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.



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


