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Define the the outer measure of a subset of the unit interval as the infimum of the sum of lengths of intervals over all such coverings by intervals. The inner measure of the same subset is one less the outer measure of the complement of the given subset. Then we have a theorem: the inner measure is less than or equal to the outer measure for any subset of the unit interval.

The proof in my book argues by contradiction. It obtains, by definition of outer measure, a covering (by intervals) of the given subset and another covering of its complement such that the sum of the sum of the lengths of the intervals (in each cover) adds less than 1. It then takes the union of such coverings and obtains the contradiction that the unit interval is contained in the new covering, yet the measure of the sums of lengths of the intervals in this new covering adds to less than 1 by the previous statement, a contradiction to a previous theorem (that for elementary sets, or intervals, if a set is contained in a union then the measure of the subset is less or equal to the sum of measures in the interval)

My problem is the theorem contradicted was applicable for only elementary sets i.e. finite disjoint unions of intervals (representation by such a union may not be unique). The union of the coverings obtained at the end of the proof is by no means an elementary set, since the coverings obtained by the def of outer measure are not necessarily disjoint. In fact my book clearly uses the measure defined on intervals for this union, and there's no reason that the union of two intervals (from the two coverings) MUST be an interval.

I understand given a finite union of intervals I can rewrite this as a disjoint union of intervals (by suitably arranging the endpoints, discarding repetitions, taking open intervals between the endpoints and the degenerate singleton sets, where we allowed such to be considered intervals). Then measure is defined for this new union. How is it that the measure of this new disjoint representation must coincide with the "measure" on the original union? The original could be bigger.

Thanks in advance for any clarification or comments, and i apologize for describing everything in plain English, if anyone is averse to that.

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    $\begingroup$ The theorem in question was not applied to the subset of the unit interval being outer and inner measured, but to the unit interval itself. The two coverings are combined to form a covering of the unit interval. Since the unit interval is simple, the theorem applies to it, so the sum over this combined covering must be greater than the measure of the unit interval. $\endgroup$ – Paul Sinclair Nov 26 '15 at 22:06
  • $\begingroup$ I understand that the measure of the unit interval must be less than the sum of measures of the intervals in any cover of the unit interval. But the cover obtained here isn't necessarily of intervals, for the union of two intervals need not be an interval. So using the measure of intervals for this cover is inappropriate, No? With this in mind, I should represent each set in the combined covering as an elementary set and then apply the theorem, or am I being too pedantic? $\endgroup$ – Nap D. Lover Nov 26 '15 at 22:18
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    $\begingroup$ A covering is a collection of sets whose union contains the set being covered. The union is not the covering. The original collection is the covering. The coverings here are collections of intervals. and their union is the collection of all the intervals in the two collections being joined. So we are covering the unit interval with a collection of intervals, not anything more compticated. And there is no requirement that they be disjoint. In fact, that is generally not possible. $\endgroup$ – Paul Sinclair Nov 26 '15 at 23:32
  • $\begingroup$ Ah thank you! I now understand where I went wrong. I thought the new cover was obtained by taking the union of each interval from the other two covers, but instead it is, as you said, the collection that includes both intervals from the original two covers, i.e. the union of the collections not the collection of the union of intervals. This helps a lot! Again, Thank you very much. $\endgroup$ – Nap D. Lover Nov 26 '15 at 23:39

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