Ultrafilters and measurability Consider a compact metric space $X$, the sigma-algebra of the boreleans of $X$, a sequence of measurable maps $f_n: X \to\Bbb R$ and an ultrafilter $U$. Take, for each $x \in X$, the $U$-limit, say $f^*(x)$, of the sequence $(f_n(x))_{n \in\Bbb N}$. 
(Under what conditions on $U$) Is $f^*$ measurable?
 A: Let me first get rid of a silly case that you probably didn't intend to include.  If $U$ is a principal ultrafilter, generated by $\{k\}$, then $f^*$ is just $f_k$, so it's measurable.
Now for the non-silly cases, where $U$ isn't principal. Here's an example of a sequence of measurable (in fact low-level Borel) functions whose $U$-limit isn't measurable.  I'll use a very nice $X$, the unit interval with the usual topology and with Lebesgue measure.  My functions $f_n$ will take only the values 0 and 1, and they're defined by $f_n(x)=$ the $n$-th bit in the binary expansion of $x$.  (The binary expansion is ambiguous when $x$ is a rational number whose denominator is a power of 2, but that's only countably many $x$'s so they won't affect measurability; resolve the ambiguity any arbitrary way you want.)  Then the $U$-limit $f^*$ of these functions sends $x$ to 1 iff the set $\{n:\text{the }n\text{-th bit in the binary expansion of }x\text{ is }1\}$ in in $U$.  In other words, if we identify $x$ via its binary expansion with a sequence of 0's and 1's and if we then regard that sequence as the characteristic function of a subset of $\mathbb N$, then $f^*$, now viewed as mapping subsets of $\mathbb N$ to $\{0,1\}$, is just the characteristic function of $U$. An old theorem of Sierpiński says that this is never Lebesgue measurable when $U$ is a non-principal ultrafilter.  
