Why do we need axiom of choice to prove that there does not a exist definition of $P(A)$, defined for all subsets $A \subset [0, 1]$

Question

I'm reading "A First Look At Rigorous Probability" by Jeffrey S. Rosenthal. On chapter one there is a proof which I can't fully understand.

Suppose, to the contrary, that $P(A)$ could be so defined for each subset $A \subseteq [0. 1]$. We will derive a contradiction to this. Define an equivalence relation on [0. 1] by: $x~y$ if and only if the difference $y - x$ is rational. This relation partitions the interval $[0. 1]$ into a disjoint union of equivalence classes. Let $H$ be a subset of $[0, 1]$ consisting of precisely one element from each equivalence class (such $H$ must exist by the Axiom of Choice). ....

My question is this: If we assume the Power Set Axiom, why do we need Axiom of Choice here. If the power set of $A$ exists, then its elements also exist. I mean, how can a set contain non-existent objects? And $H$ is clearly in the power set of $A$.

Answer 1

$P(A)$ is not the power set operator. It is the probability operator, I am guessing they mean the standard uniform one.

Indeed, Solovay proved that it is consistent that the probability measures all sets. So the axiom of choice is really needed.

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The point is that just like $\sqrt2$ or $\pi$ are not members of every field, the power set operation is not fully decided by the existence of a power set. In particular, in different set theoretic universes there will be power sets that disagree with each other.

What choice allows you, is to prove that there are non-measurable sets. The fact that there are mathematical universes where all sets are measurable means that without choice it is indeed impossible to prove such sets exists: they don't exist in some models of $\sf ZF$.

Answer 2

Yes, we don't need the Axiom of Choice in order to be able to talk about $\mathcal{P}\bigl([0,1]\bigr)$. But we do need it when, after defining an equivalence relation on $[0,1]$, you want to work with a set with one and only one element of each equivalence class. This is a straightforward application of the Axiom of Choice: we have a family of sets (the equivalence classes) and we want to have a set with one and only one element from each set of that family.