definition of $$M_{s_n}$$ and the supremum, there exists $$s'_n \leq s_n$$ and a sequence of indices $$m_n$$ such that $$|X^{m_n}_{s'_n}| > M_s+\frac{\epsilon}{2} \geq |X^{m_n}_{s}| + \frac{\epsilon}{2} \tag{1}$$

for all $$n$$, which we may rearrange and write as $$|X^{m_n}_{s'_n}| - |X^{m_n}_{s}| > \frac{\epsilon}{2} \tag{2}$$

However, clearly $$s'_n > s$$ ,otherwise it would be absorbed under the supremum in the definition of $$M_s$$ and couldn't therefore admit a value bigger than $$M_s$$. It follows by the squeeze theorem that $$s'_n \to s$$. Also note that $$|X^{m_n}_{s'_n} - X^{m_n}_s| \geq |X^{m_n}_{s'_n}| - |X^{m_n}_{s}| > \frac{\epsilon}{2} \tag{3(a)}$$

and $$|X^{m_n}_{s'_n} - X^{m_n}_s| \leq |X^{m_n}_{s'_n} - X_{s'_n}| + |X_{s'_n} - X_{s}|+ |X_s - X^{m_n}_s| \tag{3(b)}$$

Claim : $$m_n$$ isn't a sequence of bounded numbers.

Proof : Suppose that $$m_n$$ is a bounded sequence of natural numbers. Then, there exists an $$N$$ such that $$m_n \leq N$$ for all $$n$$. Now, we know that $$M^N_t$$ is a cadlag process a.s. i.e. we know that for the sequence $$s'_n$$ above, we have $$M^N_{s'_n} \to M^N_s$$. However, note that $$|X^{m_n}_{s'_n}| \leq M^N_{s'_n}$$, so we know that $$\limsup_{n \to \infty} |X^{m_n}_{s'_n}| \leq \limsup_{n \to \infty} M^N_{s'_n} = M^N_{s}$$ However, from the first inequality in $$(1)$$, we get $$\liminf_{n \to \infty} |X^{m_n}_{s'_n}| \geq M_{s}+\frac{\epsilon}{2} > M_s \geq M^N_{s} \geq \limsup_{n \to \infty} |X^{m_n}_{s'_n}|$$ This inequality above provides a contradiction, and completes the proof.$$\blacksquare$$

Therefore $$m_n$$ has a subsequence that converges to infinity. Let that convergent subsequence be $$m_n$$ itself, for the sake of easy notation.

In that case, $$X^{m_n} \to X$$ ucp. However, look now at the right hand side of $$(3(b))$$. The term $$|X_{s'_n}-X_s|$$ goes to $$0$$ as $$n$$ goes to infinity, as $$X$$ is cadlag. The terms $$|X^{m_n}_{s'_n} - X_{s'_n}|$$ and $$|X^{m_n}_{s} - X_s|$$ go to $$0$$ as $$n \to \infty$$ by the ucp convergence of $$X^{m_n}$$ to $$X$$. Finally, the entire right hand side of $$(3(b))$$ must go to $$0$$. However, this contradicts $$(3(a))$$, since $$(3(a)),(3(b))$$ combined tell you that the RHS of $$(3(b))$$ cannot go to $$0$$.

Therefore , if $$(1),(2),(3(a))$$ are to hold true on the set where $$M_{s+} \neq M_s$$, it must happen that either right-continuity or ucp convergence is violated at every element here. However, the intersection of both happens with probability $$1$$, therefore it is impossible that $$M_{s+} \neq M_s$$ be a set of non-zero probability.

Finally, $$M_t$$ must be a cadlag process.

Regarding the jumps of $$M_t$$, we see that $$M_{t-} = \sup_{n \geq 1} \sup_{s < t} |X_s^n|$$

Now, suppose that $$M_t - M_{t-} =\epsilon> 0$$. This means that the supremum in $$M_t$$ must occur at the time point $$t$$, so that means there is a sequence $$m_n$$ such that $$|X^{m_n}_t| - M_{t-} > \epsilon-\frac 1n$$ for each $$n$$. Note that $$M_{t-} \geq |X^{m_n}_{t-}|$$, so this implies via the triangle inequality that $$|\Delta X^{m_n}_t| > \epsilon-\frac 1n$$ for each $$n$$, or that $$\sup_{n} |\Delta X^n_t| \geq \epsilon = \Delta M_t = |\Delta M_t|$$ following the application of a limit and supremum on the LHS.

• Does $f$ being merely right continuous give that $\sup_{s\le t}f(s)$ is cadlag? How do we get the left limit? And why does $X_{s_n}^{m_n} \to X_s^{m_n}$ lead to a contradiction to which assumption? I don't follow why $M_{s_n}^N \to M_s^N$ implies that $X_{s_n}^{m_n} \to X_s^{m_n}$. Jan 29, 2022 at 0:43
• @nomadicmathematician Let me know if the rewrite is better. I've tried to link the statements being violated at each point. Jan 29, 2022 at 9:31
• Thank you so much for the detailed explanation. I'm sorry but maybe I don't see a few things. In the two claims you prove, the first one, what do you mean by $s_n>s$ since otherwise $g(s_n)\le g(s)$ would have been forced? I don't follow why $g(s_n)\le g(s)$ is an issue, and what is $s$ here? Finally in the second claim, why does it follow that $|X_{s_n'}^{m_n}|\to M_s^N$ because $|X_{s_n'}^{m_n}| \le M_{s_n'}^N$? Jan 29, 2022 at 10:22
• @nomadicmathematician I completed the edits. The first one had some notational confusion : in the second I had to make things a little more precise and clear. Jan 29, 2022 at 10:40
• Thanks for the wonderful answer. If you have time could you take a look at this question? It seems to require a similar argument using cadlag property, but I don't know how the approximation by just rationals work here. math.stackexchange.com/questions/4361581/… Jan 29, 2022 at 17:40