§ Integrating against ultrafilers

• Let $X$ be a set.
• Recall that a filter on $X$ is a collection of subsets $\Omega$ of $X$ that are closed under supersets and intersections (union comes for free by closure under supersets).
• Recall that an ultrafilter $\Omega$ on $X$ is a maximal filter. That is, we cannot add any more elements into the filter.
• Equivalently $\Omega$ is an ultrafilter if, for any $A \subseteq X$, either $A \in \Omega$ or $(X - A) \in \Omega$.
• Intuitively, we are considering the set of subsets of $X$ that contains a single $x \in X$.
• We can also say that ultrafilters correspond to lattice homomorphisms $2^X \to 2$.
• A lemma will show that this is equivalent to the following: Whenever $X$ is expressed as the disjoint union of three subsets $S_1, S_2, S_3 \subseteq X$, then one of then will be in $\Omega$ (there exists some $i$ such that $S_i\in \Omega$).

§ Lemma: Three picking equivalent to ultrafilter

§ Integration by ultrafilter

• Let $B$ a finite set, $X$ a set, $\Omega$ an ultrafilter on $X$.
• Given $f: X \to B$, we wish to define $\int_X f d\Omega$.
• See that the fibers of $f$ partition $X$ into disjoint subsets $f^{-1}(b_1), f^{-1}(b_2), \dots, f^{-1}(b_N)$.
• The ultrafilter $X$ picks out one of these subsets, say $f^{-1}(b_i)$ ( $i$ for "integration").
• Then we define the integral to be $b_i$.

§ What does this integral mean?

• We think of $\Omega$ as a probability measure. Subsets in $\Omega$ have measure 1, subsets outside have measure 0.
• Since we want to think of $\Omega$ as some kind of probability measure, we want that $\int_X 1 d \Omega = 1$, as would happen when we integrate a probability measure $\int d \mu = 1$.
• Next, if two functions $f, g$ are equal almost everywhere (ie, the set of points where they agree is in $\Omega$), then their integral should be the same.