Why "countability" in definition of Lebesgue measures?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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).