Is the unordered sum of measurable functions measurable?

Question

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.)

Answer 1

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.