*[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.

]]>