Lebesgue Measurability and Weak CH Let $LM$ denote "all subsets of $\Bbb{R}$ are Lebesgue measurable", and 
$WCH$ (weak continuum hypothesis) denote "every uncountable subset of $\Bbb{R}$ can be be put into 1-1 correspondence with $\Bbb{R}$". 
[Warning: in other contexts, weak CH means something totally different , i.e., it sometimes means $2^{\aleph_{0}} < 2^{\aleph_{1}}$].
We know that $LM$ and $WCH$ both hold in Solovay models. By forcing a (Ramsey) ultrafilter over a Solovay model one can also arrange $WCH+\neg LM$ (due to joint work of Di Prisco and Todorčević, who showed that the perfect set property holds in the generic extension). 
This prompts my question ($DC$ below is dependent choice).

Question: Is it known, relative to appropriate large cardinal axioms, whether there is a model of $ZF+LM+DC+\neg WCH$?

My question arose from an FOM-question of Tim Chow, and my answer to it; see also Chow's response.
 A: This is an expansion of my comments above. In the paper with Zapletal that I reference, we assume a proper class of Woodin cardinals and force over $L(\mathbb{R})$ with a partial order of countable approximations to a certain kind of MAD family (which Jindra named an "improved" MAD family). Although I have yet to write out the details, I believe that the resulting model satisfies LM (it clearly satisfies DC), and that, in this model, $\mathbb{R}$ cannot be injected into the generic MAD family. The arguments I have in mind are straightforward applications of the arguments given in the paper. 
The paper of Horowitz and Shelah referenced in my second comment works from the assumption of a strongly inaccessible cardinal and, as I understand it, adds the construction of a generic MAD family to Solovay's argument. As shown in their paper, DC + LM hold in the resulting model. I wrote to Haim and asked if $\mathbb{R}$ can be injected into the generic MAD family in this model, and he said no. He says he'll update their paper to include a proof of this (so they'll probably have a proof out before we do). 
