Let us consider signed charges, these are finitely additive signed measures (I suppose this would also work with sigma additive signed measures). We work on a measure space $(\Omega, \mathcal{A})$, where in the case of signed charges, $\mathcal{A}$ is an algebra and in case of signed measures $\mathcal{A}$ is a sigma-algebra. For breveity, I will just talk about signed charges but you can think about signed measures as well.
Now, there seem to be some conflicting definitions of what it means for such a signed charge to be bounded:
(i) A signed charge $\nu$ is bounded if $\nu(A)<\infty$ for all $A \in \mathcal{A}$. This is supposed to be equivalent to the condition $|\nu(\Omega)|<\infty$ and this in turn is supposed to be equivalent to $|\nu|(\Omega) < \infty$ (where $|\nu|$ is the total variation). Sometimes this property is also called finite instead of bounded. In Dunford and Schwartz the last condition is also called bounded variation.
(ii) A signed charge $\nu$ is bounded if $\sup\{\nu(A): A\in\mathcal{A}\}<\infty$ (Theory of charges Def.2.1.1. This should be equivalent to the existence of some $M$ such that $|\nu(A)|<M$ for all $A\in\mathcal{A}$ (This condition is used in a proof of Dunford and Schwartz for Lemma 5 in III)
(iii) A signed charge $\nu$ is bounded if it takes only values in the real numbers.
My questions are now:
(i) and (ii) are not equivalent right? But it seems that (ii) implies (i)?
Is (iii) equivalent to (ii)?
The space $ba(\mathcal{A})$ is defiend as the space of all finitely additive bounded signed measures, i.e., all bounded signed charges. which of the above Definitions of boundedness is used here?
There is an example in Theory of Charges (an Infinite Dimensional Analysis) that I am not sure I understand:
Let $\Omega = \{1,2,3,...\}$ and $\mathcal{A}=\{A\subset \Omega : A \text{ or } A^{c} \text{ finite} \}$ and $\mu$ on $\mathcal{A}$ defined by $\mu(A)=n$, if $A$ is finite and has $n$ element, $\mu(A)=-\mu(A^{c})$, if $A^{c}$ is finite. Then $\mu$ is a real charge on $\mathcal{A}$ but not bounded.
- So my take here is that the charge in this example fulfills property (i) since $\mu(\mathbb{N})=\mu(\emptyset)=0$ (and the measure is finite on all other measurable sets, too). However, it is not bounded according to Definition (ii). Is that take correct?