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This result is taken from Billiingsley, // Probability and measure //. 
====== Statement ======

There exists no probability measure $P$ such that $P(x) = 0$ for each $x \in \mathbb{R}$. Since $A(x) = 0$, this implies that it is impossible to extend $A$ to $\mathbb{R}$ at all.

====== Proof ======

The proof of this second impossibility theorem requires the well-ordering principle (equivalent to the axiom of choice) and also the continuum hypothesis. Let $S$ be the set of sequences $(s(1), s(2), \ldots)$ of positive integers. Then $S$ has the power of the continuum. (Let the $n$th partial sum of a sequence in $S$ be the position of the $n$th $1$ in the nonterminating dyadic representation of a point in $(0, 1]$; this gives a one-to-one correspondence.) By the continuum hypothesis, the elements of $S$ can be put in a one-to-one correspondence with the set of ordinals preceding the first uncountable ordinal. Carrying the well-ordering of these ordinals over to $S$ by means of the correspondence gives $S$ a well-ordering relation $<_w$ with the property that each element has only countably many predecessors.

For $s, t \in S$, write $s < t$ if $s(i) < t(i)$ for all $i > 1$. Say that $t$ rejects $s$ if $t <_w s$ and $s < t$; this is a transitive relation. Let $T$ be the set of unrejected elements of $S$. Let $V_s$ be the set of elements that reject $s$, and assume it is nonempty. If $t$ is the first element (with respect to $<_w$) of $V_s$, then $t \in T$ (if $t'$ rejects $t$, then it also rejects $s$, and since $t <_w t'$, there is a contradiction). Therefore, if $s$ is rejected at all, it is rejected by an element of $T$.

Suppose $T$ is countable, and let $t_1, t_2, \ldots$ be an enumeration of its elements. If $t^*(k) = t_k(k) + 1$, then $t^*$ is not rejected by any $t_k$ and hence lies in $T$, which is impossible because it is distinct from each $t_k$. Thus, $T$ is uncountable and must, by the continuum hypothesis, have the power of $(0, 1]$.

Let $x$ be a one-to-one map of $T$ onto $(0, 1]$; write the image of $1$ as $x_1$. Let $A_k = \{x : t(i) = k\}$ be the image under $x$ of the set of $t$ in $T$ for which $t(i) = k$. Since $t(i)$ must have some value $k$, $\bigcup_{k=1}^{\infty} A_k = (0, 1]$. Assume that $P$ is countably additive and choose $u$ in $S$ in such a way that $P\left(\bigcup_{k=1}^{\infty} A_k\right) > 1 - \frac{1}{2^i+1}$ for $i > 1$. If
\[
A = \bigcap_{k=1}^{\infty} A_k = \bigcap_{k=1}^{\infty} \{x : t(i) < u(i)\} = \{ x : t < u \},
\]
then $P(A) > 0$. If $A$ is shown to be countable, this will contradict the hypothesis that each singleton has probability $0$.

Now, there is some $t_0$ in $T$ such that $u < t_0$ (if $u \in T$, take $t_0 = u$; otherwise, $u$ is rejected by some $t_0$ in $T$). If $t < u$ for $t \in T$, then $t < t_0$, and hence $t \not\leq_w t_0$ (since otherwise $t_0$ rejects $t$). This means that $\{t : t < u\}$ is contained in the countable set $\{t : t \leq_w t_0\}$, and $A$ is indeed countable.

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