# Is the unordered sum of measurable functions measurable?

Let $$E$$ be a normed $$\mathbb R$$-vector space and $$I$$ be a nonempty set. Remember that $$(x_i)_{i\in I}\subseteq E$$ is called summable if there is a $$x\in E$$ such that for all $$\varepsilon>0$$, there is a finite $$J\subseteq I$$ such that for all finite $$K\subseteq I$$ with $$J\subseteq K$$ it holds $$\left\|x-\sum_{k\in K}x_k\right\|_E<\varepsilon\tag1.$$ In that case, we write $$x:=\sum_{i\in I}x_i$$. We can show that if $$(x_i)_{i\in I}\subseteq E$$ is summable, then $$\{i\in I:x_i\ne0\}$$ is countable.

Now let $$(\Omega,\mathcal A)$$ be a measurable space and $$X_i:\Omega\to E$$ be $$\mathcal A$$-measurable for $$i\in I$$. Assuming that $$(X_i(\omega))_{i\in I}$$ is summable for all $$\omega\in\Omega$$, are we we able to show that $$X:=\sum_{i\in I}X_i$$ is $$\mathcal A$$-measurable.

Please understand the given assumptions as open to suitable modifications. For example, I could imagine that we need to assume that each $$X_i$$ is even strongly $$\mathcal A$$-measurable (meaning that $$X_i$$ can be approximated by a sequence of $$\mathcal A$$-measurable functions of finite range) or that $$E$$ is separable.

EDIT: The claim is clearly true when $$I$$ is countable. So, this is not what I'm looking for. One particular instance, which motivated me to ask this question, is the following scenario: Given a càdlàg $$E$$-valued process $$(Y_t)_{t\ge0}$$ on $$(\Omega,\mathcal A)$$ and $$B\in\mathcal B([0,\infty)\times\mathbb R)$$, how do we see that $$\sum_{\substack{s\:\ge\:0\\\Delta Y_s\:\ne\:0}}1_B(s,\Delta Y_s)$$ is $$\mathcal A$$-measurable? (I've asked for this separately on mathematics: https://math.stackexchange.com/q/4341779/47771.)

• Your question makes it sound like it is impossible to answer this question 'no' (since the hypotheses can be modified to exclude any counterexample). Jan 3, 2022 at 20:46
• The answer is 'no' nevertheless. The suggested hypotheses do not help. Jan 3, 2022 at 20:57
• @LSpice I will try to improve the question, if I can figure out how. The desired claim is clearly true, when $I$ is countable. So, this is clearly not what I'm looking for. Please take note of my edit. Jan 3, 2022 at 21:09
• @YuvalPeres Thank you for your comment. Most probably I'm missing the right hypothesis. Please take note of my edit. I've added a motivating example. Jan 3, 2022 at 21:10
• @0xbadf00d I answered below the question as originally stated, with the suggested variations. The question on Cadlag processes is much more specific, and best stated as a separate question. If questions are modified after someone figures out an answer, this will reduce people's motivation to answer questions.) Jan 3, 2022 at 21:14

The answer is negative. Take $$E={\mathbb R}$$ and $$\Omega={\mathbb R}$$, with $$\cal A$$ the Lebesgue (or Borel) $$\sigma$$-algebra. Let $$V$$ be a subset of $${\mathbb R}$$ which is not $$\cal A$$-measurable. For each $$v \in V$$, let $$X_v:{\mathbb R} \to \{0,1\}$$ be the indicator of $$v$$, so it takes the value 1 only at $$v$$. Then each function $$X_v$$ is strongly $$\cal A$$-measurable, and $$(X_v(\omega))_{v\in V}$$ is summable for every $$\omega\in\Omega$$. However, $$\sum_{v \in V} X_v=1_V$$ is not $$\cal A$$-measurable.
• Is it true that $\sum_{v\in V}X_v=1_V$ in the sense in the question? Jan 3, 2022 at 21:14
• Yes, it is, since for each $\omega$ in $\Omega$ there is at most one $v$ such that $X_v(\omega)$ is nonzero. Jan 3, 2022 at 21:20