Impact of coining $L$ in $\mathcal L_{\omega_1, \omega}$ on which large cardinals it can satisfy?

Question

What happens to the constructible universe $L$ if we build it in the first of infinitary languages $\mathcal L_{\omega_1, \omega}$?

Would the usual limitation of $L$ not satisfying existence of a measurable cardinal still hold?

Answer 1

It seems to me that if you build the constructible universe using $\mathcal{L}_{\omega_1,\omega}$ logic, you will get the inner model $L(\mathbb{R})$. The reason is that every $\mathcal{L}_{\omega_1,\omega}$ formula is coded by a real, and so $L(\mathbb{R})$ has all the formulas and can compute truth correctly. (Note that this would not be true for $\mathcal{L}_{\omega_1,\omega_1}$, since $L(\mathbb{R})$ is not necessarily closed under $\omega$-sequences.) So the resulting model is contained within $L(\mathbb{R})$. But since every subset of $\omega$ is definable by an $\mathcal{L}_{\omega_1,\omega}$ formula specifying the members, the resulting model will contain $\mathbb{R}$ and thus contain $L(\mathbb{R})$. So they are equal.

It is indeed possible that $L(\mathbb{R})$ has a measurable cardinal, if the ambient universe has sufficiently strong large cardinals, since if AD holds in $V$, then it holds in $L(\mathbb{R})$, and this implies that $\omega_1$ is measurable, with the club filter being an ultrafilter.

If $L(\mathbb{R})$ satisfies ZFC, however, then it can have no measurable cardinals.