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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
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 ?
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.
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
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