# The Unapologetic Mathematician

## Boolean Rings

A “Boolean ring” is a commutative ring $R$ with the additional property that each and every element is idempotent. That is, for any $r\in R$ we have $r^2=r$. An immediate consequence of this axiom is that $r+r=0$, since we can calculate

\displaystyle\begin{aligned}0&=(r+r)-(r+r)\\&=(r+r)^2-(r+r)\\&=r^2+r^2+r^2+r^2-(r+r)\\&=r+r+r+r-(r+r)\\&=r+r\end{aligned}

The typical example we care about in the measure-theoretic context is a ring of subsets of some set $X$, with the operation $E\Delta F$ for addition and $E\cap F$ for multiplication. You should check that these operations satisfy the axioms of a Boolean ring. Since this is our main motivation, we will just consistently use $\Delta$ and $\cap$ to denote addition and multiplication in Boolean rings, whether they arise from a measure theoretic context or not. From here it looks a lot like set theory, but keep in mind that the objects we’re looking at may have nothing to do with sets.

We can use these operations to define the other common set-theoretic operations. Indeed

$\displaystyle E\cup F=(E\Delta F)\Delta(E\cap F)$

and

$\displaystyle E\setminus F=E\Delta(E\cap F)$

and we can then define orders in the usual manner: $E\subseteq F\Leftrightarrow E\cap F=E$.

As usual, the union of two elements is the “smallest” (with respect to this order) element above both of them, and the intersection of two elements is the “largest” element below both of them. The same goes for any finite number of elements, but if we try to move to an infinite number of elements there is no guarantee that there is any element above or below all of them, much less that such an element is unique. A “Boolean $\sigma$-ring” is a Boolean ring so that every countably infinite set of elements has a union. In this case, it is immediately true that any countably infinite set of elements has an intersection as well. The typical example, of course, is a $\sigma$-ring of subsets of a set $X$.

A “Boolean algebra” is a Boolean ring for which there is some element $X\neq0$ so that $E\subseteq X$ for all elements $E$. A “Boolean $\sigma$-algebra” is both a Boolean $\sigma$-ring and a Boolean algebra.

In the obvious way we have a full subcategory $\mathcal{B}oolean$ of the category of rings. It contains full subcategories of Boolean $\sigma$-rings, Boolean algebras, and Boolean $\sigma$-algebras.

August 4, 2010 Posted by | Algebra, Ring theory | 5 Comments

## 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

## Pulling Back and Pushing Forward Structure

Remember that we defined measurable functions in terms of inverse images, like we did for topological spaces. So it should be no surprise that we move a lot of measurable structure around between spaces by “pulling back” or “pushing forward”.

First of all, let’s say that $(Y,\mathcal{T})$ is a measurable space and consider a function $f:X\to Y$. We can always make $f$ into a measurable function by pulling back the $\sigma$-ring $\mathcal{T}$. For each measurable subset $E\subseteq Y$ we define the preimage $f^{-1}(E)=\{x\in X\vert f(x)\in E\}$ as usual, and define the pullback $f^{-1}(\mathcal{T})$ to be the collection of subsets of $X$ of the form $f^{-1}(E)$ for $E\in\mathcal{T}$. Taking preimages commutes with arbitrary setwise unions and setwise differences, and $f^{-1}(\emptyset)=\emptyset$, and so $f^{-1}(\mathcal{T})$ is itself a $\sigma$-ring. Every point $x\in X$ gives us a point $f(x)\in Y$, and every point $f(x)\in Y$ is contained in some measurable set $E\in\mathcal{T}$. Thus $x$ is contained in the set $f^{-1}(E)\in f^{-1}(\mathcal{T})$, and so we find that $(X,f^{-1}(\mathcal{T}))$ is a measurable space. Clearly, $f^{-1}(\mathcal{T})$ contains the preimage of every measurable set $E\in\mathcal{T}$, and so $f:(X,f^{-1}(\mathcal{T}))\to(Y,\mathcal{T})$ is measurable.

Measures, on the other hand, go the other way. Say that $(X,\mathcal{S},\mu)$ is a measure space and $f:(X,\mathcal{S})\to(Y,\mathcal{T})$ is a measurable function between measurable spaces, then we can define a new measure $\nu$ on $Y$ by “pushing forward” the measure $\mu$. Given a measurable set $E\subseteq Y$, we know that its preimage $f^{-1}(E)\subseteq X$ is also measurable, and so we can define $\nu(E)=\mu(f^{-1}(E))$. It should be clear that this satisfies the definition of a measure. We’ll write $\nu=f(\mu)$ for this measure.

If $f:X\to Y$ is a measurable function, and if $\mu$ is a measure on $X$, then we have the equality

$\displaystyle\int g\,d(f(\mu))=\int(g\circ f)\,d\mu$

in the sense that if either integral exists, then the other one does too, and their values are equal. As usual, it is sufficient to prove this for the case of $g=\chi_E$ for a measurable set $E\subseteq Y$. Linear combinations will extend it to simple functions, the monotone convergence theorem extends to non-negative measurable functions, and general functions can be decomposed into positive and negative parts.

Now, if $\chi_E$ is the characteristic function of $E$, then $\left[\chi_E\circ f\right](x)=1$ if $f(x)\in E$ — that is, if $x\in f^{-1}(E)$ — and $0$ otherwise. That is, $\chi_E\circ f=\chi_{f^{-1}(E)}$. We can then calculate

$\displaystyle\int\chi_E\,d(f(\mu))=\left[f(\mu)\right](E)=\mu(f^{-1}(E))=\int\chi_{f^{-1}(E)}\,d\mu=\int(\chi_E\circ f)\,d\mu$

As a particular case, applying the previous result to the function $g\chi_E$ shows us that

\displaystyle\begin{aligned}\int\limits_Eg(y)\,d\left[f(\mu)\right](y)&=\int\limits_Eg\,d(f(\mu))\\&=\int g\chi_E\,d(f(\mu))\\&=\int(g\circ f)(\chi_E\circ f)\,d\mu\\&=\int(g\circ f)\chi_{f^{-1}(E)}\,d\mu\\&=\int\limits_{f^{-1}(E)}(g\circ f)\,d\mu\\&=\int\limits_{f^{-1}(E)}g(f(x))\,d\mu(x)\end{aligned}

We can go back and forth between either side of this equation by the formal substitution $y=f(x)$.

Finally, we can combine this with the Radon-Nikodym theorem. If $f:X\to Y$ is a measurable function from a measure space $(X,\mathcal{S},\mu)$ to a totally $\sigma$-finite measure space $(Y,\mathcal{T},\nu)$ so that the pushed-forward measure $f(\mu)$ is absolutely continuous with respect to $\nu$. Then we can select a non-negative measurable function

$\displaystyle\phi=\frac{d(f(\mu)}{d\nu}:Y\to\mathbb{R}$

so that

$\displaystyle\int g(f(x))\,d\mu(x)=\int g(y)\phi(y)\,d\nu(y)$

again, in the sense that if one of these integrals exists then so does the other, and their values are equal. The function $\phi$ plays the role of the absolute value of the Jacobian determinant.

August 2, 2010 Posted by | Analysis, Measure Theory | 4 Comments