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Talk:Hahn decomposition theorem

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A precision

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It seems that the proof of the theorem relies on the following result. Is it true ?
if is finite then the set is bounded.

More specifically, I think that the mistake is in the claim
Also, ε1 is finite since 0 > μ(A) > −∞.
I will fix it now. Oded (talk) 16:39, 10 June 2008 (UTC)[reply]

Use of min/max in proof

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I don't understand the use of min(t_n/2,1) instead of just t_n, nor the use of max(s_n/2,-1) instead of s_n/2 in the proof of the Hahn decomposition theorem. — Preceding unsigned comment added by 77.8.166.226 (talk) 15:41, 7 November 2014 (UTC)[reply]


Why the sum is  ?

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The end of the proof says that . I don't see why it's true.

Each is negative, but they are expected to be increasing because they are infimum of a set which become smaller and smaller since is increasing. What prevents having something like  ? In that case whose series converges.


The same is at many other places:

- The italian wikipedia : https://it.wikipedia.org/wiki/Teorema_di_decomposizione_di_Hahn

- In an french course : https://www.imo.universite-paris-saclay.fr/~joel.merker/Enseignement/Integration/abstraite-integration.pdf


For me, a correct way to write that part of the proof is:

Since , we have

.

The is of no use anywhere.


Laurent.Claessens (talk) 06:09, 23 July 2023 (UTC)[reply]

I agree that the maximum isn't used anywhere.
For the other question, I'm not sure I understand what your problem is. The article does exactly what you present as a correct proof. (up to the max thing) GreatLeaderMayonnaise (talk) 18:39, 4 January 2024 (UTC)[reply]

Proof of the Hahn decomposition theorem

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The proof seemed to rely on the assumption that if μ takes on the value -∞, it doesn't take the value +∞. Why is this assumption justifiable? Kerry (talk) 16:08, 2 October 2015 (UTC)[reply]

It is because it would violate the additivity of signed measure. Suppose and are measurable sets with and . Observe that and are mutually disjoint. Consider three cases. (Case: ) We have implying by additivity. Note that the equality can never hold no matter what is. (Case: ) The argument is similar as before. (Case: ) We have implying by additivity. As a result, . Now which is undefined. Since all cases lead to a contradiction, we conclude that a signed measure cannot take both and as values. Alexvong1995 (talk) 12:01, 22 December 2018 (UTC)[reply]