Talk:Hahn decomposition theorem

A precision
It seems that the proof of the theorem relies on the following result. Is it true ?

if $$\mu(A)$$ is finite then the set $$M=\{\mu(E)\;|\;E\in\Sigma \;\; E\subseteq A\}$$ is bounded.


 * More specifically, I think that the mistake is in the claim


 * Also, &epsilon;1 is finite since 0 > &mu;(A) > &minus;&infin;.


 * I will fix it now. Oded (talk) 16:39, 10 June 2008 (UTC)

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)

Why the sum is $$-\infty$$ ?
The end of the proof says that $$\sum_{n=0}^{\infty}\max(s_n/2,-1)=-\infty$$. I don't see why it's true.

Each $$s_n$$ is negative, but they are expected to be increasing because they are infimum of a set which become smaller and smaller since $$N_n$$ is increasing. What prevents having something like $$s_n=-1/n^2$$ ? In that case $$max(s_n/2,-1)=s_n/2=-1/n^2$$ 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 $$ \mu(D_n)\leq \frac{s_n}{2}\leq \frac{\mu(D)}{2} $$, we have

$$\mu(N)\leq \sum_{n\geq 0}\mu(D_n) \leq \sum_{n\geq 0}\frac{\mu(D)}{2}=-\infty$$.

The $$max(s_n/2,-1)$$ is of no use anywhere.

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


 * 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)

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)


 * It is because it would violate the additivity of signed measure. Suppose $$A$$ and $$B$$ are measurable sets with $$\mu(A) = -\infty$$ and $$\mu(B) = +\infty$$. Observe that $$A \setminus B, B \setminus A$$ and $$A \cap B$$ are mutually disjoint. Consider three cases. (Case: $$\mu(A \cap B) = +\infty$$) We have $$\mu(A) = \mu((A \setminus B) \cup (A \cap B))$$ implying $$-\infty = \mu(A \setminus B) + \infty$$ by additivity. Note that the equality can never hold no matter what $$\mu(A \setminus B)$$ is. (Case: $$\mu(A \cap B) = -\infty$$) The argument is similar as before. (Case: $$\mu(A \cap B) = m \in \mathbb{R}$$) We have $$\mu(A) = \mu((A \setminus B) \cup (A \cap B))$$ implying $$-\infty = \mu(A \setminus B) + m$$ by additivity. As a result, $$\mu(A \setminus B) = -\infty$$. Now $$\mu(A \cup B) = \mu((A \setminus B) \cup B) = -\infty + \infty$$ which is undefined. Since all cases lead to a contradiction, we conclude that a signed measure cannot take both $$-\infty$$ and $$+\infty$$ as values. Alexvong1995 (talk) 12:01, 22 December 2018 (UTC)