Why do we use open sets for outer measure?

Question

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.

Answer 1

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 $m^i(B)+m^o(A\backslash B)=l(A)$.

Answer 2

It may have something to do with the fact that an open set is the union of countably many open intervals, and thus the outer measure can be defined as the infimum of the measures of open supersets, whereas a closed set is not necessarily the union of countably many closed intervals and points (the Cantor set is a counterexample).