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!



A question on ultraproducts of $L_{p}(\mu)$-spaces
As for the first question the answer is **no** even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.


When $p=1$ we may use the following reasoning. The space $L\_1(\mu)$ is [weakly compactly generated](https://www.encyclopediaofmath.org/index.php/WCG_space) because the inclusion map ${\rm id}\colon L\_2(\mu)\to L\_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell\_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L\_1[0,1]$.


When $p\in (1,2)$, then $\ell\_p(\Gamma)$ does not embed into $L\_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:



> 
> P. Enflo, H. P. Rosenthal, Some results concerning $L\_p(\mu)$-spaces. *J. Functional Analysis* **14** (1973), 325–348.
> 
> 
> 


This was extended to $p\in (2,\infty)$ by [Bill Johnson](https://mathoverflow.net/users/2554/bill-johnson) and Gideon Schechtman (see [this paper](https://arxiv.org/pdf/1301.4086.pdf)).


As for the second question, should such construction exist, it is most likely covered in one of Fremlin's *Measure theory* volumes.


