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This question is motivated by the question https://math.stackexchange.com/questions/4644235/ on Math Stack Exchange. First, I need to define a notion of transfinite summability that I have not seen elsewhere.


Transfinite summability of series.

Let $(S,\leq)$ be a totally ordered space. In this post,

  • an initial segment of $(S,\leq)$ is a set $I \subset S$ such that for every $\alpha \in I$, we have $\{\beta \in S : \beta \leq \alpha\} \subset I$;
  • an interval of $(S,\leq)$ is a set $I \subset S$ such that for every $\alpha,\beta \in I$ with $\alpha \leq \beta$, we have $\{\gamma \in S : \alpha \leq \gamma \leq \beta \} \subset I$.

Also, for any partition $\mathcal{P}$ of $S$ and any $J \subset S$, we write $\mathcal{P}(J) \subset \mathcal{P}$ for the set of elements of $\mathcal{P}$ that contain at least one element of $J$.

Given a pairwise-disjoint collection $\mathcal{C}$ of non-empty intervals of $(S,\leq)$, we write $\leq_\mathcal{C}$ for the total ordering of $\mathcal{C}$ that is naturally inherited from $\leq$.

A nice partition of $(S,\leq)$ is a partition $\mathcal{P}$ of $S$ such that every element of $\mathcal{P}$ is a non-empty interval of $(S,\leq)$ and the total ordering $\leq_\mathcal{P}$ is a well-ordering.

Now let $V$ be a Banach space (or just a complete normed abelian group). For any $\mathbf{x}=(x_\alpha)_{\alpha \in S} \in V^S$, an $\mathbf{x}$-nice partition of $(S,\leq)$ is a nice partition $\mathcal{P}$ of $(S,\leq)$ such that for each $I \in \mathcal{P}$, $$ \sum_{\alpha \in I} \| x_\alpha \| < \infty. $$ Given an $\mathbf{x}$-nice partition $\mathcal{P}$ of $(S,\leq)$, we define for each initial segment $J$ of $(S,\leq)$ a value $$ \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha \, \in \, V \cup \{\text{NaN}\} $$ by the following recursive procedure:

  • In the case that $J=\emptyset$: Define $$ \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha = 0. $$
  • In the case that $\mathcal{P}(J)$ has a maximum element $I$ with respect to $\leq_\mathcal{P}$: Define $$ \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha = \left( \sum_{\alpha \in \bigcup(\mathcal{P}(J) \setminus \{I\})}^{\mathcal{P},\leq} x_\alpha \right) + \sum_{\alpha \in I \cap J} x_\alpha $$ if $$ \sum_{\alpha \in \bigcup(\mathcal{P}(J) \setminus \{I\})}^{\mathcal{P},\leq} x_\alpha \in V; $$ otherwise, define $$ \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha = \text{NaN}. $$
  • In the case that $J \neq \emptyset$ and $\mathcal{P}(J)$ has no maximum element with respect to $\leq_\mathcal{P}$ (and hence $J=\bigcup(\mathcal{P}(J))$): Suppose there exists $x \in V$ with the property that for every neighbourhood $U \subset V$ of $x$, there exists $\gamma_U \in J$ such that for every $\beta \in J$ with $\gamma_U \leq \beta$, $$ \sum_{\alpha \leq \beta}^{\mathcal{P},\leq} x_\alpha \in U; $$ then we define $$ \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha = x. $$ But if no such $x \in V$ exists, then we define $$ \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha = \text{NaN}. $$

Proposition. Given two $\mathbf{x}$-nice partitions $\mathcal{P}_1$ and $\mathcal{P}_2$ of $(S,\leq)$, we have that for every initial segment $J$ of $(S,\leq)$, $$ \sum_{\alpha \in J}^{\mathcal{P}_1,\leq} x_\alpha = \sum_{\alpha \in J}^{\mathcal{P}_2,\leq} x_\alpha. $$

The proof is a straightforward transfinite induction on the initial segments of $(\mathcal{P},\leq_\mathcal{P})$ with $\mathcal{P}=\{I_1 \cap I_2 : I_1 \in \mathcal{P}_1,\, I_2 \in \mathcal{P}_2\} \setminus \{\emptyset\}$.

So then, given $\mathbf{x} \in V^S$, if there exists an $\mathbf{x}$-nice partition of $(S,\leq)$, then for each initial segment $J \subset S$ we can simply define $$ \sum_{\alpha \in J}^\leq x_\alpha $$ without specifying the $\mathbf{x}$-nice partition $\mathcal{P}$. Note that, by construction, for initial segments $J_1,J_2 \subset S$ with $J_1 \subset J_2$, we have $$ \sum_{\alpha \in J_2}^\leq x_\alpha = \sum_{\alpha \in J_2}^{\leq|_{J_1}} x_\alpha $$ where $\leq|_{J_1}$ denotes the inherited ordering on $J_1$ from $(S,\leq)$.

Definition. We say that $\mathbf{x} \in V^S$ is summable if there exists an $\mathbf{x}$-nice partition of $(S,\leq)$ and $$ \sum_{\alpha \in S}^\leq x_\alpha \neq \text{NaN}. $$

Note that, by construction, if $\mathbf{x}$ is summable then every initial segment $J \subset S$ has $$ \sum_{\alpha \in J}^\leq x_\alpha \neq \text{NaN}. $$

From now on, we will omit the $\leq$ above the $\sum$ sign.


My questions.

Given a càdlàg function $f \colon [0,1] \to \mathbb{R}$, we write $\mathcal{D}(f) \subset (0,1]$ for the set of discontinuity points of $f$.

  • Has the above notion of summability been studied before? Does it have a name?
  • Given a càdlàg function $f \colon [0,1] \to \mathbb{R}$, is $(f(t)-f(t-))_{t \in \mathcal{D}(f)}$ necessarily summable?
  • Given a càdlàg function $f \colon [0,1] \to \mathbb{R}$ for which $(f(t)-f(t-))_{t \in \mathcal{D}(f)}$ is summable, do we necessarily have that the function $\tilde{f} \colon [0,1] \to \mathbb{R}$ given by $$ \tilde{f}(t) = f(t) - \sum_{s \in \mathcal{D}(f) \cap [0,t]} (f(s)-f(s-)) $$ is continuous?

Update. Anthony Quas has provided a negative answer to my second question (and hence the third question is arguably not very interesting).

But forgetting about jumps of càdlàg functions, I'm still quite interested in my first question: Has my above notion of summability been studied before? It feels like a very natural generalisation of $\sum_{n=1}^\infty x_n$ for non-$l^1$ sequences $(x_n)_{n \in \mathbb{N}}$ to the more general case of $S$-indexed families of numbers $(x_\alpha)_{\alpha \in S}$ for totally ordered sets $S$ besides $\mathbb{N}$.

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  • $\begingroup$ I'm having a bit of a hard time parsing your definitions but here is a construction that might yield a non-summable cadlag function. Let $S$ be the set of triadic rationals, so that any $a\in S$ can be written as $\frac j{3^n}$ for some $j$ and $n$. Define $g(\frac j{3^n})$ to be $1/2^n$ if $j\equiv 1\pmod 3$ and $-1/2^n$ if $j\equiv 2\pmod 3$. Then build a function $f(x)=\sum_{a\in S\,a\le x}g(a)$. $\endgroup$ Commented Aug 1, 2023 at 23:47
  • $\begingroup$ Your definition of the sum is quite unclear. In particular, if $\mathcal P$ is the partition into the singleton sets and $J=S=\mathcal D(f)\ni1$, then how is $\sum_{\alpha \in \bigcup(\mathcal{P}(J) \setminus \{I\})}^{\mathcal{P},\leq} x_\alpha$ defined, for $x_t:=f(t)-f(t-)$? $\endgroup$ Commented Aug 1, 2023 at 23:54
  • $\begingroup$ @IosifPinelis I'm defining by transfinite recursion: unless I've made a mistake, there will exist a unique $\left( \sum_{\alpha \in J}^{\mathcal{P},\leq} x_\alpha \, : \, \text{initial segment } J \subset S \right)$ fulfilling the three bullet points in my definition. [Note that in the case that you've asked about, taking $\mathcal{P}$ to be the partition into singleton sets is only possible if $\mathcal{D}(f)$ is well-ordered.] $\endgroup$ Commented Aug 2, 2023 at 0:04
  • $\begingroup$ @AnthonyQuas Forgive the probably stupid question, but how are you defining $\sum_{a \in S} g(a)$ given that [presumably as the whole point of your example] $\sum_{a \in S} |g(a)|=\infty$? $\endgroup$ Commented Aug 2, 2023 at 0:18
  • $\begingroup$ @AnthonyQuas Okay, I guess you're defining $\sum_{a \in S,\ a\leq x} g(a)$ as $\lim_{n \to \infty} \sum_{a \in S_n,\ a\leq x} g(a)$ with $S_n=\{\frac{j}{3^n} : 1 \leq j \leq 3^n\}$. Nice idea - it probably works! $\endgroup$ Commented Aug 2, 2023 at 0:35

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Anthony Quas has provided an example of a càdlàg function for which the jumps are not summable:

As in https://math.stackexchange.com/questions/10257/, for any $S \subset (0,1]$ and $(x_t)_{t \in S} \in \mathbb{R}^S$ with $\#\{t \in S : |x_t|>\varepsilon\}<\infty$ for all $\varepsilon>0$, there exists a càdlàg function $f \colon [0,1] \to \mathbb{R}$ that is continuous on $[0,1] \setminus S$ and has $f(t)-f(t-)=x_t$ for all $t \in S$.

So take $S=\{\frac{j}{3^n} : n \geq 1, \ 1 \leq j \leq 3^n\}$, and for each $t \in S$, expressing $t$ in reduced-fraction form as $t=\frac{j}{3^n}$ with $j \equiv 1 \text{ or } -1 \ \mathrm{mod}\, 3$, let $$ x_t = \begin{cases} \tfrac{1}{2^n} & j \equiv 1 \\ -\tfrac{1}{2^n} & j \equiv -1. \end{cases} $$ And now construct $f$ accordingly (so $S=\mathcal{D}(f)$). The only non-empty intervals $I$ of $S$ for which $\sum_{t \in I} |x_t| < \infty$ are singletons, and $S$ is not well-ordered. Hence there does not exist an $(x_t)_{t \in S}$-nice partition of $S$.

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  • $\begingroup$ Actually, since my definition works in a general Banach space and hence does not distinguish $\mathrm{NaN}$ from $\pm\infty$, much simpler counterexamples exist, such as the counterexample in my answer to math.stackexchange.com/questions/4644235. But I'm pretty sure that one can modify my definition from its current formulation in terms of a Banach space $V$ to a formulation in terms of the extended reals $\bar{\mathbb{R}}$, and then Anthony Quas's idea will probably be the simplest counterexample. $\endgroup$ Commented Aug 8, 2023 at 19:38

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