MA makes unions of fewer than continuum many null sets null

Let $P_{\epsilon }$ be the set of open subsets $U$ of $\mathbb{R}$ with $m(U)<\epsilon$, ordered by reverse inclusion: $U\le V$ means $U\supseteq V$, so a larger open set is a stronger condition in the convention of [F3]. For every $\alpha$, the subcollection $D_{\alpha }=\{U:N_{\alpha }\subseteq U\}$ is dense: given $U$, choose by F1 an open cover $O$ of the null set $N_{\alpha }\setminus U$ with $m(O)<\epsilon -m(U)$; then $U\cup O\le U$ lies in $D_{\alpha }$.

The order is ccc. Enumerate the rational open intervals whose closures lie in a condition $U$; their union is $U$. The increasing finite unions $W_j\subseteq U$ therefore have union $U$, so F2 supplies some finite rational-interval union $W=W_j$ with $m(U\setminus W)<(\epsilon -m(U))\mathrm{/}2$. If two conditions $U,V$ share this $W$, then, labelling so that $m(U)\le m(V)$, $m(U\cup V)\le m(W)+m(U\setminus W)+m(V\setminus W)<m(U)+\frac{\epsilon -m(U)}{2}+\frac{\epsilon -m(U)}{2}=\epsilon ,$ so $U\cup V$ is a condition below both in the declared reverse-inclusion order. Only countably many sets $W$ occur, so no antichain of conditions is uncountable.

MA gives a filter meeting all $D_{\alpha }$. Its union $U$ covers every $N_{\alpha }$. Directedness toward stronger conditions in the reverse-inclusion order makes every finite subunion of filter members a subset of one stronger member, so it has measure below $\epsilon$. The rational base gives a countable cofinal subfamily for the union, so continuity from below yields $m(U)\le \epsilon$. Repeating for $\epsilon =2^{-n}$ proves the total union has outer measure zero, and completeness gives nullity.

A singleton has measure zero. Applying the first clause to the family of singletons indexed by any set of reals of size below $\mathfrak{c}$ gives the second. AC was used to enumerate/cardinalize the family and choose all covers and approximations.