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.
