The countably additive part of ba is ell-one

Let $\nu$ be countably additive and put $a_n=\nu (\{n\})$. For every $N$, [given] the singleton partition of $\{0,\dots ,N\}$ gives ${\sum }_{n=0}^N\mathrm{\mid }{a}_n\mathrm{\mid }\le \mathrm{\mid }\nu \mathrm{\mid }(\mathbb{N})$. Thus $(a_n)\in {\mathrm{\ell }}^1$. Countable additivity applied to $A={⨆}_{n\in A}\{n\}$ gives the displayed formula.

Conversely, if $(a_n)\in {\mathrm{\ell }}^1$, absolute convergence makes [given, step 1.1] ${\nu }_a(A)={\sum }_{n\in A}{a}_n$ independent of enumeration and countably additive; also $\mathrm{\mid }{\nu }_a\mathrm{\mid }(\mathbb{N})={\sum }_n\mathrm{\mid }{a}_n\mathrm{\mid }<\mathrm{\infty }$. This proves the equivalence and linearity of the subspace.

Use [A1] exactly through [L1], and let $L$ be the resulting mean. By [L2], [given, A1, L1, L2, step 2.1] $\nu (A):=L({\mathbf{1}}_A)$ is a charge. Shift invariance makes all singleton masses equal because $S{\mathbf{1}}_{\{n+1\}}={\mathbf{1}}_{\{n\}}$. Moreover $S{\mathbf{1}}_{\{0\}}=0$, so shift invariance and linearity give $\nu (\{0\})=L({\mathbf{1}}_{\{0\}})=L(0)=0$. Hence $\nu (\{n\})=0$ for every $n$ but $\nu (\mathbb{N})=L(\mathbf{1})=1$, so it is not countably additive. The subspace is proper.