First , the definitions I work with –
A set $A$ has measure zero if for any $ϵ>0$ there are open sets ${S_i},i∈\mathbb{N}$ such that $A\subset \bigcup ^∞ _{i=0}S_i $ and $∑^∞_{i=0}vol(S_i)<ϵ$.
A set $A$ has a volume zero if the there is such a finite cover.
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I need to prove that compact set $A \subset \mathbb{R}^n$ has measure zero iff it has volume zero.
My attempt-
If $A$ has measure zero, then there exsits such countable cover.Also, $A$ is compact,so there is a finite cover for the mentioned above cover,which is the required cover for volume zero.
If $A$ has a volume zero, then there is a finite cover, which is countable, and we are done.
Is it correct???
Best Answer
Yes, your proof is correct. One implication is trivial (a finite cover is countable). For the other one, if $(S_i)_{i \in \Bbb N}$ is an open cover with $\sum _{i \in \Bbb N} vol (S_i) < \epsilon$, then by compactness we may extract a finite subcover $(S_{i_j})_{1 \le j \le n}$ with $\sum _{i =1} ^n vol (S_{i_j}) < \sum _{i \in \Bbb N} vol (S_i) < \epsilon$.