*[Finite choice can be proven without AC by induction on the cardinality of . -T.]*

Also, to define inner measure in a reasonable way in this fashion, one would have to insist that the are disjoint or almost disjoint (if one could re-use the same box over and over again, then any set with non-trivial interior would have infinite inner measure.)

]]>For any X < \sum_{n=1}^\infty |B_n|, one can find N such that \sum_{n=1}^N |B_n| \geq X. For instance, if \sum_{n=1}^\infty |B_n| is finite, one can find N such that \sum_{n=1}^N |B_n| \geq \sum_{n=1}^\infty |B_n| – \vvarepsilon.

]]>In the elementary and Jordan measure theory (and Riemann integration), one can see the symmetry such as inner and outer measure, lower and upper integral and the terms “measurable”, “integrable” are defined when the “symmetric” concepts are equal to each other. Can one say that in Lebesgue measure and integration theory, such symmetry does not exist any more (one does have the correct version of the Lebesgue inner measure though, things are note defined as “when lower and upper or inner and outer stuffs match with each other…”) and it is this asymmetry that makes the Lebesgue theory more “powerful” than the Riemann integration theory?

]]>Let for any be the naive definition. I’m trying to show as you said that

.

The direction is obvious since one can replace the finite union with the infinite one using empty sets. I don’t see why is true. I thought Lemma 7 might be useful since one could have the estimate

if one has the elementary set such that

. But I don’t see why such exists.

Could you explain why one can achieve the second inequality?

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