Working in $\sf ZF$, is there a natural definition of an algebra $\Sigma$ with the following properties:
- $\Sigma$ is a $\sigma$-algebra on $\Bbb R$, i.e. it is closed under complement and countable unions.
- ${\cal B}\subseteq\Sigma\subseteq{\cal L}$, where $\cal B$ is the Borel $\sigma$-algebra on $\Bbb R$ and $\cal L$ is the collection of Lebesgue measurable sets (which in this context is an algebra but not a $\sigma$-algebra).
- If countable choice (CC) is assumed (or more specifically if $\cal L$ is already a $\sigma$-algebra), then $\Sigma=\cal L$.
So far the best I can come up with is $\Sigma=\begin{cases}{\cal L}&{\cal L}\mbox{ is a }\sigma\mbox{-alg.}\\{\cal B}&o.w.\\\end{cases}$, which is not a very "nice" solution (if I had to point at something specific, I would say it unnecessarily invokes the law of excluded middle, i.e. it is nonconstructive). I tried using a maximal $\sigma$-algebra below $\cal L$, but I don't think one necessarily exists.
Note: this proof is incorrect, but is left up for its instructional value.
As Asaf points out in the comments, $\cal L$ is already a $\sigma$-algebra in $\sf ZF$, so the problem is trivial and satisfied by $\Sigma=\cal L$. (Edit: This statement is followed by a long proof, which reminds me of a certain joke.)
We start with the definition of $\mu^*(A)$, the Lebesgue outer measure, defined as $$\mu^*(A)=\inf\left\{\sum_{n=1}^\infty|I_n|:(I_n)_n\mbox{ is a sequence of open intervals with }A\subseteq\bigcup_{n=1}^\infty I_n\right\}.$$
Let $I_X$ be the set of open intervals with endpoints in $X$. Since $I_\Bbb Q$ is countable, fix a wellorder $\prec$ of $I_\Bbb Q$. Then we can define a function $T_\epsilon:I_\Bbb R\to I_\Bbb Q$ by $$T_\epsilon(I)=\min_\prec\{J\in I_\Bbb Q:I\subseteq J\land|J|\le|I|+\epsilon\}.$$ Note that the minimum is well-defined because given $I=(a,b)$ we can pick $r,s\in\Bbb Q$ with $a-\epsilon/2<r<a$ and $b<s<b+\epsilon/2$. Thus, given an interval covering $(I_n)$ and $\epsilon>0$, we can set $J_n=T_{\epsilon/2^n}(I_n)$ to get $\sum_{n=1}^\infty|J_n|\le\sum_{n=1}^\infty|I_n|+\epsilon$ and $\bigcup_{n=1}^\infty I_n\subseteq \bigcup_{n=1}^\infty J_n$. Thus we have shown that
$$\mu^*(A)=\inf\left\{\sum_{n=1}^\infty|I_n|:(I_n)_n\mbox{ is a sequence in $I_\Bbb Q$ with }A\subseteq\bigcup_{n=1}^\infty I_n\right\}.$$
But we're not done yet. Given $A$ and $\epsilon>0$, we can recursively construct a suitable interval covering of $A$: Assuming $I_k$ exists for all $k<n$, let $I_n$ be the $\prec$-minimal interval such that there exists an infinite sequence $(J_n)_n\in I_\Bbb Q^\Bbb N$ extending $(I_k)_{k=1}^n$, and satisfying the conditions $A\subseteq\bigcup_n J_n$, $\sum_n|J_n|\le\mu^*(A)+\epsilon$. (This is just an application of König's lemma, where choice is not needed because the set of children of each node is canonically ordered.)
Thus, we have
proven[Edit: see below, this is false]:The important property of this theorem is that it gives a canonical representative for each pair $(A,\epsilon)$. Thus, turning to our original goal, given a sequence $(A_n)_n$ of sets, we can take $I_{nk}=K(A_n,\epsilon2^{-n})_k$ to get a countable sequence of intervals covering $\bigcup_nA_n$ with $\sum_{n,k}|I_{nk}|\le\sum_n\mu^*(A_n)+\epsilon$; thus $\mu^*$ is countably subadditive, $\mu$ is countably additive, and $\cal L$ is a $\sigma$-algebra.
Edit: The reason this proof fails is because the function $K$ will not necessarily satisfy $A\subseteq\bigcup_n I_n$, because the sequence could get "distracted" enumerating only a countable subset of $A$ and forget to get around to the rest of it. Under the König's lemma interpretation of an infinite path through a tree, just because every finite path of an infinite tree extends to a "good" infinite path does not mean that every infinite path is good. For example, $\Bbb Z$ is a null set, so any finite sequence $(I_n)_{n<k}$ with $\sum_{n<k}|I_n|<\epsilon$ can be extended to an infinite sequence covering $\Bbb Z$ with $\sum_{n=1}^\infty|I_n|<\epsilon$. Nevertheless, there are infinite sequences that don't cover $\Bbb Z$ and satisfy the assumption, such as $I_n=(n-\epsilon2^{-n},n+\epsilon2^{-n})$ which only covers $\Bbb N$.
In fact, as stated the $\prec$-first $I_n$ to be chosen will probably be an empty interval (which can always be extended with more intervals to satisfy the assumptions) so that $K$ always yields a sequence of empty intervals.