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 $\varepsilon$.

  • $\begingroup$ 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

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.


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.