Let $X_i$, $i=1,2\ldots$ be finite sets of integers such that $X_i\rightarrow \infty$. Let $X$ be an ultraproduct of $X_i$. Let $\mu$ the Loeb measure associated with the the normalized counting measures on $X_i$. Let $B$ be the completion w.r.t $\mu$ of the algebra of internal sets. How to prove $(X,B,\mu)$ is not separable? Thank you for your help.

2$\begingroup$ As phrased, this sounds like an exercise you have been assigned; if not, could you please provide some more details about where you've seen the result claimed? $\endgroup$ – Yemon Choi Feb 17 '11 at 1:19

$\begingroup$ You need to assume the ultrafilter used in the ultraproduct is nonprincipal. $\endgroup$ – Joel David Hamkins Feb 17 '11 at 3:44

1$\begingroup$ 1. The ultrafilter used in the ultraproduct is nonprincipal. 2. I haven't seen this claimed. I hypothesize that it is true but in order to be more accurate I change the question to "Is $(X,B,\mu)$ non separable? $\endgroup$ – John Smith Feb 17 '11 at 10:14
The background is, I suspect, the recent papers of Elek and Szegedy, in which the nonseparability of this space place a large (and confounding) role.
I strongly prefer the language of nonstandard analysis, so I'll phrase this answer in those terms. The last paragraph has some comments about how to "downgrade" to the ultraproduct lingo.
Let $N$ be an unlimited integer, $[N]=\{0,1,2,\dots,N\}$ the hyperfinite initial segment of the natural numbers, and for internal sets $A\subseteq[N]$ set $\mu(A) = st(A/N)$, the standard part of the relative density of $A$. The Loeb counting measure is the completion of $\mu$.
Now, let $E_1, E_2,\dots$ be a countable collection of measurable sets, and without loss of generality we can assume that the $E_i$ are internal (in a Loeb measure, each measurable $E$ has an internal $I$ with $\mu(E \bigtriangleup I)=0)$, and that they form a $\pi$system (are closed under finite intersections). Here's an outline for the rest of the argument:
 Almost all internal subsets $Z\subseteq[N]$ are independent of each $E_i$, and have measure $1/2$;
 and so BorelCantelli tells us that almost all internal $Z\subseteq[N]$ are independent of all of the $E_i$ (this is where we use countability);
 if $Z$ is independent of each element in a $\pi$system, then it is independent of each element of the $\sigma$algebra generated by the system;
 If $Z$ is independent of itself, it has measure 0 or 1;
 therefore, almost all $Z$ are not in the $\sigma$algebra by the $E_i$.
The first bullet can be accomplished by counting, but is more naturally seen as a probability statement: let $Z$ contain $k\in[N]$ with probability $1/2$, and suppose that each $k$ is independent of each other. The Central Limit Theorem now says that $\mu(Z)=1/2$ with probability 1, and applies similarly to the independenceof$E_i$ claim. The second bullet should be in any measure theory book that discusses probability. The third bullet should be more widely known that it is; I learned of the $\pi$system approach to proving things about $\sigma$algebras from Williams' gemofabook "Probability with Martingales". The last two are undergrad points.
If you are uncomfortable with language of nonstandard analysis, it is a powerful paradigm for the situation when you are looking at a single nonprincipal ultraproduct. The conversion is this: an unlimited $N$ corresponds to a sequence that goes to infinity, an internal set is the ultraproduct of sets, and functions can be understood as the pointwise application of the function to each term of a sequence.