# Is ultrafilter convolution a special case of convolution of probability measures?

If $$G$$ is a semigroup and $$\beta G$$ is the set of ultrafilters on $$G$$, then $$\beta G$$ is also a semigroup: given $$p,q\in \beta G$$, we define $$p*q := \{E\subseteq G : \{g\in G : g^{-1}A\in q\}\in p\}.$$ This is a strange definition at first, but indeed it is an associative binary operation on $$\beta G$$, and we have an embedding of semigroups $$G\hookrightarrow \beta G$$ by identifying each element of $$G$$ with its point-mass.

An ultrafilter $$p$$ can be identified with its indicator function $$\chi_p:\mathcal{P}(G)\rightarrow \{0,1\}$$, where $$\mathcal{P}(G)$$ is the power set of $$G$$. This is a finitely-additive probability measure on $$G$$.

Thus let $$\mathrm{Pr}^0(G)$$ be the space of all finitely-additive probability measures $$\mu:\mathcal{P}(G)\rightarrow [0,1]$$, so that $$\beta G\subseteq \mathrm{Pr}^0(G)$$. (Note that we consider every subset of $$G$$ to be measurable.) Given two probability measures $$\mu,\nu\in \mathrm{Pr}^0(G)$$, we should be able to convolve them as follows:

1. Construct the product probability measure $$\mu\times \nu$$, determined uniquely by the requirement that $$(\mu\times\nu)(A\times B)=\mu(A)\nu(B)$$ for all $$A,B\subseteq G$$.

2. Let the convolution $$\mu*\nu$$ be the pushforward of $$\mu\times \nu$$ via the multiplication map $$G\times G\rightarrow G$$. Thus $$(\mu*\nu)(E) := (\mu\times \nu)\left(\{(x,y)\in G\times G: xy\in E\}\right).$$

Is this correct? This should be an associative binary operation on $$\mathrm{Pr}^0(G)$$, making it into a semigroup.

Viewing $$\beta G$$ as a subset of $$\mathrm{Pr}^0(G)$$ by $$p\mapsto \chi_p$$, we can ask: is ultrafilter convolution the same as convolution of the corresponding probability measures?

Question: Does $$p\mapsto \chi_p$$ define a semigroup homomorphism $$\beta G \rightarrow \mathrm{Pr}^0(G)$$?

• You have defined $\mu\times\nu$ only on sets of the form $A\times B$. There is no reason to expect this to extend uniquely to all subsets of $G\times G$. Mar 6, 2020 at 20:53
• Well, I wouldn't say that there is no reason --- it is indeed true for countably-additive probability measures. Does it fail if we replace "countably-additive" with "finitely-additive"? Mar 6, 2020 at 20:55
• No, even then you can only extend to the $\sigma$-algebra generated by rectangles. Mar 6, 2020 at 20:56
• It is reasonable to expect your definition $\mu\times\nu$ to extend to a unique finitely additive probability measure on some algebra of subsets of $G\times G$. But there is no reason that algebra would contain the sets $\{(x,y)\in G\times G:xy\in E\}$ you need to define $\mu*\nu$. Mar 6, 2020 at 20:58
• In the case of commutative $G$, your convolution, if it exists, would also be commutative (just by inspection of the definition). But that's not the case for the product of ultrafilters. Mar 6, 2020 at 23:38