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?