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 $x:=\sum_{i\in I}$. 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.