Let

- $ (\Omega,\mathcal A)$ be a measurable space
- $ E$ be a $ \mathbb R$ -Banach space
- $ \mu:\mathcal A\to E$ with $ \mu(\emptyset)=0$ and $ $ \mu\left(\biguplus_{n\in\mathbb N}A_n\right)=\sum_{n\in\mathbb N}\mu(A_n)\tag1$ $ for all disjoint $ (A_n)_{n\in\mathbb N}\subseteq\mathcal A$

Now, let $ $ |\mu|(A):=\sup\left\{\sum_{i=1}^n\left\|\mu(A_i)\right\|_E:n\in\mathbb N\text{ and }A_1,\ldots,A_n\in\mathcal A\text{ are disjoint with }\biguplus_{i=1}^nA_i\subseteq A\right\}$ $ for $ A\in\mathcal A$ . I’ve read in a lecture note that $ |\mu|$ is a measure on $ (\Omega,\mathcal A)$ , if $ |\mu|(\Omega)<\infty$ . What goes wrong if $ |\mu|(\Omega)=\infty$ ?

Let $ (A_n)_{n\in\mathbb N}\subseteq\mathcal A$ be disjoint and $ A:=\biguplus_{n\in\mathbb N}A_n$ . First of all, it’s easy to see (and that’s even stated in that lecture note) that $ $ \sum_{n\in\mathbb N}|\mu|(A_n)\le|\mu|(A)\tag2\;.$ $ So, the problem must occur in the proof of the other inequality:

- Let $ k\in\mathbb N$ and $ B_1,\ldots,B_k\in\mathcal A$ be disjoint with $ $ \biguplus_{i=1}^kB_i\subseteq A\tag3$ $
- Then, $ (A_n\cap B_i)_{n\in\mathbb N}$ is disjoint with $ $ \biguplus_{n\in\mathbb N}(A_n\cap B_i)=B_i\tag4$ $ for all $ i\in\left\{1,\ldots,k\right\}$ and $ A_n\cap B_1,\ldots,A_n\cap B_k$ are disjoint with $ $ \biguplus_{i=1}^k(A_n\cap B_i)\subseteq A_n\tag5$ $ for all $ n\in\mathbb N$
- Thus, \begin{equation}\begin{split}\sum_{i=1}^k\left\|\mu(B_i)\right\|_E&=\sum_{i=1}^k\left\|\mu\left(\biguplus_{n\in\mathbb N}(A_n\cap B_i)\right)\right\|_E=\sum_{i=1}^k\left\|\sum_{n\in\mathbb N}\mu(A_n\cap B_i)\right\|_E\&\le\sum_{i=1}^k\sum_{n\in\mathbb N}\left\|\mu(A_n\cap B_i)\right\|_E\&=\sum_{n\in\mathbb N}\sum_{i=1}^k\left\|\mu(A_n\cap B_i)\right\|_E\le\sum_{n\in\mathbb N}|\mu|(A_n)\end{split}\tag6\end{equation}
- $ (6)$ should immediately yield $ $ |\mu|(A)\le\sum_{n\in\mathbb N}|\mu|(A_n)\tag7$ $

By $ (2)$ and $ (7)$ we obtain the $ \sigma$ -additivity of $ |\mu|$ . Hence, $ |\mu|$ is a measure (clearly, not a finite one, but that wasn’t claimed). So, is there anything wrong in my proof?