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Let $\{a_i\mid i\in I\}$ and $\{b_i\mid i\in I\}$ be a two sets of non-negative real numbers. If $a_i\leq b_i$ for every $i$, and $\sum_{i\in I}b_i=1$, then when $\sum_{i\in I}a_i$ converges? If $I$ is countable, then it converges. But I want to know that is there a more general result?

khers
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An uncountable series converges only if a countable subset of its elements are non-zero. Therefore if $\sum_{i \in I} b_i$ convergences then there exists a countable subset $J$ s.t. if $i \notin J \implies b_i = 0$ and since $a_i \leq b_i$ this implies that $a_i = 0 \: \forall i \notin J$. Therefore the uncountable series $\sum_{i \in I} a_i$ reduces to a countable one and you can apply comparison test of countable series to conclude the series converges.

Samir
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  • Thanks so much! I also find the proof of the assertion which you used https://math.stackexchange.com/questions/1573710/divergence-of-a-positive-series-over-a-uncountable-set – khers Feb 05 '24 at 12:23