Finite Content with Atomfreeness on Infinite Algebra

Question

For an algebra $\mathcal{A}$ (not necessarily a $\sigma$-Algebra) over $X =[0,1]$ we define a finite content to be a function $\mu \colon \mathcal{A} \rightarrow [0,a]$ for some $a \in \mathbb{R}^{>0}$ with

1. $\mu(\emptyset) = 0$, $\mu(X) = a$ and
2. for all $A, B \in \mathcal{A}$ it holds that $\mu(A \cup B) = \mu(A) + \mu(B) - \mu(A \cap B)$.

Furthermore, $\mu$ is said to be atom-free, if for all $x \in X$ it holds that $\mu(\{x\}) = 0$.

Obviously, the power set $\mathcal{P}(X)$ is an algebra over $X$. Given these definitions, I am wondering whether there is any formal argument why either

1. there is no finite, atom-free content $\nu \colon \mathcal{P}(X) \rightarrow [0,1]$ or
2. why one cannot write down such a content $\nu \colon \mathcal{P}(X) \rightarrow [0,1]$ explicitly in a finite amount of space?

Note that I do not demand $\nu$ to satisfy translation invariance.

Answer 1

Given an infinite set $X$, we can pick an arbitrary free ultrafilter $\mathcal F$ on $X$ and define a required content putting for each subset $A$ of $X$ put $\mu(A)=a$, if $A\in\mathcal F$, and $\mu(A)=0$, otherwise.

I think whether the required content can be “constructed” explicitly is a hard question. A similar question for countable $X$ was considered at MO. According to Stefan Geschke

In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$ (giving all singletons measure 0 and $\mathbb N$ measure 1). The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property, and such a set cannot be constructed without some help of the axiom of choice. In other words, there is no explicit "construction" of a non-trivial finitely additive measure.

I guess the same conclusion holds for your question, but I am to weak in set theory to provide less vague and more rigorous claims.