I am seriously struggling in properly getting various steps in the proof of Theorem 1.12.40 (Ulam's Theorem) in Bogachev's first volume on measure theory which concerns the classical statement that assuming the CH we cannot get a 'well-behaved' measure that is countably additive. Incidentally, I feel the points I am not getting are showing a deep misunderstanding of various important issues. Thus, below the theorem with the proof.
Let $X$ be a set of cardinality $\aleph_1$, i.e., $X$ is equipotent to the set of all ordinal numbers that are smaller than the first uncountable ordinal number. Note that $X$ is uncountable and can be well-ordered in such a way that every element is preceded by an at most countable set of elements. The following theorem is due to Ulam [967].
1.12.40. Theorem. If a finite countably additive measure $\mu$ is defined on all subsets of the set $X$ of cardinality $\aleph_1$ and vanishes on all singletons, then it is identically zero.
Proof: It suffices to consider only nonnegative measures. By hypothesis, $X$ can be well-ordered in such a way that, for every $y$, the set $\{x: x < y\}$ is at most countable. There is an injective mapping $x \mapsto f(x,y)$ of this set into $\mathbb{N}$. Thus, for every pair $(x,y)$ with $x < y$ one has a natural number $f(x,y)$. For every $x \in X$ and every natural $n$, we have the set
$$A^{n}_x = \{ y \in X : x<y, \ f(x,y)=n \}.$$
For fixed $n$, the sets $A^{n}_x$, $x \in X$, are pairwise disjoint. Indeed, let $y \in A^{n}_x \cap A^{n}_z$, where $x \neq z$. We may assume that $x < z$. This is, however, impossible, since $x < y$, $z < y$ and hence $f(x,y) \neq f(z,y)$ by the injectivity of the function $f(\cdot,y)$. Therefore, by the countable additivity of the measure, for every $n$, there can be an at most countable set of points $x$ such that $\mu (A^{n}_x) > 0$. Since $X$ is uncountable, there exists a point $x \in X$ such that $\mu(A^{n}_x) = 0$ for all $n$. Hence $A = \bigcup^{\infty}_{n=1} A^{n}_x$ has measure zero. It remains to observe that the set $X \setminus A$ is at most countable, since it is contained in the set $\{y: y \leq x\}$, which is at most countable by hypothesis. Indeed, if $y > x$, then $y \in A^{n}_x$, where $n = f(x,y)$. Therefore, $\mu (X \setminus A) = 0$, which completes the proof
Here there are the problems I have.
Please, observe that in general – for concreteness – I am thinking about this statement with $X := [0,1]$.
1) "By hypothesis, $X$ can be well-ordered in such a way that, for every $y$, the set $\{x: x < y\}$ is at most countable."
Now, take $y = 1$. As a result, $\{ x : x < 1\} = [0 ,1)$, which isn't countable, right?
More in general, which one is the set-theoretic result Bogachev has in mind?
2) "Therefore, by the countable additivity of the measure, for every $n$, there can be an at most countable set of points $x$ such that $\mu (A^{n}_x) > 0$."
I do not understand this statement at all!
Those this mean that actually assuming countable additivity forces the setting to just have countable sets of a certain nature?
(Seriously, here I am completely at loss.)
3) "Since $X$ is uncountable, there exists a point $x \in X$ such that $\mu(A^{n}_x) = 0$ for all $n$."
I see this comes from the previous step, but I rather see that for every $n \in \mathbb{N}$ there exists an $x \in X$ such and such, which is of course weaker.
4) "It remains to observe that the set $X \setminus A$ is at most countable, since it is contained in the set $\{y: y \leq x\}$, which is at most countable by hypothesis."
Now, this should come from the paragraph I wrote down that comes right before the statement of the proof. The question is: where can I find a proper statement of this remark made by Bogachev?
Sorry for the vivisection of this – I can imagine – rather straightforward proof, but I have the feeling that it is quite important to catch every step here, since it could really help me improve my understanding of measure theory (and set theory).
Thank you for your time and patience.
As always, any feedback is most welcome!
