# The Unapologetic Mathematician

## Stone’s Representation Theorem I

Today we start in on the representation theorem proved by Marshall Stone: every boolean ring $\mathcal{B}$ is isomorphic (as a ring) to a ring of subsets of some set $X$. That is, no matter what $B$ looks like, we can find some space $X$ and a ring $\mathcal{S}$ of subsets of $X$ so that $\mathcal{B}\cong\mathcal{S}$ as rings.

We start by defining the “Stone space” $S(\mathcal{B})$ of a Boolean ring $\mathcal{B}$. This is a representable functor, and the representing object is the two-point Boolean ring $\mathcal{B}_0$. That is, $S(\mathcal{B})=\hom_\mathbf{Rng}(\mathcal{B},\mathcal{B}_0)$, the set of (boolean) ring homomorphisms from $\mathcal{B}$ to $\mathcal{B}_0$. To be clear, $\mathcal{B}_0$ consists of the two points $\{0,1\}$, with the operations $\Delta$ for addition and $\cap$ for multiplication, and the obvious definitions of these operations. This is a contravariant functor — if we have a homomorphism of Boolean rings $f:\mathcal{B}\to\hat{\mathcal{B}}$ we get a function between the Stone spaces $S(f):S(\hat{\mathcal{B}})\to S(\mathcal{B})$, which takes a function $\lambda:\hat{\mathcal{B}}\to\mathcal{B}_0$ to the function $\lambda\circ f:\mathcal{B}\to\mathcal{B}_0$.

The Stone space isn’t just a set, though; it’s a topological space. We define the topology by giving a base of open sets. That is, we’ll give a collection of sets — closed under intersections — which we declare to be open, and we define the collection of all open sets to be given by unions of these sets. For each element $b\in\mathcal{B}$, we define a basic set like so:

$\displaystyle s(b)=\left\{\lambda\in S(\mathcal{B})\vert\lambda(b)=1\right\}$

To see that this collection of sets is closed under intersection, consider two such sets $s(b)$ and $s(b')$. I say that the intersection of these sets is the set $s(b\cap b')$. Indeed, if $\lambda\in s(b)$ and $\lambda\in s(b')$, then

$\displaystyle\lambda(b\cap b')=\lambda(b)\cap\lambda(b')=1\cap1=1$

Conversely, if $\lambda\in s(b\cap b')$, then $b\cap b'\subseteq b$. Thus

$\displaystyle1=\lambda(b\cap b')\subseteq\lambda(b)$

and so $\lambda(b)=1$, and $\lambda\in s(b)$. Similarly, $\lambda\in s(b')$. Thus we see that $s(b)\cap s(b')=s(b\cap b')$.

In fact, this map from $\mathcal{B}$ to the basic sets is exactly the mapping we’re looking for! We’ve already seen that our base is closed under intersection, and that the map $s$ preserves intersections. I say that we also have $s(b\Delta b')=s(b)\Delta s(b')$. If $\lambda\in s(b)$ but $\lambda\notin s(b')$, then $\lambda(b)=1$ and $\lambda(b')=0$. Then

$\displaystyle\lambda(b\Delta b')=\lambda(b)\Delta\lambda(b')=1\Delta0=1$

and similarly if $\lambda\in s(b')$ but $\lambda\notin s(b)$. Thus $s(b)\Delta s(b')\subseteq s(b\Delta b')$. Conversely, if $\lambda\in s(b\Delta b')$, then

$\displaystyle\lambda(b)\Delta\lambda(b')=\lambda(b\Delta b')=1$

and so either $\lambda(b)=1$ and $\lambda(b')=0$ or vice versa. Thus $s(b\Delta b')=s(b)\Delta s(b')$.

So we know that $s$ is a homomorphism of (boolean) rings. However, we don’t know yet that it’s an isomorphism. Indeed, it’s possible that $s$ has a nontrivial kernel — $s(b)$ could be $\emptyset\subseteq S(\mathcal{B})$ for some $b$. We must show that given any $b$ there is some $\lambda:\mathcal{B}\to\mathcal{B}_0$ so that $\lambda(b)=1$.

For a finite boolean ring $\mathcal{B}$ this is easy: we pick some minimal element $b'\subseteq b$ and define $\lambda(x)=1$ if and only if $b'\subseteq x$. Such a $b'$ exists because there’s at least one element below $b$$b$ itself is one — and there can only be finitely many so we can just take their intersection. Clearly $\lambda(b)=1$ by definition, and it’s straightforward to verify that $\lambda$ is a homomorphism of boolean rings using the fact that $b'$ is an atom of $\mathcal{B}$.

For an infinite boolean ring, things are trickier. We define the set $X^*$ of all functions $\mathcal{B}\to\mathcal{B}_0$, not just the ring homomorphisms. This is the product of one copy of $\mathcal{B}_0$ for every element of $\mathcal{B}$. Since each copy of $\mathcal{B}_0$ is a compact Hausdorff space, Tychonoff’s theorem tells us that $X^*$ is a compact Hausdorff space. If $\tilde{\mathcal{B}}$ is any finite subring of $\mathcal{B}$ containing $b$, let $X^*(\tilde{\mathcal{B}})$ be the collection of those functions $\lambda^*\in X^*$ which are homomorphisms when restricted to $\tilde{\mathcal{B}}$ and for which $\lambda^*(b)=1$.

I say that the class of sets of the form $X^*(\tilde{\mathcal{B}})$ has the finite intersection property. That is, if we have some finite collection of finite subrings $\tilde{\mathcal{B}}_1,\dots,\tilde{\mathcal{B}}_n$ and the finite subring $\tilde{\mathcal{B}}$ they generate, then we have the relation

$\displaystyle X^*(\tilde{\mathcal{B}})\subseteq\bigcap\limits_{i=1}^nX^*(\tilde{\mathcal{B}}_i)$

Indeed, $b$ is clearly contained in the generated ring $\tilde{\mathcal{B}}$. Further, if $\lambda^*$ is a homomorphism on $\tilde{\mathcal{B}}$ then it’s a homomorphism on each subring $\tilde{\mathcal{B}}_i$.

Okay, so since $\tilde{\mathcal{B}}$ is a finite boolean ring, the proof given above for the finite case shows that $X^*(\tilde{\mathcal{B}})$ is nonempty. Thus the intersection of any finite collection of sets $\{X^*(\tilde{\mathcal{B}}_i)\}$ is nonempty. And thus, since $X^*$ is compact, the intersection of all of the $\{X^*(\tilde{\mathcal{B}})\}$ is nonempty.

That is, there is some function $\lambda^*:\mathcal{B}\to\mathcal{B}_0$ which is a homomorphism of boolean rings on any finite boolean subring containing $b$, and with $\lambda^*(b)=1$. Given any other two points $b_1$ and $b_2$ there is some finite boolean subring containing $b$, $b_1$, and $b_2$, and so we must have $\lambda^*(b_1\cap b_2)=\lambda^*(b_1)\cap\lambda^*(b_2)$ and $\lambda^*(b_1\Delta b_2)=\lambda^*(b_1)\Delta\lambda^*(b_2)$ within this subring, and thus within the whole ring. Thus $\lambda^*$ is a homomorphism of boolean rings sending $b$ to $1$, which shows that $s(b)\neq\emptyset$.

Therefore, the map $s$ is a homomorphism sending the boolean ring $\mathcal{B}$ isomorphically onto the identified base of the Stone space $S(\mathcal{B})$.

August 18, 2010 Posted by | Analysis, Measure Theory | 6 Comments

## Associated Metric Spaces and Absolutely Continuous Measures II

Yesterday, we saw that an absolutely continuous finite signed measure $\nu$ on a measure space $(X,\mathcal{S},\mu)$ defines a continuous function on the associated metric space $\mathfrak{S}$, and that a sequence of such finite signed measures that converges pointwise is actually uniformly absolutely continuous with respect to $\mu$.

We’re going to need to assume that $\nu$ is nonnegative. We’d usually do this by breaking $\nu$ into its positive and negative parts, but it’s not so easy to get ahold of the positive and negative parts of $\nu$ in this case. However, we can break each $\nu_n$ into $\nu_n^+$ and $\nu_n^-$. Then we can take the limits $\nu^{\geq0}(E)=\lim_n\nu_v^+(E)$ and $\nu^{\leq0}(E)=\lim_n\nu_v^-(E)$, which will still satisfy $\nu(E)=\nu^{\geq0}(E)-\nu^{\leq0}$. The only difference between this decomposition and the positive and negative parts is that this pair of set functions might have some redundancy that gets cancelled off in this subtraction. And so, without loss of generality, we will assume that all the $\nu_n$ are nonnegative, and that their limit $\nu$ is as well.

Now, given such a sequence, define the limit function $\nu(E)=\lim_n\nu_n(E)$. I say that $\nu$ is itself a finite signed measure, and that $\nu\ll\mu$. Indeed, $\nu(E)$ is finite by assumption, and additivity is easy to check. As for absolute continuity, if $\mu(E)=0$, then each $\nu_n(E)=0$ since $\nu_n\ll\mu$, and so $\nu(E)=0$ as the limit of the constant zero sequence.

What we need to check is continuity. We know that it suffices to show that $\nu$ is continuous from above at $\emptyset$. So, let $\{E_m\}$ be a decreasing sequence of measurable sets whose limit is $\emptyset$. We must show that the limit of $\nu(E_m)$ is zero. But we know that the limit of $\mu(E_m)$ is zero, and thus for a large enough $m$ we can make $\mu(E_m)<\delta$ for any given $\delta$. And since $\nu\ll\mu$ we know that for any $\epsilon$ there is some $\delta$ so that if $\mu(E)<\delta$ then $\nu(E)<\epsilon$. Thus we can always find a large enough $m$ to guarantee that $\nu(E_m)<\epsilon$, and so the limit is zero, as asserted.

Finally, what happens if we remove the absolute continuity requirement from the $\nu_n$? That is: what can we say if $\{\nu_n\}$ is a sequence of finite signed measures on $X$ so that $\nu(E)=\lim_n\nu_n(E)$ exists and is finite for each $E\in\mathcal{S}$. I say that $\nu(E)$ is a signed measure. What we need is to find some measure $\mu$ so that all the $\nu_n\ll\mu$, and then we can use the above result.

Since $\nu_n$ is a finite signed measure, we can pick some upper bound $c_n\geq\lvert\nu_n(E)\rvert$. Then we define

$\displaystyle\mu(E)=\sum\limits_{n=1}^\infty\frac{1}{2^nc_n}\lvert\nu_n\rvert(E)$

If any $\lvert\nu\rvert(E)\neq0$, then $\mu(E)\neq0$, and so $\lvert\nu_n\rvert\ll\mu$. And thus $\nu_n\ll\mu$ for all $n$, as desired.

August 17, 2010

## Associated Metric Spaces and Absolutely Continuous Measures I

If $\mathfrak{S}$ is the metric space associated to a measure space $(X,\mathcal{S},\mu)$, and if $\nu$ is a finite signed measure that is absolutely continuous with respect to $\mu$. Then $\nu$ defines a continuous function on $\mathfrak{S}$.

Indeed, if $E\in\mathcal{S}$ is any set with $\mu(E)<\infty$, then $E$ represents a point of $\mathfrak{S}$, and $\nu(E)$ defines the value of our function at this point. If $F\subseteq\mathcal{S}$ is another set representing the same point, then $\mu(E\Delta F)=0$. By absolute continuity, $\nu(E\Delta F)=0$ as well, and so $\nu(F)=\nu(E)$. Thus our function doesn’t depend on the representative we use.

As for continuity at a point $E$, given an $\epsilon>0$, we want to find a $\delta>0$ so that if $\mu(E\Delta F)<\delta$ then $\lvert\nu(E)-\nu(F)\rvert<\epsilon$. We calculate

\displaystyle\begin{aligned}\lvert\nu(E)-\nu(F)\rvert&=\lvert\nu(E\setminus F)-\nu(F\setminus E)\rvert\\&\leq\lvert\nu(E\setminus F)\rvert+\lvert\nu(F\setminus E)\rvert\\&\leq\lvert\nu(E\Delta F)\rvert+\lvert\nu(E\Delta F)\rvert\\&=2\lvert\nu(E\Delta F)\rvert\\&\leq2\lvert\nu\rvert(E\Delta F)\end{aligned}

Since $\nu$ is finite, we know that for every $\epsilon>0$ there is a $\delta>0$ so that if $\lvert\mu\rvert(E\Delta F)=\mu(E\Delta F)<\delta$ then $\lvert\nu\rvert(E\Delta F)<\epsilon$. Using this $\delta$, our assertion of continuity follows.

Now, if $\{\nu_n\}$ is a sequence of finite signed measures on $X$ that are all absolutely continuous with respect to $\mu$, and if the limit $\lim_n\nu_n(E)$ exists and is finite for each $E\in\mathcal{S}$, then the sequence is uniformly absolutely continuous with respect to $\mu$. That is,

For any $\epsilon>0$ we can define the set

$\displaystyle\mathfrak{E}_k=\bigcap\limits_{m=k}^\infty\bigcap\limits_{n=k}^\infty\left\{E\in\mathfrak{S}\bigg\vert\lvert\nu_n(E)-\nu_m(E)\rvert\leq\frac{\epsilon}{3}\right\}$

Since each $\nu_n$ is continuous as a function on $\mathfrak{S}$, each of these $\mathfrak{E}_k$ is a closed set. Since the sequence $\{\nu_n(E)\}$ always converges to a finite limit, it must be Cauchy for each $E$, and so the union of all the $\mathfrak{E}_k$ is all of $\mathfrak{S}$. Thus the countable union of these closed subsets has an interior point. But since $\mathfrak{S}$ is a complete metric space, it is a Baire space as well. And thus one of the $\mathfrak{E}_k$ must have an interior point as well.

Thus there is some $k_0$, some radius $r_0$, and some set $E_0$ so that the ball $\{E\in\mathfrak{S}\vert\rho(E,E_0) is contained in $\mathfrak{E}_{k_0}$. Let $\delta$ be a positive number with $\delta, and so that $\lvert\nu_n(E)\rvert<\frac{\epsilon}{3}$ whenever $\mu(E)<\delta$ and $1\leq n\leq k_0$ This $\delta$ will suffice (by definition) for all $n$ up to $k_0$. We will show that it works for higher $n$ as well. Note that if $\mu(E)<\delta$, then

\displaystyle\begin{aligned}\rho(E_0\setminus E,E_0)=\mu\left((E_0\setminus E)\Delta E_0\right)&=\mu(E_0\cap E)\leq\mu(E)<\delta

so $E_0\setminus E$ and $E_0\cup E$ are both inside $\mathfrak{E}_{k_0}$. And so we calculate

$\displaystyle\lvert\nu_n(E)\rvert\leq\lvert\nu_{k_0}(E)\rvert+\lvert\nu_n(E_0\cup E)-\nu_{k_0}(E_0\cup E)\rvert+\lvert\nu_n(E_0\setminus E)-\nu_{k_0}(E_0\setminus E)\rvert$

The first term is less than $\frac{\epsilon}{3}$ by the definition of $\delta$. The second and third terms are less than or equal to $\frac{\epsilon}{3}$ because $E_0\cup E$ and $E_0\setminus E$ are in $\mathfrak{E}_{k_0}$. Since the same $\delta$ works for all $n$, the absolute continuity is uniform.

August 16, 2010 Posted by | Analysis, Measure Theory | 1 Comment

## Baire Sets

Looking over my notes from topology it seems I completely skipped over Baire sets. This was always one of those annoying topics that I never had much use for, partly because I didn’t do point-set topology or analysis. Also, even in my day the usual approach was a very classical and awkward one. Today I’m going to do a much more modern and streamlined one, and I can motivate it better from a measure-theoretic context to boot!

Basically, the idea of a Baire set is one that can’t be filled up by “negligible” sets. We’ve used that term in measure theory to denote a subset of a set of measure zero. But in topology we don’t have a “measure” to work with. Instead, we use the idea of a closed “nowhere dense” set — one for which there is no open set on which it is dense. The original motivation was a set like a boundary of a region; in the context of Jordan content we saw that such a set was negligible.

Clearly such a set has no interior — no open set completely contained inside — and any finite union of them is still nowhere dense. However, if we add up countably infinitely many we might have enough points to be dense on some open set. However, we don’t want to be able to actually fill such an open set. In the measure-theoretic context, this corresponds to the way any countable union of negligible sets is still negligible.

So, let’s be more specific: a “Baire set” is one for which the interior of every countable union of closed, nowhere dense sets is empty. Equivalently, we can characterize Baire sets in complementary terms: every countable intersection of dense open sets is dense. We can also use the contrapositive of the original definition: if a countable union of closed sets has an interior point, then one of the sets must itself have an interior point.

We’re interested in part of the famous “Baire category theorem” — the name is an artifact of the old, awkward approach and has nothing to do with category theory — which tells us that every complete metric space $X$ is a Baire space. Let $\{U_n\}$ be a countable collection of open dense subsets of $X$. We will show that their intersection is dense by showing that any nonempty open set $W$ has some point $x$ — the same point — in common with all the $U_n$.

Okay, Since $U_1$ is dense, then $U_1\cap W$ is nonempty, and it contains a point $x_1$. As the intersection of two open sets, it’s open, and so it contains an open neighborhood of $x_1$ which set can take to be an open metric ball of radius $r_1>0$. But then $B(x_1,r_1)$ is an open set, which will intersect $U_2$. This process will continue, and for every $n$ we will find a point $x_n$ and a radius $r_n$ so that $B(x_n,r_n)\subseteq B(x_{n-1},r_{n-1})\cap U_n$. We can also at each step pick $r_n<\frac{1}{n}$.

And so we come up with a sequence of points $\{x_n\}$. At each step, the ball $B(x_n,r_n)$ contains the whole tail of the sequence past $x_n$, and so all of these points are within $r_n$ of each other. Since $r_n$ gets arbitrarily small, this shows that $x_n$ is Cauchy, and since $X$ is complete, the sequence must converge to a limit $x$. This point $x$ will be in each set $U_n$, since $x\in B(x_n,r_n)\subseteq U_n$, and it’s obviously in $W$, as desired.

The other part of the Baire category theorem says that any locally compact Hausdorff space is a Baire space. In this case the proof proceeds very similarly, but with the finite intersection property for compact spaces standing in for completeness.

August 13, 2010

## Properties of Metric Spaces of Measure Rings

Today we collect a few facts about the metric space $\mathfrak{S}$ associated to a measure ring $(\mathcal{S},\mu)$.

First of all, the metric $\rho$ on $\mathfrak{S}$ is translation-invariant. That is, if $E$, $F$, and $G$ are sets in $\mathcal{S}$ with finite measure, then $\rho(E\Delta G,F\Delta G)=\rho(E,F)$. Indeed, we calculate

\displaystyle\begin{aligned}\rho(E\Delta G,F\Delta G)&=\mu\left((E\Delta G)\Delta(F\Delta G)\right)\\&=\mu\left((E\Delta F)\Delta(G\Delta G)\right)\\&=\mu\left((E\Delta F)\Delta\emptyset\right)\\&=\mu(E\Delta F)\\&=\rho(E,F)\end{aligned}

Next we need to define an “atom” of a measure ring to be a minimal nonzero element. That is, it’s an element $\emptyset\neq E\in\mathcal{S}$ so that if $F\subseteq E$ then either $F=\emptyset$ or $F=E$. We also define a measure ring to be “non-atomic” if it has no atoms. As a quick result: a totally $\sigma$-finite measure ring can have at most countably many atoms, since $X$ must contain all of them, and no two of them can have a nontrivial intersection — if there were uncountably many of them, any decomposition of $X$ could only cover countably many of them.

On the other hand, we define a metric space to be “convex” if for any two distinct elements $E$ and $F$ there is an element $G$ between them. That is, $G$ is neither $E$ nor $F$, and it satisfies the equation $\rho(E,F)=\rho(E,G)+\rho(G,F)$. We assert that the metric space of a $\sigma$-finite measure ring is convex if and only if the measure ring is non-atomic.

Let $E$ and $F$ be elements of $\mathfrak{S}$. Without loss of generality we can assume that $F=\emptyset$ by using the translation-invariance of $\rho$. Indeed, we can replace $E$ with $E\Delta F$ and $F$ with $F\Delta F=\emptyset$. There will be an element $G$ between $E$ and $F$ if and only if $G\Delta F$ is between $E\Delta F$ and $\emptyset$.

So for any $E$ is there an element $G$ between $E$ and $\emptyset$? Such a $G$ will satisfy

\displaystyle\begin{aligned}\mu(E)&=\rho(E,\emptyset)\\&=\rho(E,G)+\rho(G,\emptyset)\\&=\mu(E\Delta G)+\mu(G)\\&=\mu(E\setminus G)+\mu(G\setminus E)+\mu(G\setminus E)+\mu(G\cap E)\\&=\mu(E)+2\mu(G\setminus E)\end{aligned}

This is only possible if $\mu(G\setminus E)=0$, which means that $G\setminus E=\emptyset$, and so $G\subseteq E$. But for $G$ to be between $E$ and $\emptyset$ it cannot be equal to either of them, which means that $E$ cannot be an atom for any $E$ with $\mu(E)<\infty$. Since we can decompose any $E\in\mathcal{S}$ into a countable union of elements of finite measure, no element of infinite measure can be an atom either.

August 12, 2010 Posted by | Analysis, Measure Theory | 5 Comments

## Specifying Morphisms Between Boolean Rings

It will be useful to have other ways of showing that a given function $f:\mathcal{S}\to\mathcal{T}$ between boolean rings is a morphism. The definition, of course, is that $f(E\Delta F)=f(E)\Delta f(F)$ and $f(E\cap F)=f(E)\cap f(F)$, since $\Delta$ and $\cap$ here denote our addition and multiplication, respectively.

It’s also sufficient to show that $f(E\setminus F)=f(E)\setminus f(F)$ and $f(E\cup F)=f(E)\cup f(F)$. Indeed, we can build both $\Delta$ and $\cap$ from $\setminus$ and $\cup$:

\displaystyle\begin{aligned}E\Delta F&=(E\setminus F)\cup(F\setminus E)\\E\cap F&=E\setminus(E\setminus F)=F\setminus(E\setminus F)\end{aligned}

So if $\setminus$ and $\cup$ are preserved, then so are $\Delta$ and $\cap$.

A little more subtle is the fact that for surjections we can preserve order in both directions. That is, if $E\subseteq F$ is equivalent to $f(E)\subseteq f(F)$ for a surjection $f$ from one underlying set onto the other, then $f$ is a homomorphism of Boolean rings. We first show that $f$ preserves $\cup$ by using its characterization as a least upper bound: $E\subseteq E\cup F$, $F\subseteq E\cup F$, and if $E\subseteq G$ and $F\subseteq G$ then $E\cup F\subseteq G$.

So, since $f$ preserves order, we know that $f(E)\subseteq f(E\cup F)$, and that $f(F)\subseteq f(E\cup F)$. We can conclude from this that $f(E)\cup f(F)\subseteq f(E\cup F)$, and we are left to short that $f(E\cup F)\subseteq f(E)\cup f(F)$. But f(E)\cup f(F)=f(G)\$ for some $G\in\mathcal{S}$, since $f$ is surjective. Since $f(E)\subseteq f(G)$, we must have $E\subseteq G$, and similarly $F\subseteq G$. But then $E\cup F\subseteq G$, and so $f(E\cup F)\subseteq f(G)=f(E)\cup f(F)$, as we wanted to show.

We thus know that $f$ preserves the operation $\cup$, and we can similarly show that $f$ preserves $\cap$, using the dual universal property. In order to build $\setminus$ and $\Delta$ from $\cup$ and $\cap$, we need complements. But complements satisfy a universal property of their own: $E\cap E^c=\emptyset$, and if $E\cap F=\emptyset$ then $F\subseteq E^c$.

First, I want to show that $f(\emptyset)=\emptyset$ by showing it is below all elements of $\mathcal{T}$. Indeed, by the surjectivity of $f$, every element of $\mathcal{T}$ is the image of some element $E\in\mathcal{S}$. Thus we want to show that $f(\emptyset)\subseteq f(E)$. But this is true because $\emptyset\subseteq E$ for all $E\in\mathcal{S}$, and so $f(\emptyset)=\emptyset$

Now given a set $E$ and its complement $E^c=X\setminus E$, we know that $E\cap E^c=\emptyset$. Since $f$ preserves $\cap$, we must have $f(E)\cap f(E^c)=f(\emptyset)=\emptyset$. Let’s say $G\in\mathcal{T}$ is another set so that $f(E)\cap G=\emptyset$. By the surjectivity of $f$, we have $G=f(F)$ for some $F\in\mathcal{S}$. Thus $f(E\cap F)=f(E)\cap f(F)=\emptyset=f(\emptyset)$, and thus $E\cap F=\emptyset$. This tells us that $F\subseteq E^c$, and thus $G=f(F)\subseteq f(E^c)$. And so $f(E^c)=f(E)^c$.

Therefore, since $f$ preserves $\cup$, $\cap$, and complements, it preserves $\cup$ and $\setminus$, and thus is a morphism of Boolean rings.

## Completeness of Boolean Rings

We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite the same thing. We say that a Boolean ring $\mathcal{R}$ is complete if every set of elements of $\mathcal{R}$ has a union.

A complete Boolean ring is clearly a Boolean $\sigma$-algebra. It’s an algebra, because we can get a top element $X$ by taking the union of all the elements of $\mathcal{R}$ together, and it’s a $\sigma$-ring because countable unions are included under all unions. The converse is a little hairier.

First off, every totally finite measure algebra $(\mathcal{S},\mu)$ is complete. That is, if $\mu(X)<\infty$ — and thus $\mu(E)<\infty$ for all $E\in\mathcal{S}$, then every collection $\mathcal{E}$ of elements of $\mathcal{S}$ will have a union.

We let $\hat{\mathcal{E}}$ be the collection of all finite unions of elements in $\mathcal{E}$, all of which exist since $\mathcal{S}$ is a Boolean ring. We let $\alpha$ be the (finite) supremum of the measures of all elements in $\hat{\mathcal{E}}$. Since this is a finite supremum, there must be some increasing sequence $\{E_n\}$ so that $\mu(E_n)$ increases to $\alpha$. I say that the union $E$ of the sequence $\{E_n\}$ — which exists because $\mathcal{S}$ is a $\sigma$-ring — is the union of all of $\mathcal{E}$.

Indeed, we must have $\mu(E)=\alpha$ by the continuity of the measure $\mu$. Take any set $E_0\in\mathcal{E}$ and consider the difference $D=E_0\setminus E$, which is the amount by which $E_0$ extends past $E$. Our assertion is that this is always $\emptyset$. Define the sets $D_n=E_n\uplus D$, which is a disjoint union since $D$ is disjoint from $E$. Each of the $D_n$ is in $\hat{\mathcal{E}}$, and the limit of the sequence is $E\uplus D$. This set has measure

$\displaystyle\mu(E\uplus D)=\mu(E)+\mu(D)=\alpha+\mu(D)$

but since each of the $\mu(D_n)$ is bounded above by $\alpha$ then so is their limit. Thus $\mu(D)=0$, and thus $D=\emptyset$, since we consider all negligible sets to be the same.

The same is true for totally $\sigma$-finite measure algebras. Indeed, we can break such a measure algebra up into a countable collection of finite measure algebras by breaking $X$ into a countable number of elements $X_n$ so that $\mu(X_n)<\infty$. We define the finite measure algebra $\mathcal{S}_n$ to consist of the intersections of elements of $\mathcal{S}$ with $X_n$. Then given any collection $\mathcal{E}\subseteq\mathcal{S}$ we consider its image under such intersections to get $\mathcal{E}_n\subseteq\mathcal{S}_n$. What we said above shows that each of these collections has an intersection $E_n\in\mathcal{S}_n$, and their union in $\mathcal{S}$ is the union of the original collection $\mathcal{E}$.

## Completeness of the Metric Space of a Measure Space

Our first result today is that the metric space associated to the measure ring of a measure space $(X,\mathcal{S},\mu)$ is complete.

To see this, let $\{E_n\}$ be a Cauchy sequence in the metric space $\mathfrak{S}$. That is, for every $\epsilon>0$ there is some $N$ so that $\rho(E_m,E_n)<\epsilon$ for all $m,n>N$. Unpacking our definitions, each $E_n$ must be an element of the measure ring $(\mathcal{S},\mu)$ with $\mu(E_n)<0$, and thus must be (represented by) a measurable subset $E_n\subseteq X$ of finite measure. On the side of the distance function, we must have $\mu(E_m\Delta E_n)<\epsilon$ for sufficiently large $m$ and $n$.

Let’s recast this in terms of the characteristic functions $\chi_{E_n}$ of the sets in our sequence. Indeed, we find that $\chi_{E_m\Delta E_n}=\lvert\chi_{E_m}-\chi_{E_n}\rvert$, and so

$\displaystyle\mu(E_m\Delta E_n)=\int\chi_{E_m\Delta E_n}\,d\mu=\int\lvert\chi_{E_m}-\chi_{E_n}\rvert\,d\mu$

that is, a sequence $\{E_n\}$ of sets is Cauchy in $\mathfrak{S}(\mu)$ if and only if its sequence of characteristic functions $\left\{\chi_{E_n}\right\}$ is mean Cauchy. Since mean convergence is complete, the sequence of characteristic functions must converge in mean to some function $f$. But mean convergence implies convergence in measure, which is equivalent to a.e. convergence on sets of finite measure, which is what we’re dealing with.

Thus the limiting function $f$ must — like the characteristic functions in the sequence — take the value $0$ or $1$ almost everywhere. Thus it is (equivalent to) the characteristic function of some set. Since $f$ must be measurable — as the limit of a sequence of measurable functions — it’s the characteristic function of a measurable set, which must have finite measure since its measure is the limit of the Cauchy sequence $\{\mu(E_n)\}$. That is, $f=\chi_E$, where $E\in\mathfrak{S}(\mu)$, and $E$ is the limit of $\{E_n\}$ under the metric of $\mathfrak{S}(\mu)$. Thus $\mathfrak{S}(\mu)$ is complete as a metric space.

August 9, 2010 Posted by | Analysis, Measure Theory | 1 Comment

## The Metric Space of a Measure Ring

Let $(\mathcal{S},\mu)$ 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, since we’ve required that $\mu$ be positive — that $\mu(E)=0$ only if $E=\emptyset$ — we don’t need to worry about negligible elements.

And so we write $\mathfrak{S}=\mathfrak{S}(\mu)$ for the metric space consisting of the elements $E\in\mathcal{S}$ with $\mu(E)<\infty$. This has a distance function $\rho$ defined by $\rho(E,F)=\mu(E\Delta F)$. We also write $\mathfrak{S}=\mathfrak{S}(\mu)$ for the metric associated with the measure algebra associated with the measure space $(X,\mathcal{S},\mu)$. We say that a measure space or a measure algebra is “separable” if the associated metric space is separable.

Now, if we set

\displaystyle\begin{aligned}f(E,F)&=E\cup F\\g(E,F)&=E\cap F\end{aligned}

then $f:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}$, $g:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}$, and $\mu:\mathfrak{S}\to\mathbb{R}$ itself are all uniformly continuous.

Indeed, if we take two pairs of sets $E_1$, $E_2$, $F_1$, and $F_2$, we calculate

\displaystyle\begin{aligned}\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cap E_2^c\cap F_2^c)\cup(F_1\cap E_2^c\cap F_2^c)\right)\\&\leq\mu(E_1\cap E_2^c\cap F_2^c)+\mu(F_1\cap E_2^c\cap F_2^c)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)\end{aligned}

Similarly, we find that $\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\leq\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)$. And thus

\displaystyle\begin{aligned}\rho\left((E_1\cup F_1),(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\Delta(E_2\cup F_2)\right)\\&=\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)+\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)+\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)\\&=\mu(E_1\Delta E_2)+\mu(F_1\Delta F_2)\\&=\rho(E_1,E_2)+\rho(F_1,F_2)\end{aligned}

And so if we have control over the distance between $E_1$ and $E_2$, and the distance between $F_1$ and $F_2$, then we have control over the distance between $E_1\cup F_1$ and $E_2\cup F_2$. The bounds we need on the inputs uniform, and so $f$ is uniformly continuous. The proof for $g$ proceeds similarly.

To see that $\mu$ is uniformly continuous, we calculate

\displaystyle\begin{aligned}\left\lvert\mu(E)-\mu(F)\right\rvert&=\left\lvert\left(\mu(E)-\mu(E\cap F)\right)-\left(\mu(F)-\mu(F\cap E)\right)\right\rvert\\&=\left\lvert\mu\left(E\setminus(E\cap F)\right)-\mu\left(F\setminus(F\cap E)\right)\right\rvert\\&=\left\lvert\mu(E\setminus F)-\mu(F\setminus E)\right\rvert\\&\leq\mu(E\setminus F)+\mu(F\setminus E)\\&=\mu(E\Delta F)\\&=\rho(E,F)\end{aligned}

Now if $(X,\mathcal{S},\mu)$ is a $\sigma$-finite measure space so that the $\sigma$-ring $\mathcal{S}$ has a countable set of generators, then $\mathfrak{S}(\mu)$ is separable. Indeed, if $\{E_n\}$ is a countable sequence of sets that generate $\mathcal{S}$, then we may assume (by $\sigma$-finiteness) that $\mu(E_n)<\infty$ for all $n$. The ring generated by the $\{E_n\}$ is itself countable, and so we may assume that $\{E_n\}$ is itself a ring. But then we know that for every $E\in\mathfrak{S}(\mu)$ and for every positive $\epsilon$ we can find some ring element $E_n$ so that $\rho(E,E_n)=\mu(E\Delta E_n)<\epsilon$. Thus $\{E_n\}$ is a countable dense set in $\mathfrak{S}(\mu)$, which is thus separable.

August 6, 2010 Posted by | Analysis, Measure Theory | 5 Comments

## 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