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.)

  • 1
    $\begingroup$ Your question makes it sound like it is impossible to answer this question 'no' (since the hypotheses can be modified to exclude any counterexample). $\endgroup$
    – LSpice
    Jan 3, 2022 at 20:46
  • $\begingroup$ The answer is 'no' nevertheless. The suggested hypotheses do not help. $\endgroup$ Jan 3, 2022 at 20:57
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Jan 3, 2022 at 21:09
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Jan 3, 2022 at 21:10
  • 4
    $\begingroup$ @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.) $\endgroup$ Jan 3, 2022 at 21:14

1 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.

  • $\begingroup$ Is it true that $\sum_{v\in V}X_v=1_V$ in the sense in the question? $\endgroup$ Jan 3, 2022 at 21:14
  • $\begingroup$ Yes, it is, since for each $\omega$ in $\Omega$ there is at most one $v$ such that $X_v(\omega)$ is nonzero. $\endgroup$ Jan 3, 2022 at 21:20
  • $\begingroup$ Oh right. I see $\endgroup$ Jan 3, 2022 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.