# 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 -

1. [...] Metric Space of a Measure Ring Let be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, [...]

Pingback by The Metric Space of a Measure Ring « The Unapologetic Mathematician | August 6, 2010 | Reply

2. [...] Space of a Measure Space Our first result today is that the metric space associated to the measure ring of a measure space is [...]

Pingback by Completeness of the Metric Space of a Measure Space « The Unapologetic Mathematician | August 9, 2010 | Reply

3. [...] off, every totally finite measure algebra is complete. That is, if — and thus for all , then every collection of elements of will [...]

Pingback by Completeness of Boolean Rings « The Unapologetic Mathematician | August 10, 2010 | Reply

4. [...] Spaces of Measure Rings Today we collect a few facts about the metric space associated to a measure ring [...]

Pingback by Properties of Metric Spaces of Measure Rings « The Unapologetic Mathematician | August 12, 2010 | Reply

5. [...] we first dealt with measure rings, we quotiented out by the ideal of negligible sets. We don’t have a [...]

Pingback by Stone’s Representation Theorem III « The Unapologetic Mathematician | August 20, 2010 | Reply

6. [...] in Measure Algebras Let be a totally finite measure algebra, and write for the maximal element. Without loss of generality, we can assume that is normalized [...]

Pingback by Partitions in Measure Algebras « The Unapologetic Mathematician | August 24, 2010 | Reply

7. [...] on , and let be Lebesgue measure. If is a sequence of partitions of the maximal element of the measure algebra into intervals, and if the limit of the sequence of norms is zero, then is [...]

Pingback by The Measure Algebra of the Unit Interval « The Unapologetic Mathematician | August 25, 2010 | Reply