$\sum_{H \in \mathscr H} 1_H(\omega) = 1$

Question

ETA: Nothing to see here. I was thinking of this: https://math.stackexchange.com/revisions/863937/3

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For any (countable?) $\sigma$-algebra $\mathscr H$, do we have $$\sum_{H \in \mathscr H} 1_H(\omega) = 1 \ \forall \ \omega \ \in \Omega$$?

Answer 1

If $\mathscr{H} = \{\emptyset, \Omega\}$, then the result clearly holds.

If $\mathscr{H}$ is non-trivial, then for all $\omega \in \Omega$, there are at least two measurable subsets of $\Omega$ containing $\omega$. Therefore, the sum in question is greater than or equal to $2$.

So the result holds if and only if $\mathscr{H}$ is trivial.