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Let $A\subset\Bbb{R}$. I already proved that if $A$ has zero measure, then every proper subset $B\subset A$ has zero measure. My question is if the reciprocal proposition holds. I would be like: If every subset $B\subset A$ has zero measure, then $A$ has zero measure. (I think maybe it holds, but I couldn't think a good proof). I want to use this to prove that if $m^*(A)>0$, then exists a subset $B\subset A$ such that $m^*(B)>0$, where $m^*$ is the Lebesque outer measure. My definition of $A$ having zero measure means that , for every $\varepsilon>0$ exists a denumerable collection of closed intervals such that the collection covers $A$ and the sum of all the lenghts of the closed intervals if less than $\varepsilon$.

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  • $\begingroup$ Hint: Measure is countably additive, and a singleton has measure 0. $\endgroup$ Nov 11 '19 at 23:04
  • $\begingroup$ I assume you mean every strict subset otherwise the answer is fairly trivial :P $\endgroup$ Nov 11 '19 at 23:05
  • $\begingroup$ Just a subtlety (likely about notation): Not necessarily every subset $B \subseteq A$ has zero measure as those don't have to be measurable. $\endgroup$
    – Qi Zhu
    Nov 11 '19 at 23:09
  • $\begingroup$ @QiZhu The assumption on $A$ is exactly that all (proper) subsets of $A$ have measure $0$. This does not imply that that subset is measurable always. $\endgroup$ Nov 11 '19 at 23:11
  • $\begingroup$ Are you considering any measure? If so, consider the counting measure in $\mathbb{R}$ and consider $A=\{ 0 \} $. Any proper subset of $A$ is the empty set and so has measure 0, but $A$ has measure 1. $\endgroup$
    – Ramiro
    Nov 11 '19 at 23:37
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Any set $A$ is a subset of itself, so this holds. If you rephrase to ask about proper subsets, take $A \setminus \{*\}$ and use subadditivity. A similar result holds in any measure space (or outer measure) where points have 0 measure.

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