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, ε₁ is finite since 0 > μ(A) > −∞.

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)

Without the maximum, it is harder to see that the $D_{n}$ exist when $s_{n}=-\infty$ 18:30, 18 May 2025 (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)