# If any subset of a set has zero measure, then the set has zero measure?

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 $$\varepsilon$$.

• Hint: Measure is countably additive, and a singleton has measure 0. Nov 11 '19 at 23:04
• I assume you mean every strict subset otherwise the answer is fairly trivial :P Nov 11 '19 at 23:05
• Just a subtlety (likely about notation): Not necessarily every subset $B \subseteq A$ has zero measure as those don't have to be measurable. Nov 11 '19 at 23:09
• @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. Nov 11 '19 at 23:11
• 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. 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.