In this question, it has been asked whether the fact that the countable union of Lebesgue null sets is null, uses the Axiom of Choice or not. My opinion is that it is not needed. Here's a proof of this fact taken from my class notes:
Take a sequence $A_1,A_2,\ldots$ of Lebesgue null sets. Let $\varepsilon>0$. Define $$A:=\bigcup_{n=1}^\infty A_n.$$ Since the sets $A_n$ are null, for each $n\in\mathbb{N}$ there exists a sequence of intervals $\{I_{nm}\}_m$ such that $$A_n\subset \bigcup_{m=1}^\infty I_{nm}\quad\textit{and}\quad\sum_{m=1}^{\infty}|I_{nm}|<\frac{\varepsilon}{2^n}.$$ ($|\cdot |$ means measure of the interval). Now, one could just say that the family of intervals $\{I_{nm}\}_{n,m}$ covers $A$ and has sum of measures less than $\varepsilon$, to get the desired result. But in this case one will be using the Axiom of Choice to prove that $\{I_{nm}\}_{n,m}$ is countable! So instead we proceed as follows:
Consider a bijective function $\varphi:\mathbb{N}\longrightarrow \mathbb{N}\times\mathbb{N}$ such that if $\varphi(k)=(n_k,m_k)$ then $k\geq n_k$ (the function $\varphi(k)=(n_k,m_k)$, where $k=2^{n_k-1}(2m_k-1)$ does the job). Therefore, $$\bigcup_{k=1}^nI_{\varphi(k)}\subset \bigcup_{k=1}^n\bigcup_{m=1}^{\infty}I_{km}.$$ It follows that $$\sum_{k=1}^n|I_{\varphi(k)}|< \sum_{k=1}^n\frac{\varepsilon}{2^k}=\varepsilon\left(1-\frac{1}{2^n}\right).$$ Thus, $$\sum_{k=1}^\infty |I_{\varphi(k)}|<\varepsilon.$$ Finally, $\{I_{\varphi(k)}\}_k$ is a countable family of intervals and $$A\subset\bigcup_{k=1}^\infty I_{\varphi(k)}.\qquad \blacksquare$$
Am I right in thinking that this proof doesn't rely on the Axiom of Choice?
It is consistent with $\sf ZF$ that $\Bbb R$ is the countable union of countable sets.
It is not hard to prove, without any appeal to choice, that a countable set is null in the given definition. We can simply use intervals with rational endpoints, or even just intervals centred around each point in the set.
So, in models such as that, the countable union of null sets can be equal to the entire real line, which is certainly not null.
(On the other hand, if we require the the measure is $\sigma$-additive, then by definition the union of null sets is null. Incidentally, in the case above, every set of reals is null, which makes the measure kind of useless.)