Non-measurability of ultrafilter on $\omega$ It is well-known that any non-principal ultrafilter on $\omega$ is non-measurable regarded as a subset of $2^{\omega}$. My question is "how well-known" is this fact?
Here is the only proof I know: Let $\mathcal{U}$ be a non-principal ultrafilter on $\omega$. If $\mathcal{U}$ were measurable, then, by Kolmogorov 0-1 law, it would have either measure $0$ or $1$ since it is closed under finite changes containing the Fréchet filter. On the other hand, the bit flipping operation is measure preserving and it maps the ultrafilter to its complement, so $\mathcal{U}$ would have measure $1/2$. Contradiction.
What are some essentially different proofs of this fact? Is there a proof that avoids 0-1 laws?
 A: After seeing Henno’s comment I tracked down the relevant paper by Sierpiński.
Wacław Sierpiński, ‘Fonctions additives non complètement additives et fonctions non mesurables’, Fundamenta Mathematicae $\mathbf{30}$, $96$-$99$ ($1938$), essentially proves the result. Specifically, he shows that if there is a finitely additive, $\{0,1\}$-valued function on $\wp(\Bbb Z^+)$ that is not countably additive, then there is a $\{0,1\}$-valued function on $\Bbb R$ that is not Lebesgue measurable. Note that a function $f:\wp(\Bbb Z^+)\to\{0,1\}$ that is finitely additive but not countably additive defines a free ultrafilter $\mathscr{U}_f=\{U\subseteq\Bbb Z^+:f(U)=1\}$ on $\Bbb Z^+$, and conversely, a free ultrafilter $\mathscr{U}$ on $\Bbb Z^+$ defines such a function $f_{\mathscr{U}}$ by $f_{\mathscr{U}}(U)=1$ iff $U\in\mathscr{U}$. Thus, he shows in effect that if there is a free ultrafilter on $\Bbb Z^+$, then there is a non-measurable $\{0,1\}$-valued function on $\Bbb R$, and it follows easily from the construction that $\mathscr{U}$ is not measurable in the Cantor space.
His argument, translated into English and modern notation, is as follows.

Suppose that $f:\wp(\Bbb Z^+)\to\{0,1\}$ is finitely additive but not countably additive. Suppose that $f(\{p\})=1$ for some $p\in\Bbb Z^+$. If $p\in E\subseteq\Bbb Z^+$, then
$$f(E)=f(\{p\})+f(E\setminus\{p\})=1+f(E\setminus\{p\})\;,$$
so $f(E)=1$. If $p\notin E\subseteq\Bbb Z^+$, then
$$1=f(E\cup\{p\})=f(E)+f(\{p\})=f(E)+1\;,$$
so $f(E)=0$. But then $f$ is countably additive, contrary to hypothesis, so $f(\{n\})=0$ for each $n\in\Bbb Z^+$. Similarly, we must have $f(\Bbb Z^+)=1$, as otherwise $f$ is identically $0$ and hence countably additive.
Now define $\varphi:\Bbb R\to\{0,1\}$ as follows. Each $x\in\Bbb R$ has a unique representation
$$x=\lceil x\rceil-1+\sum_{k\in\Bbb Z^+}\frac{b_k(x)}{2^k}$$
such that each $b_k(x)\in\{0,1\}$, and $B_x=\{k\in\Bbb Z^+:b_k(x)=1\}$ is infinite, and we set $\varphi(x)=f(B_x)$. Suppose that $\varphi$ is measurable. For $i\in\{0,1\}$ let $E_i=\{x\in\Bbb R:\varphi(x)=i\}$; each $E_i$ is measurable, and $\Bbb R=E_0\cup E_1$, so at least one of $E_0$ and $E_1$ must have positive Lebesgue measure (possibly infinite).
Let $x\in\Bbb R$, and let
$$y=-(\lceil x\rceil-1)+\sum_{k\in\Bbb Z^+\setminus B_x}\frac1{2^k}=1-x\;;$$
clearly $y=1-x$. If $\Bbb Z^+\setminus B_x$ is infinite, then $\varphi(y)=f(\Bbb Z^+\setminus B_x)=1-f(B_x)=1-\varphi(x)$. Thus, $\varphi(1-x)=1-\varphi(x)$ for each $x\in\Bbb R$ such that $1-x$ is not a dyadic rational. Thus, $E_1$ differs from $1-E_0$ by a countable set, so $E_0$ and $E_1$ must both have positive Lebesgue measure.
Now let $x\in\Bbb R$ be arbitrary, and let $r$ be a dyadic rational. Then the symmetric difference $B_x\mathrel{\triangle}B_{x+r}$ is finite, and since $f$ vanishes on singletons, $f(B_{x+r})=f(B_x)$, and therefore $\varphi(x)=\varphi(x+r)$. Thus, the sets $E_0$ and $E_1$ are invariant under translation by a dyadic rational. Since they also have positive measure, it follows that $E_0\cap[0,1]$ and $E_1\cap[0,1]$ both have measure $1$, which is impossible, since they are complementary. Thus, $\varphi$ cannot be Lebesgue measurable.

As you can see, it’s basically the same proof, but without explicit appeal to the zero-one law.
