# Large cardinals and measurability in $L(A)$

Under suitable large-cardinal assumptions, in the inner model $$L(\mathbb R)$$ one can have $$\omega_1$$ and $$\omega_2$$ measurable (this follows from determinacy).

I was wondering if it is possible to reconcile measurability of $$\omega_1$$ (or other cardinals) with the model $$L( \mathbb R^{\omega_1} )$$ or objects alike?

Here $$\mathbb R^{\omega_1}$$ is the product space of $$\omega_1$$-many copies of the real line.

• The notations $L[A]$ and $L(A)$ switched roles at some point. $L[\Bbb R]$ is a model generated by adding the predicate for the set of real numbers into your language. It is a model of ZFC, whereas you seem to mean $L(\Bbb R)$, which is the smallest model containing all the reals... Sep 10, 2022 at 12:47
• In fact, $L[\mathbb{R}]$ is just $L$ (since "is a real" is a sufficiently low-complexity property). Sep 10, 2022 at 17:17

For each $$\alpha<\omega_1$$, choose a real $$r_\alpha$$ that codes the ordertype $$\alpha$$. This sequence $$\langle r_\alpha : \alpha < \omega_1 \rangle$$ then codes a sequence of surjections from $$\omega$$ to each countable ordinal. From this, you can then run the Ulam matrix construction and show that there is no countably complete ultrafilter on $$\omega_1$$.
Let $$F$$ be a countably complete filter on $$\omega_1$$ in $$L(\mathbb R^{\omega_1})$$, and let $$\sigma_\alpha : \omega \to \alpha$$ be the surjection coded by $$r_\alpha$$. For $$n<\omega$$ and $$\beta<\omega_1$$, let $$S^\beta_n = \{ \alpha > \beta : \sigma_\alpha(n) = \beta \}$$. Since $$\bigcup_{n<\omega} S_n^\beta$$ is the tail interval $$(\beta,\omega_1)$$, we can take $$n_\beta$$ to be the least $$n$$ such that $$S^\beta_n$$ is $$F$$-positive, by countable completeness. Let $$T_n = \{ \beta : n_\beta = n \}$$. Since $$\omega_1$$ is uncountable, there is $$n$$ such that $$|T_n| > 1$$. So for such $$T_n$$, if $$\beta<\gamma$$ are in $$T_n$$ and $$\alpha \in S_n^\beta \cap S_n^\gamma$$, then $$\sigma_\alpha(n) = \beta = \gamma$$, which is impossible. Thus $$S_n^\beta$$ and $$S_n^\gamma$$ are disjoint $$F$$-positive sets, and so $$F$$ is not an ultrafilter.