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According to Wikipedia, the definition of the Lebesgue outer measure of a set $E$ is as follows:

$$ \lambda^*(E) = \operatorname{inf} \left\{\sum_{k=1}^\infty l(I_k) : {(I_k)_{k \in \mathbb N}} \text{ is a sequence of open intervals with } E\subseteq \bigcup_{k=1}^\infty I_k\right\} $$

Suppose we remove the countability requirement: $$ \lambda^*(E) = \operatorname{inf} \left\{\sum_{I \in X} l(I) : X \text{ is a set of open intervals with } E\subseteq \bigcup_{I \in X} I\right\} $$

(Where we could just define $\sum_{I \in X} l(I)$ to be $\operatorname{sup} \left\{\sum_{I \in Y} l(I) : Y \text{ is a finite subset of } X \right\}$.)

Why would this not work? What would break if we remove the countability requirement?

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If $\sum_{I \in X} l(I)=\infty$, then there is a countable subset of $X$ that sum up to infinity too.

If $\sum_{I \in X} l(I)<\infty$, then only countably many terms of $l(I)$ is non-zero.

In either case, it'd just be the same as allowing only countable infinite terms.

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    $\begingroup$ Do you have a proof of why your second claim is true? $\endgroup$ – user98404 Aug 25 '16 at 14:56
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    $\begingroup$ Suppose $\Sigma_{I \in X} l(I) = S < \infty$. There are at most 2 terms of the sum with $l(I) \ge S/2$. There are at most $3$ terms with $l(I) \ge S/3$. There are at most $4$ terms with $l(I) \ge S/4$. And so on... $\endgroup$ – Lee Mosher Aug 25 '16 at 15:05
  • $\begingroup$ @chell Yes I do. It's kind of a common knowledge but I'll find you the link. $\endgroup$ – BigbearZzz Aug 25 '16 at 15:05
  • $\begingroup$ @chell Here's some more detailed answers to that question of yours math.stackexchange.com/questions/70194/… and math.stackexchange.com/questions/1413874/…. $\endgroup$ – BigbearZzz Aug 25 '16 at 15:08
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It will work. The thing is, if $X$ is uncountable the sum is necessarily infinite. Thus only countable families $X$ will give approximations to the infimum, if it is indeed finite.

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As long as you are working with the reals, keep in mind that there is a countable basis of open sets for the topology, specifically the collection of open intervals with rational endpoints. Any open set in the topology can be written as a union of sets from this basis. So your set $\cup_{I\in X} I$ can be written as a countable union (since there are only countably many sets in the basis) of open intervals (probably different from the intervals in $X$) from the basis. In the definition, the $\inf$ must be realized by a countable union of open intervals, because any union of open intervals (in fact any open set) is equivalent to a union of open intervals from this countable basis, so necessarily a countable union. As others have already pointed out, this doesn't work in more general spaces, hence the need to be specific in the definition, but for the reals, there is no need for more than a countable collection of intervals.

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    $\begingroup$ Note that your proposal doesn't always work in every second-countable spaces because while it is true that every open set can be written as a union of countable bases, the bases need not be disjoint. $\endgroup$ – BigbearZzz Aug 25 '16 at 15:23
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The answer is much simpler that given above. Firstly, the sum of a collection of nonnegative reals (even uncountable) is defined as the supremum of its finite `sub-sums.' However, in your comment:

(Where we could just define $\sum_{I \in X} l(I)$ to be $\sup\{\sum_{I \in Y} l(I):\ Y \text{ is a finite subset of } X\}$.)

the point is that the collection $\{I \in Y\}$ in general does not cover the set $E$, so, the corresponding finite sum does not appear in ``your'' version of an outer measure.

Secondly, if the set $E$ is unbounded, the only way you can cover it by a finite number of interval is when one of them is unbounded - but then the sum of their lengths will equal $\infty$. So, all unbounded sets (e.g. the integers) have infinite your outer measure.

Finally, it is not too difficult to verify that for a bounded set $E$, your outer measure of $E$ coincides with the ('classical') Lebesgue measure of the topological closure of $E$ (this closure is compact in that case).

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  • $\begingroup$ Copied from an "answer": Sorry, my previous answer deals with the version of `finite' collections of intervals - I haven't seen earlier that the discussion is about arbitrary collections. Of course, the above answers are correct. I'm sorry for my oversight. $\endgroup$ – quid Aug 25 '16 at 19:23

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