Is the Jordan content a pre-measure? I am currently dealing with the theory of the Jordan content $\iota: \mathcal{J}(\mathbb{R}^n) \rightarrow [0,\infty]$ where $\mathcal{J}(\mathbb{R}^n)$ denotes the ring of Jordan-measurable sets. I asked myself the question, whether it is a pre-measure. Let $(A_k) \in \mathcal{J}(\mathbb{R}^n)^{\mathbb{N}}$ be a disjoint set-sequence such that $\biguplus_{k=1}^\infty A_k \in \mathcal{J}(\mathbb{R}^n)$.
From finite additivity and monotonicity of the Jordan content, we obtain:
$$\sum_{k=1}^n \iota(A_k)=\iota \left( \biguplus_{k=1}^n A_k \right) \leq \iota \left( \biguplus_{k=1}^\infty A_k \right)$$
and thus $\sum_{k=1}^\infty \iota(A_k) \leq \iota( \biguplus_{k=1}^\infty A_k)$.
Unfortunately I have no clue how to prove the other ineqality (or provide a counterexample) and would appreciate any hint you could give me.
 A: Recall that to have a pre-measure, you must have two structures: a Boolean algebra (which is like $\sigma$-algebra, but only with closedness under finite unions) and a pre-measure itself.
Definition (Pre measure) Let $\mathcal{B}_0$ be a Boolean algebra. A pre-measure $\mu_0$ on $\mathcal{B}_0$ is a finitely additive measure
$$\mu_0: \mathcal{B}_0 \to [0,+\infty]$$
with the property that


*$\mu_0 \left(\bigcup_{n=1}^{\infty} E_n\right) = \sum_{n=1}^{\infty} \mu_0(E_n)$ whenever $(E_n)_{n\in\mathbb{N}}$ are disjoint sets in $\mathcal{B}$ such that $\bigcup_{n=1}^{\infty} E_n$ is in $\mathcal{B}_0$.

Now for Jordan content we have a Jordan algebra, i.e., the set of all Jordan measurable sets $\mathcal{J}$ satisfies our Boolean algebra axioms. But is Jordan content $\sigma$-additive, if the Boolean algebra allows for it?
Consider the set of rational numbers $q \in \mathbb{Q}\cap [0,1]$. Then each $q$ is Jordan measurable, yet the whole set is too entangled with $[0,1]$ so that the Jordan measure digresses to 1 instead of 0 thus damaging the $\sigma$-additivity (see this answer for a more rigorous treatment).
Thus, although Jordan measurable sets define a Boolean algebra, the Jordan measure itself fails to be $\sigma$-additive.
