# The Unapologetic Mathematician

## Functions on Boolean Rings and Measure Rings

We’re not just interested in Boolean rings as algebraic structures, but we’re also interested in real-valued functions on them. Given a function $\mu:\mathcal{R}\to\mathbb{R}$ on a Boolean ring $\mathcal{R}$, we say that $\mu$ is additive, or a measure, $\sigma$-finite (on $\sigma$-rings), and so on analogously to the same concepts for set functions. We also say that a measure $\mu$ on a Boolean ring is “positive” if it is zero only for the zero element of the Boolean ring.

Now, if $\mathcal{S}$ is the Boolean $\sigma$-ring that comes from a measurable space $(X,\mathcal{S})$, then usually a measure $\mu$ on $\mathcal{S}$ is not positive under this definition, since there exist sets of measure zero. However, remember that in measure theory we usually talk about things that happen almost everywhere. That is, we consider two sets — two elements of $\mathcal{S}$ — to be “the same” if their difference is negligible — if the value of $\mu$ takes the value zero on this difference. If we let $\mathcal{N}=\mathcal{N}(\mu)\subseteq\mathcal{S}$ be the collection of $\mu$-negligible sets, it turns out that $\mathcal{N}$ is an ideal in the Boolean $\sigma$-ring $\mathcal{S}$. Indeed, if $M$ and $N$ are negligible, then so is $M\Delta N$, so $\mathcal{N}$ is an Abelian subgroup. Further, if $N\in\mathcal{N}$ and $E\in\mathcal{S}$, then $E\cap N\in\mathcal{N}$, so $\mathcal{N}$ is an ideal.

So we can form the quotient ring $\mathcal{S}/\mathcal{N}(\mu)$, which consists of the equivalence classes of elements which differ by an element of measure zero. This is equivalent to our old rhetorical trick of only considering properties up to “almost everywhere”. Using this new definition of “equals zero”, any measure $\mu$ on a Boolean $\sigma$-ring $\mathcal{S}$ gives rise to a positive measure on the quotient $\sigma$-ring $\mathcal{S}/\mathcal{N}(\mu)$. In particular, given a measure space $(X,\mathcal{S},\mu)$, we write $\mathcal{S}(\mu)=\mathcal{S}/\mathcal{N}(\mu)$ for the Boolean $\sigma$-ring it gives rise to.

We say that a “measure ring” $(\mathcal{S},\mu)$ is a Boolean $\sigma$-ring $\mathcal{S}$ together with a positive measure $\mu$ on $\mathcal{S}$. For instance, if $(X,\mathcal{S},\mu)$ is a measure space, then $(\mathcal{S}(\mu),\mu)$ is a measure ring.. If $\mathcal{S}$ is a Boolean $\sigma$-algebra we say that $(\mathcal{S},\mu)$ is a measure algebra. We say that measure rings and algebras are (totally) finite or $\sigma$-finite the same as for measure spaces. Measure rings, of course, form a category; a morphism $f:(\mathcal{S},\mu)\to(\mathcal{T},\nu)$ from one measure algebra to another is a morphism of boolean $\sigma$-algebras $f:\mathcal{S}\to\mathcal{T}$ so that $\mu(E)=\nu(f(E))$ for all $E\in\mathcal{S}$.

I say that the mapping which sends a measure space $(X,\mathcal{S},\mu)$ to its associated measure algebra $(\mathcal{S}(\mu),\mu)$ is a contravariant functor. Indeed, let $f:(X,\mathcal{S},\mu)\to(Y,\mathcal{T},\nu)$ be a morphism of measure spaces. That is, $f$ is a measurable function from $X$ to $Y$, so $\mathcal{S}$ contains the pulled-back $\sigma$-algebra $f^{-1}(\mathcal{T})$. This pull-back defines a map $f^{-1}:\mathcal{T}\to\mathcal{S}$. Further, since $f$ is a morphism of measure spaces it must push forward $\mu$ to $\nu$. That is, $\nu=f(\mu)$, or in other words $\nu(E)=\mu(f^{-1}(E))$. And so if $\nu(E)=0$ then $\mu(f^{-1}(E))=0$, thus the ideal $\mathcal{N}(\nu)\subseteq\mathcal{T}$ is sent to the ideal $\mathcal{N}(\mu)\subseteq\mathcal{S}$, and so $f^{-1}$ descends to a homomorphism between the quotient rings: $f^{-1}:\mathcal{T}(\nu)\to\mathcal{S}(\mu)$. As we just said, $\nu(E)=\mu(f^{-1}(E))$, and thus we have a morphism of measure algebras $f^{-1}:(\mathcal{T}(\nu),\nu)\to(\mathcal{S}(\mu),\mu)$. It’s straightforward to confirm that this assignment preserves identities and compositions.

August 5, 2010