# The Unapologetic Mathematician

## Stone’s Representation Theorem III

We conclude our coverage of Stone’s representation theorem with a version for Boolean $\sigma$-algebra. Each such algebra $\mathcal{B}$ is isomorphic to a $\sigma$-algebra of subsets of some space, modulo a $\sigma$-ideal.

We start as we did for any Boolean algebra, by using the map $s$ sending $\mathcal{B}$ to the Boolean algebra of clopen subsets of the Stone space $S(\mathcal{B})$. This algebra is, of course, our identified base. Let $\mathcal{S}$ be the $\sigma$-ring generated by this base.

When we first dealt with measure rings, we quotiented out by the ideal of negligible sets. We don’t have a measure on $S(\mathcal{B})$, so we don’t have measurable sets, but we do have something almost as good. Just as when we discussed Baire spaces, we can use nowhere-denseness as a topological stand-in for negligibility. In fact, we’ll define a “meager” set to be any countable union of nowhere-dense sets, and we’ll let $\mathcal{N}$ be the collection of all meager sets in $\mathcal{S}$. It is straightforward to verify that $\mathcal{N}$ is a $\sigma$-ideal in $\mathcal{S}$.

A quick side note: classically, meager sets were said to be “of the first category”. Any other sets were said to be “of the second category”. This is the root of the term “category” in the Baire category theorem.

Now if $\{E_n\}$ is a sequence of clopen sets, then we can find a unique preimage $s^{-1}(E_n)\in\mathcal{B}$. Since $\mathcal{B}$ is a $\sigma$-algebra, we can take the countable union of these elements, and then apply $s$ to the result. That is, we can define the set

$\displaystyle E=s\left(\bigcup\limits_{n=1}^\infty s^{-1}(E_n)\right)$

Now, of course, we can also take the countable union of the sets themselves. If $s$ were an isomorphism of $\sigma$-algebras, then this union would be exactly $E$ itself; but we aren’t usually so lucky. Instead, I say that the difference

$\displaystyle E\setminus\bigcup\limits_{n=1}^\infty E_n$

is nowhere-dense, and thus countable unions of clopen sets are clopen modulo meager sets.

Indeed, the countable union of the (open) sets $E_n$ is open, and so its complement is closed. The difference above is the intersection of the (closed) set $E$ with the (closed) complement of the union, and is thus closed. So if it were dense on some nonempty open set $U$ it would have to actually contain $U$. But since $U$ is open it must be the union of some collection of basic clopen sets, and we can take $U$ to be one of these sets. That is, $U\subseteq E$ and $U\cap E_n=\emptyset$ for each $n$. Since these relations only involve clopen sets, we find that $s^{-1}(U)\subseteq s^{-1}(E)$ and $s^{-1}(U)\cap s^{-1}(E_n)=\emptyset$. But there’s no way this can happen if $s^{-1}(E)$ is the union of the $s^{-1}(E_n)$!

So the map $s$ takes elements of $\mathcal{B}$ to clopen sets in $\mathcal{S}$, and then on to their equivalence classes in $\mathcal{S}/\mathcal{N}$, and this map commutes with countable unions. All that remains to show that this is an isomorphism $\mathcal{B}\cong\mathcal{S}/\mathcal{N}$ is to show that no two clopen sets in $\mathcal{S}$ represent the same equivalence class. That is, if $U_1$ and $U_2$ are distinct clopen sets, then $U_1\Delta U_2$ cannot be meager. Equivalently, no nonempty clopen set is meager.

But this is just a consequence of the Baire category theorem, here in its formulation for compact Hausdorff spaces. Indeed, if a clopen set in any Baire space — and a compact Hausdorff space is Baire — were the countable union of closed nowhere-dense sets, then its interior would be empty. But since it’s open its interior is itself, and thus is would have to be empty. Thus if $U_1$ and $U_2$ are clopen sets representing the same equivalence class modulo $\mathcal{N}$ then their (clopen) symmetric difference $U_1\Delta U_2$ is meager, and thus empty. That is, $U_1=U_2$.

August 20, 2010

## Stone’s Representation Theorem II

We can extend yesterday’s result in the case that $\mathcal{B}$ is a Boolean algebra. Now as a ring, $\mathcal{B}$ has a unit. We adjust our definition of the Stone space to define

$\displaystyle S(\mathcal{B})=\hom_\mathbf{Ring}(\mathcal{B},\mathcal{B}_0)$

That is, we insist that the ring homomorphisms preserve the identity, sending the top element of $\mathcal{B}$ to $1\in\mathcal{B}_0$. This doesn’t really change anything we said yesterday, and it all goes through as before.

What is new is that the image of $s$ — the identified base for the topology on $S(\mathcal{B})$ — consists of all the subsets of $S(\mathcal{B})$ which are clopen — both open and closed. That is, elements of $\mathcal{B}$ correspond to unions of connected components of $S(\mathcal{B})$.

First, we must show that $s(b)$ is closed, since we already know that it’s open by definition. I say that the complement $s(b)^c$ is actually the open basic set $s(b^c)$. Indeed, $b^c=b\Delta1$, and we calculate

\displaystyle\begin{aligned}s\left(b^c\right)&=s(b\Delta1)\\&=s(b)\Delta s(1)\\&=s(b)\Delta1\\&=s(b)^c\end{aligned}

Thus each set $s(b)$ is the complement of an open set, and is thus closed as well.

It also happens that our base is closed under finite unions. Indeed, we use DeMorgan’s laws and calculate

\displaystyle\begin{aligned}s(b)\cup s(b')&=\left(s(b)^c\cap s(b')^c\right)^c\\&=\left(s\left(b^c\right)\cap s\left(b'^c\right)\right)^c\\&=s\left(b^c\cap b'^c\right)^c\\&=s\left(\left(b^c\cap b'^c\right)^c\right)\\&=s(b\cup b')\end{aligned}

And from there we can extend to any finite unions we want.

Now I say that if a base of clopen sets in a compact space is closed under finite unions, then it contains every clopen set in the space. Indeed, such a clopen set can be written as a union of sets in the base since it’s open. This union gives an open covering of the set. Since the set is closed, it is compact. And so the open covering we just found has a finite subcover. That is, we can write our clopen set as a finite union of basic sets, and so it is itself in the base by assumption.

Thus in our particular case, our base of $S(\mathcal{B})$ consists of all the clopen sets in the Stone space, as we asserted!

August 19, 2010 Posted by | Analysis, Measure Theory | 2 Comments

## 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

## 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

## 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

## Measurable Functions on Pulled-Back Measurable Spaces

We start today with a possibly surprising result; pulling back a $\sigma$-ring puts significant restrictions on measurable functions. If $f:X\to Y$ is a function from a set into a measurable space $(Y,\mathcal{T})$, and if $g:X\to\mathbb{R}$ is measurable with respect to the $\sigma$-ring $f^{-1}(\mathcal{T})$ on $X$, then $g(x_1)=g(x_2)$ whenever $f(x_1)=f(x_2)$.

To see this fix a point $x_1\in X$, and let $F_1\subseteq Y$ be a measurable set containing $f(x_1)$. Its preimage $f^{-1}(F_1)\subseteq X$ is then a measurable set containing $x_1$. We can also define the level set $\{x\in X\vert g(x)=g(x_1)\}$, which is a measurable set since $g$ is a measurable function. Thus the intersection

$\displaystyle\{x\in X\vert g(x)=g(x_1)\}\cap f^{-1}(F_1)\subseteq X$

is measurable. That is, it’s in $f^{-1}(\mathcal{T})$, and so there exists some measurable $F\subseteq Y$ so that $f^{-1}(F)$ is this intersection. Clearly $f(x_1)\in F$, and so $f(x_2)$ is as well, by assumption. But then $x_2\in f^{-1}(F)\subseteq\{x\in X\vert g(x)=g(x_1)\}$, and we conclude that $g(x_2)=g(x_1)$.

From this result follows another interesting property. If $f:X\twoheadrightarrow Y$ is a mapping from a set $X$ onto a measurable space $(Y,\mathcal{T})$, and if $g:(X,f^{-1}(\mathcal{T})\to(Z,\mathcal{U})$ is a measurable function, then there is a unique measurable function $h:(Y,\mathcal{T})\to(Z,\mathcal{U})$ so that $g=h\circ f$. That is, any function that is measurable with respect to a measurable structure pulled back along a surjection factors uniquely through the surjection.

Indeed, since $f$ is surjective, for every $y\in Y$ we have some $x\in X$ so that $f(x)=y$. Then we define $h(y)=g(x)$, so that $g(x)=h(f(x))$, as desired. There is no ambiguity about the choice of which preimage $x$ of $y$ to use, since the above result shows that any other choice would lead to the same value of $g(x)$. What’s not immediately apparent is that $h$ is itself measurable. But given a set $M\in\mathcal{U}$ we can consider its preimage $\{y\in Y\vert h(y)\in M\}$, and the preimage of this set:

\displaystyle\begin{aligned}f^{-1}\left(\{y\in Y\vert h(y)\in M\}\right)&=\left\{x\in X\big\vert f(x)\in\{y\in Y\vert h(y)\in M\}\right\}\\&=\{x\in X\vert h(f(x))\in M\}\\&=\{x\in X\vert g(x)\in M\}\\&=g^{-1}(M)\end{aligned}

which is measurable since $g$ is a measurable function. But then this set must be the preimage of some measurable subset of $Y$, which shows that the preimage $h^{-1}(M)\subseteq Y$ is measurable.

It should be noted that this doesn’t quite work out for functions $f$ that are not surjective, because we cannot uniquely determine $h(y)$ if $y$ has no preimage under $f$.

August 3, 2010