@PaulBlainLevy That's implied by Vopenka's principle, since SCH holds above a strongly compact cardinal.

I'm pretty sure KM proves that for every $X$ there is a satisfaction class for the language expanded by $X.$ It certainly holds for $(V_{\kappa}, V_{\kappa+1}),$ $\kappa$ inaccessible.

To remove ambiguity about what choice, reflection, and large cardinal principles are, it might be easier to work in an ambient model of ZFC and ask about models of the form $(V_{\kappa}, V_{\kappa+1}),$ where $\kappa$ is assumed to have large cardinal properties.

ZF + DC + "$\omega_1$ doesn't inject into $\mathbb{R}$" has consistency strength an inaccessible since it implies $\omega_1$ is inaccessible in $L.$ In particular, it's not a theorem of "all sets of reals have property of Baire."

It's consistent with ZF that $\aleph(\mathbb{R})=\aleph_{\omega}.$ Ref: "A model of Z-F set theory with $\aleph(2^{\omega}) = \aleph_{\omega}$" by Derrick and Drake.

@Reflecting_Ordinal $i_0$ is the sup of countable ordinals which are definable in $L$ from $V$-cardinals (equivalently, from $\omega_n^V$'s).

@EmilJeřábek I got confirmation from Gao that it's a ZF theorem.

I think their proof yields an explicit construction in this case, but I'll email them to check.

This answer has been significantly overhauled. The above discussion refers to a previous argument which has been generalized to the (3) $\rightarrow$ (4) in the current answer.

My comment was referring to my pre-edit definition, which isn’t even closed under complements.

@EmilJeřábek Suppose $X$ is measurable against bounded intervals. Assume $Y$ has finite outer measure (otherwise trivial). Fix an open cover $C$ of $Y$ of measure less than $\lambda^*(Y)+\epsilon.$ Finitely many intervals $\{I_k\}_1^n$ in $C$ contain all but $\frac{\epsilon}{2}$ of the measure. For each $k \le n,$ choose covers $I_k \cap X \subset C_{1, k}$ and $I_k \setminus X \subset C_{2, k}$ such that $\lambda(C_{1,k})+\lambda(C_{2,k})<\lambda(I_k)+\frac{\epsilon}{2n}.$ The rest is clear.

@FrançoisG.Dorais I think your earlier suggestion is reasonable. That if $\mathbb{R}$ is a countable union of countable sets, then there is no total isometry-invariant finitely additive probability measure on $\mathcal{P}([0,1])$ which has the Lebesgue density property.