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Does there exist a $\sigma$-algebra $A \supset M$ (with a measure $m$ defined on it) such that

  1. $m(I) = L(I)$ where $I$ is any interval in $\Bbb{R}$ and $L(I)$ means the length of the interval,
  2. $m$ is $\sigma$-additive.
  3. $m$ is translation invariant,

where $M$ is the set of all measurable sets with respect to Lebesgue outer measure (i.e. the usual class of Lebesgue-measurable sets).

PhoemueX
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Sushil
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  • If you've already developed Lebesgue measure on the Lebesgue $\sigma$-algebra, have you tried showing that it satisfies the desired properties? Or are you asking for $A$ to be strictly finer than $M$? – Josh Keneda Sep 10 '14 at 15:43
  • @JoshKeneda Yes I have seen the proof that it satisfies all these properties. – Sushil Sep 10 '14 at 15:44
  • An excellent expository paper on this is: Andrew Michael Bruckner and Jack Gary Ceder, On improving Lebesgue measure, Nordisk Matematisk Tidskrift 23 #2 (1975), 59-68. – Dave L. Renfro Sep 10 '14 at 15:46
  • @DaveL.Renfro I am not able to find the article. May you please help. – Sushil Sep 10 '14 at 16:02
  • I don't have time to research and write up an answer now (I'm at work), but this is something I'm sure someone else will answer anyway. I gave the reference because it's not very well know and in my opinion should be better known. Most every university library (in the U.S., at least) has this journal, and it's probably even more readily available in Europe. Unfortunately, I don't have digital access to this journal and I only have an off-print copy of the paper (which is at home). In any event, you should find more than you need by googling the phrase "extending Lebesgue measure". – Dave L. Renfro Sep 10 '14 at 16:11
  • A similar question was asked here: https://math.stackexchange.com/questions/209532/extension-of-the-lebesgue-measurable-sets

    The top answer affirms the existence of such an $A$.

     – Josh Keneda Sep 10 '14 at 18:24
  • I had a few moments free just now, so I thought I would look here to see if others have added anything. Also, just now I also noticed that you are in India, so I guess my comments about U.S. and European libraries does not help much! :) If you are really interested in this topic, then I can send you a photocopy of the Bruckner/Ceder paper by postal mail. If you want me to send it to you, send me your postal address to my email, which is at my profile. I do not have easy access to a scanner, so photocopy + postal mail is easiest for me. – Dave L. Renfro Sep 10 '14 at 18:40
  • @JoshKeneda thanks for reference it cleared a lot of doubts. – Sushil Sep 11 '14 at 15:13

1 Answers1

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This was first done by Szpilrajn (1935) and later strengthened to non separable extensions by Kakutani and Kodaira (1950). Ciesielski gives a short proof of the non existence of any maximal such extension here.

hot_queen
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