# Is the proof of the Hahn decomposition on Wikipedia valid?

I am trying to understand the proof of the Hahn decomposition theorem on Wikipedia.

The proof has two steps. First proving the given claim on the site. There it says that $$t_n=\sup\{\mu(B): B\in\Sigma,\, B\subset A_n\}$$ might a priori be $\infty$. Why does this still give us a set $\mu(B_{n})$ which satisfies the condition $$\mu(B_n)\ge \min\{1,t_n/2\}$$ and $$\mu(B_n) < \infty$$?

• As far as I can see, WP only claims that $\mu(B_n)\ge \min\{1,t_n/2\}$, but not that $\mu(B_n)<\infty$ – Hagen von Eitzen May 6 '17 at 11:25
• Okay, but then still, how could such a $\mu(B_{n}$ exist? I know that every bounded set in $\mathbb{R}$ has a convergent subsequence to its supremum. But if here $t_{n}= \infty$ the set is not bounded so how should such a $B_{n}$ exist? – vaoy May 6 '17 at 11:29
If $t_n=\sup=\infty,$ then the set of real numbers $\{\mu(B):\,\ldots\}$ is not bounded from above and so for every $k\in \mathbb{N}$ you can find an admissible set $C_k\subset A_n$ such that $\mu(C_k)\ge k$. If you take $B_n=\cup_k C_k$, then $B_n\subset A_n$ and $\mu(B_n)\ge \mu(C_k)\ge k$ for every $k$ so $\mu(B_n)=\infty\ge \min\{1,\infty\}=1$. Note that $\mu(D)$ is finite so what WP writes after is fine if the series diverges.
• But dovergig to $- \infnty$ is not allowed. That is the contradiction they conclude at the end of the proof of the claim. – vaoy May 6 '17 at 15:45
• $\mu$ cannot take value $-\infty$ but can take value $\infty$. – Gio67 May 6 '17 at 16:01
• signed measures cannot take both values $-\infty$ and $\infty$ otherwise $\mu(A)+\mu(B)$ would make no sense. You either allow them to take value $-\infty$, in which case $\infty$ is excluded, or viceversa. If you see at the very beginning of the proof in WP, it excludes the case $-\infty$, while $\infty$ is allowed. – Gio67 May 6 '17 at 18:00