TITLE: Is there a $\sigma$-algebra on $\mathbb{R}$ strictly between the Borel and Lebesgue algebras?
QUESTION [40 upvotes]: So, after proving that $\mathfrak{B}(\mathbb{R})\subset \mathfrak{L}(\mathbb{R})$, I asked myself, and now asking you, is there a set $\mathfrak{S}(\mathbb{R})$, which satisfies:
$$\mathfrak{B}(\mathbb{R} )\subset \mathfrak{S}(\mathbb{R})\subset \mathfrak{L}(\mathbb{R})$$
($\mathfrak{B}(\mathbb{R})$ is the Borel's set on $\mathbb{R}$ ; $\mathfrak{L}(\mathbb{R})$ is family of sets which are Lebesgue's measurable - which is a $\sigma$ algebra.)

REPLY [2 votes]: Bumping an old question, I think this is a situation where the language of elementary submodels is convenient.
It's easy to construct a set $M$ with the following properties (e.g. via repeated Lowenheim-Skolem to build an elementary chain of length $\omega_1$):

$M\prec V$ (or $M\prec H_\kappa$ for "large enough" $\kappa$ if one prefers),

$M$ is closed under $\omega$-sequences, and

$\vert M\vert = \mathfrak{c}$ (of course $\vert M\vert<2^\mathfrak{c}$ is really all we need).


These properties ensure that $M$ contains every real and every Borel set, does not contain every Lebesgue measurable set, and satisfies enough of the true theory of the universe for us to relativize basic concepts and facts to $M$. In particular, we automatically get the following:

$\mathfrak{L}(\mathbb{R})^M=\mathfrak{L}(\mathbb{R})\cap M$ is a $\sigma$-algebra strictly between $\mathfrak{B}(\mathbb{R})$ and $\mathfrak{L}(\mathbb{R})$.

Note that it would not be enough to replace condition $(2)$ by merely "$\mathfrak{B}(\mathbb{R})\subseteq M$" - we also used $\omega$-closedness to ensure that anything $M$ thinks is a $\sigma$-algebra really is a $\sigma$-algebra.
Admittedly, this isn't really that different from the existing answers since we're hiding the "close under appropriate operations" bit in the construction of $M$ itself. However, I do think it makes things simpler to think about, and yields intermediate $\sigma$-algebras of a somewhat different flavor.