From this question and this question (and their answers) I gather that it is consistent with ZF without The Axiom of Choice to assume that there exist countable sets $A_n$, $n\in \mathbb N$, such that $\mathbb R=\bigcup_{n\in \mathbb N} A_n$. Given any positive measure $\mu$ on the Borel $\sigma$-algebra of $\mathbb R$ which satisfies $\mu(\{x\})=0$ for all singletons $\{x\}$, this implies $$ \mu(\mathbb R)=\sum_{n\in \mathbb N} \mu(A_n)=0. $$ In particular Lebesgue measure cannot exist. However, when I looked through a construction of Lebesgue measure, I was not able to pinpoint any particular argument that (could not be refined to an argument that) required any choice. This has left me a bit puzzled
My question is if I have misunderstood the situation, or if indeed some amount of choice is necessary to construct Lebesgue measure on $\mathbb R$. In the second case, I would be very interested in knowing 'how much' choice is needed.
That is correct, if the real numbers are a countable union of countable sets, then measure theory as we know it goes out the window. The same can be said if the real numbers are a countable union of countable union of countable sets, and so on. Essentially for the same reason.
To your question, the Lebesgue measure is the completion of the Borel measure. So we first need to ensure that the Borel measure is non-trivial. But if each singleton gets measure $0$, and the countable union of measure zero sets is a measure zero set, then it is impossible to have any set of positive measure which is:
and of course the list can proceed, but in the case $\Bbb R$ already satisfies the third criteria, there's no point really.