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Can one show that Lebesgue measure is $\sigma$-additive on using only its definition which is $\lambda^n ([a,b))= \prod_{i=1}^n(b_i-a_i)$ and the fact that the set of semi-open boxes form a semi-ring? I saw a proof in a book which relies way too heavily on handwavery which isn't particularly helpful as I'm self studying the material.

user63697
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  • σ-additivity is more like an axiom which we want our measure to obey. The definition you quote only defines the Lebesgue measure of $n$-boxes whose sides are parallel to the coordinate axes, as it seems. But the requirement that our measure shall be σ-additive leads to the measure being defined on a much richer category of shapes, for example solid polyhedra and balls, and all open sets, and more. – Jeppe Stig Nielsen Jan 07 '14 at 15:28
  • @JeppeStigNielsen the book I'm going through uses Caratheodory's theorem to extend lebesgue measure, so all one needs to achieve that is to show $\lambda$ is a pre-measure and the set of $n$-boxes is a semi-ring . The theorem then guaranties the existence of an extension. – user63697 Jan 07 '14 at 15:35

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One defines $\lambda^n$ on intervals as shown, then extends by finite additivity to finite disjoint unions of intervals, which form a semi-ring. To continue the extension, we need to show that result is "countably additive"..., that is, if an element $B$ of the sigma-ring happens to equal a countable disjoint union of other elements $A_k$, then the series converges to the value: $\lambda^n(B) = \sum_k \lambda^n(A_k)$. So first reduce to proving the case where $B$ is an interval and the sets $A_k$ are intervals. Then the proof will proceed (using the Heine-Borel theorem) by reducing to the case of a finite union. And that last case has to be done by some combinatorics.

GEdgar
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  • The proof you are mentioning is the one I read however it made no sense and I felt like it had many gaps, here is the question I posed about it:http://math.stackexchange.com/questions/630262/about-the-proof-that-lebesgue-measure-is-a-premeasure Is the proof you are talking about the same one? – user63697 Jan 07 '14 at 16:10
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    That's the one. To write it out in enough detail for beginners takes a few pages. And the Heine-Borel step is important, since all the rest would work in $\mathbb Q^n$, where this result fails. – GEdgar Jan 07 '14 at 16:21
  • Do you happen to know a place were I could find it(a detailed proof)? I've searched several times without any results. – user63697 Jan 07 '14 at 16:29
  • Maybe Royden, Real Analysis. It's in its 4th edition, so maybe has the arguments in smoothed out form. – GEdgar Jan 07 '14 at 17:12
  • Found it! Section 20.2. Thank you! – user63697 Jan 07 '14 at 17:30
  • Minor remark: To obtain a semi-ring, you do not actually have to extend to the finite disjoint unions of intervals; it suffices to take all intervals of the form $[a,b)$. – PhoemueX Feb 24 '15 at 19:54