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.


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???

  • 1
    $\begingroup$ By cover do you mean open cover? $\endgroup$
    – Umberto P.
    Jan 1 '17 at 20:33
  • 1
    $\begingroup$ What are the $S_i?$ $\endgroup$
    – zhw.
    Jan 1 '17 at 20:51
  • $\begingroup$ @UmbertoP. a cover with open sets $\endgroup$
    – ChikChak
    Jan 1 '17 at 21:01
  • $\begingroup$ @zhw. Question edited $\endgroup$
    – ChikChak
    Jan 1 '17 at 21:03

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 $\sum _{i =1} ^n vol (S_{i_j}) < \sum _{i \in \Bbb N} vol (S_i) < \epsilon$.


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