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Let $I$ be a set, $\mathcal{U}$ be an ultrafilter on $I$ and $1\leq p<\infty$. Let $X_{i}=L_{p}(\mu_{i})(i\in I)$, where $\mu_{i}$ is a probability measure for each $i\in I$. Relying on standard results from Banach lattice theory, we can prove that the ultraproduct $(X_{i})_{\mathcal{U}}$ of $(X_{i})_{i\in I}$ is isometrically isomorphic to $L_{p}(\mu)$ for some measure $\mu$. But I want to know whether the measure $\mu$ can be taken to be finite or even a probability measure. Thank you!

Let $I$ be a set, $\mathcal{U}$ be an ultrafilter on $I$ and $1\leq p<\infty$. Let $X_{i}=L_{p}(\mu_{i})(i\in I)$, where $\mu_{i}$ is a probability measure for each $i\in I$. Relying on standard results from Banach lattice theory, we can prove that the ultraproduct $(X_{i})_{\mathcal{U}}$ of $(X_{i})_{i\in I}$ is isometrically isomorphic to $L_{p}(\mu)$ for some measure $\mu$.

Question 1: Can the above measure $\mu$ be taken to be finite or even a probability measure?

Question 2: Is there a direct construction of the measure $\mu$ without relying on Banach lattice theory?

Thank you!