The standard definition for outer measure of a set of real numbers $A$ is:

$$ m*(A) = inf {\Large \{} \sum_{k=1}^{\infty} \ell(I_k) \; {\Large |} \; A \subseteq \bigcup_{k=1}^{\infty} I_k {\Large \}} $$

(as, for example, in Royden, Real Analysis, 4th ed., p. 31) where the $I_k$ are required to be nonempty, open, bounded intervals.

My question is: Would it amount to the same thing if we instead required that the $I_k$ are nonempty, closed, bounded intervals? It feels like there should be some reason why inner measure uses closed sets and outer measure uses open sets, but I can't think of an example set $A$ where $m*(A)$ would differ from $m**(A)$ defined exactly as above except that each $I_k$ is required to be closed.


If we replace $I_k$ with $\bar I_k$ then $l(I_k)=l(\bar I_k).$ On the other hand if $ G=\{J_k :k\in N\}$ is a set of closed bounded intervals with $A\subset \cup G,$ then for any $e>0$ we can cover each $J_k$ with a bounded open interval $I_k$ with $l(I_k)<e 2^{-k}+l(J_k)$. So $m*=m**$. It is a matter of convenience. For example with the usual def'ns of outer measure $m^o$ and inner measure $m^i$ it is fairly obvious that if $A$ is a bounded interval and $B\subset A$ then $m^i(B)+m^o(A\backslash B)=l(A)$.

  • $\begingroup$ Your argument shows that my $m*(A)$ and $m**(A)$ take the same values, so that's a big part of the answer I was looking for. I guess using open for outer and closed for inner makes it more difficult to fit. So, it becomes non-trivial that a countable set of points has measure 0. If we used closed sets for outer measure, it would be immediate because we just take each point $a$ as covered by $[a, a]$. And it would follow immediately that the rationals have measure 0. Also not sure whether it's relevant, but Heine-Borel gives finite subcovers only for collections of open sets. $\endgroup$ – Marshall Farrier Dec 19 '15 at 3:26

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