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The following question is exercise 134Xi in D. Fremlin's Measure Theory volume 1. It is a "Basic exercise". Usually, basic exercises are basic indeed and not too difficult. I stumble on this one, however.

Recall that a set $A \subset [0,1]$ has full outer measure if any (Lebesgue-)measurable $F \subset [0,1] \setminus A$ is negligible (i.e. the outer Lebesgue measure of $A$ is 1). Prove that there is a disjoint sequence $(A_n)$ of subsets of $[0,1]$, all of full outer measure.

Note that 134D (in the book quoted above) is the statement that there is some subset $C$ of the real line such that both $C$ and and its complement are of full outer measure. Of course we are trying to improve on this. I am not able to modify the proof of 134D in order to achive this, though.

[Also, I am aware of the fact that we can actually find a non-countable family of disjoint subsets of outer measure, but I would like to see an elementary (i.e. "basic") proof of the weaker result above.]

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  • $\begingroup$ I don't know what would be appropriate background to use, but you can find the countably infinite case discussed in the last answer (and its comments) in the math overflow question Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure?. $\endgroup$ – Dave L. Renfro May 21 '14 at 21:18
  • $\begingroup$ Incidentally, I believe the earliest mention in the literature that it is possible to partition $[0,1]$ into countably infinitely many pairwise disjoint sets, each with outer Lebesgue measure $1,$ is pp. 243-244 of Edward B. Van Vleck's 1908 Trans. Amer. Math. Soc. paper On non-measurable sets of points, with an example. $\endgroup$ – Dave L. Renfro May 21 '14 at 21:24
  • $\begingroup$ Yes I've read that MO thread and comments. According to Gerald Edgar, the Vitali construction may be used to answer the problem but I don't see how. This article of Van Vleck looks interesting but not quite easy to read. Anyway, thank you Mr Renfro. $\endgroup$ – timofei May 22 '14 at 20:03
  • $\begingroup$ I looked some more and found this math overflow question -- Vitali sets of full outer measure -- by googling "vitali set" and "full measure" together. $\endgroup$ – Dave L. Renfro May 22 '14 at 20:12
  • $\begingroup$ This construction is not really elementary but maybe Fremlin has higher expectations on his reader than I first thought. I'll try to understand that. Thank you. $\endgroup$ – timofei May 23 '14 at 15:48

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