# The Unapologetic Mathematician

## A Counterexample

I’ve been thinking about yesterday’s post about the product of measurable spaces, and I’m not really satisfied. I’d like to present a concrete example of what can go wrong when one or the other factor isn’t a total measurable space — that is, when the underlying set of that factor is not a measurable subset of itself.

Let $X=\mathbb{R}$ be the usual real line. However, instead of the Borel or Lebesgue measurable sets, let $\mathcal{S}$ be the collection of all countable subsets of $X$. We check that this is a $\sigma$-ring: it’s clearly closed under unions and differences, making it a ring, and the union of a countable collection of countable sets is still countable, so it’s monotone, and thus a $\sigma$-ring. And every point $x\in X$ is in the measurable set $\{x\}\in\mathcal{S}$, so $(X,\mathcal{S})$ is a measurable space. We’ll let $(Y,\mathcal{T})=(Z,\mathcal{U})=(X\mathcal{S})$ be two more copies of the same measurable space.

Now we define the “product” space $(X\times Y,\mathcal{S}\times\mathcal{T})$. A subset of $X\times Y$ is in $\mathcal{S}\times\mathcal{T}$ if it can be written as the union of a countable number of measurable rectangles $A\times B$ with $A\in\mathcal{S}$ and $B\in\mathcal{T}$. But if $A$ and $B$ are both countable, then $A\times B$ is also countable, and thus any measurable subset of $X\times Y$ is countable. On the other hand, if a subset of $X\times Y$ is countable, it’s the countable union of all of its singletons $\{(a,b)\}=\{a\}\times\{b\}$, each of which is a measurable rectangle. That is, if a subset of $X\times Y$ is countable, then it is measurable. That is, $\mathcal{S}\times\mathcal{T}$ is the collection of all the countable subsets of $X\times Y$.

Next we define the function $f:Z\to X\times Y$ by setting $f(z)=(z,0)$ for any real number $z$. We check that it’s measurable: given a measurable set $C\subseteq X\times Y$ we calculate

$\displaystyle f^{-1}(C)=\{z\in Z\vert f(z)\in C\}=\{z\in Z\vert (z,0)\in C\}$

but if $C$ is measurable, then it’s countable, and it can contain only countably many points of the form $(z,0)$. The preimage $f^{-1}(C)$ must then be countable, and thus measurable, and thus $f$ is a measurable function.

That said, the component function $f_2=\pi_2\circ f$ is not measurable. Indeed, we have $f_2(z)=0$ for all real numbers $z$. The set $\{0\}\subseteq Y$ is countable — and thus measurable — but its preimage $f^{-1}(\{0\})=Z$ is the entire real line, which is uncountable and is not measurable. This is exactly the counterintuitive result we were worried about.

July 16, 2010

## Product Measurable Spaces

Now we return to the category of measurable spaces and measurable functions, and we discuss product spaces. Given two spaces $(X,\mathcal{S})$ and $(Y,\mathcal{T})$, we want to define a $\sigma$-ring of measurable sets on the product $X\times Y$.

In fact, we’ve seen that given rings $\mathcal{S}$ and $\mathcal{T}$ we can define the product $\mathcal{S}\times\mathcal{T}$ as the collection of finite disjoint unions of sets $A\times B$, where $A\in\mathcal{S}$ and $B\in\mathcal{T}$. If $\mathcal{S}$ and $\mathcal{T}$ are $\sigma$-rings, then we define $\mathcal{S}\times\mathcal{T}$ to be the smallest monotone class containing this collection, which will then be a $\sigma$-ring. If $\mathcal{S}$ and $\mathcal{T}$ are $\sigma$-algebras — that is, if $X\in\mathcal{S}$ and $Y\in\mathcal{T}$ — then clearly $X\times Y\in\mathcal{S}\times\mathcal{T}$, which is thus another $\sigma$-algebra.

Indeed, $(X\times Y,\mathcal{S}\times\mathcal{T})$ is a measurable space. The collection $\mathcal{S}\times\mathcal{T}$ is a $\sigma$-algebra, and every point is in some one of these sets. If $(x,y)\in X\times Y$, then $x\in A\in\mathcal{S}$ and $y\in B\in\mathcal{T}$, so $(x,y)\in A\times B\in\mathcal{S}\times\mathcal{T}$. But is this really the $\sigma$-algebra we want?

Most approaches to measure theory simply define this to be the product of two measurable spaces, but we have a broader perspective here. Indeed, we should be asking if this is a product object in the category of measurable spaces! That is, the underlying space $X\times Y$ comes equipped with projection functions $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. We must ask if these projections are measurable for our choice of $\sigma$-algebra, and also that they satisfy the universal property of a product object.

Checking measurability is pretty straightforward. It’s the same for either projector, so we’ll consider $\pi_1:X\times Y\to X$ explicitly. Given a measurable set $A\in\mathcal{S}$, the preimage $\pi_1^{-1}(A)=A\times Y$ must be measurable as well. And here we run into a snag: we know that $A$ is measurable, and so for any measurable $B\in\mathcal{T}$ the subset $A\times B\subseteq A\times Y$ is measurable. In particular, if $\mathcal{T}$ is a $\sigma$-algebra then $A\times Y$ is measurable right off. However, if $Y$ is not itself measurable then $Y$ is not the limit of any increasing countable sequence of measurable sets either (since $\sigma$-rings are monotonic classes). Thus $A\times Y$ is not the limit of any increasing sequence of measurable products $A\times B$, and so $A\times Y$ can’t be in the smallest monotonic class generated by such products, and thus can’t be measurable!

So in order for $(X\times Y,\mathcal{S}\times\mathcal{T})$ to be a product object, it’s necessary that $\mathcal{T}$ be a $\sigma$-algebra. Similarly, $\mathcal{S}$ must be a $\sigma$-algebra as well. What would happen if this condition fails? Consider a measurable function $f:Z\to X\times Y$ defined by $f(z)=(f_1(z),f_2(z))$. We can write $f_1=\pi_1\circ f$, but since $\pi_1$ is not measurable we have no guarantee that $f_1$ will be measurable!

On the other hand, all is not lost; if $f_1:Z\to X$ and $f_2:Z\to Y$ are both measurable, and if $A\in\mathcal{S}$ and $B\in\mathcal{T}$, then we calculate the preimage

\displaystyle\begin{aligned}f^{-1}(A\times B)&=\left\{z\in Z\vert f(z)\in A\times B\right\}\\&=\left\{z\in Z\vert (f_1(z),f_2(z))\in A\times B\right\}\\&=\left\{z\in Z\vert f_1(z)\in A\textrm{ and }f_2(z)\in B\right\}\\&=\left\{z\in Z\vert f_1(z)\in A\right\}\cap\left\{z\in Z\vert f_2(z)\in B\right\}\\&=f_1^{-1}(A)\cap f_2^{-1}(B)\end{aligned}

which is thus measurable. Monotone limits of finite disjoint unions of such sets are easily handled. Thus if both components of $f$ are measurable, then $f$ is measurable.

Okay, so back to the case where $\mathcal{S}$ and $\mathcal{T}$ are both $\sigma$-algebras, and so $\pi_1$ and $\pi_2$ are both measurable. We still need to show that the universal property holds. That is, given two measurable functions $f_1:Z\to X$ and $f_2:Z\to Y$, we must show that there exists a unique measurable function $f:Z\to X\times Y$ so that $f_1=\pi_1\circ f$ and $f_2=\pi_2\circ f$. There’s obviously a unique function on the underlying set: $z\mapsto(f_1(z),f_2(z))$. And the previous paragraph shows that this function must be measurable!

So, the uniqueness property always holds, but with the one caveat that the projectors may not themselves be measurable. That is, the full subcategory of total measurable spaces has product objects, but the category of measurable spaces overall does not. However, we’ll still talk about the “product” space $X\times Y$ with the $\sigma$-ring $\mathcal{S}\times\mathcal{T}$ understood.

A couple of notes are in order. First of all, this is the first time that we’ve actually used the requirement that every point in a measurable space be a member of some measurable set or another. It will become more important as we go on. Secondly, we define a “measurable rectangle” in $X\times Y$ to be a set $A\times B$ so that $A\in\mathcal{S}$ and $B\in\mathcal{T}$ — that is, one for which both “sides” are measurable. The class of all measurable sets $\mathcal{S}\times\mathcal{T}$ is the $\sigma$-ring generated by all the measurable rectangles.

July 15, 2010 Posted by | Analysis, Measure Theory | 5 Comments

## Lebesgue Decomposition

As we’ve said before, singularity and absolute continuity are diametrically opposed. And so it’s not entirely surprising that if we have two totally $\sigma$-finite signed measures $\mu$ and $\nu$, then we can break $\nu$ into two uniquely-defined pieces, $\nu_c$ and $\nu_s$, so that $\nu_c\ll\mu$, $\nu_s\perp\mu$, and $\nu=\nu_c+\nu_s$. We call such a pair the “Lebesgue decomposition” of $\nu$ with respect to $\mu$.

Since a signed measure is absolutely continuous or singular with respect to $\mu$ if and only if it’s absolutely continuous or singular with respect to $\lvert\mu\rvert$, we may as well assume that $\mu$ is a measure. And since in both cases $\nu$ is absolutely continuous or singular with respect to $\mu$ if and only if $\nu^+$ and $\nu^-$ both are, we may as well assume that $\nu$ is also a measure. And, as usual, we can break our measurable space $X$ down into the disjoint union of countably many subspaces on which both $\mu$ and $\nu$ are both totally finite. We can assemble the Lebesgue decompositions on a collection such subspaces into the Lebesgue decomposition on the whole, and so we can assume that $\mu$ and $\nu$ are totally finite.

Now that we can move on to the proof itself, the core is based on our very first result about absolute continuity: $\nu\ll\mu+\nu$. Thus the Radon-Nikodym theorem tells us that there exists a function $f$ so that

$\displaystyle\nu(E)=\int\limits_Ef\,d(\mu+\nu)=\int\limits_Ef\,d\mu+\int\limits_Ef\,d\nu$

for every measurable set $E$. Since $0\leq\nu(E)\leq\mu(E)+\nu(E)$, we must have $0\leq f\leq1$ $(\mu+\nu)$-a.e., and thus $0\leq f\leq1$ $\nu$-a.e. as well. Since

Let us define $A=\{x\in X\vert f(x)=1\}$ and $B=\{x\in X\vert0\leq f(x)<1\}$. Then we calculate

$\displaystyle\nu(A)=\int\limits_A\,d\mu+\int\limits_A\,d\nu=\mu(A)+\nu(A)$

and thus (by the finiteness of $\nu$), $\mu(A)=0$. Defining $\nu_s(E)=\nu(E\cap A)$ and $\nu_c(E)=\nu(E\cap B)$, then it’s clear that $\nu_s\perp\mu$. We still must prove that $\nu_c\ll\mu$.

If $\mu(E)=0$, then we calculate

$\displaystyle\int\limits_{E\cap B}\,d\nu=\nu(E\cap B)=\int\limits_{E\cap B}f\,d\mu+\int\limits_{E\cap B}f\,d\nu=\int\limits_{E\cap B}f\,d\nu$

and, therefore

$\displaystyle\int\limits_{E\cap B}(1-f)\,d\mu=0$

But $1-f\geq0$ $\nu$-a.e., which means that we must have $\nu_c(E)=\nu(E\cap B)=0$, and thus $\nu_c\ll\mu$.

Now, suppose $\nu=\nu_s+\nu_c$ and $\nu=\bar{\nu}_s+\bar{\nu}_c$ are two Lebesgue decompositions of $\nu$ with respect to $\mu$. Then $\nu_s-\bar{\nu}_s=\bar{\nu}_c-\nu_c$. We know that both singularity and absolute continuity pass to sums, so $\nu_s-\bar{\nu}_s$ is singular with respect to $\mu$, while $\bar{\nu}_c-\nu_c$ is absolutely continuous with respect to $\mu$. But the only way for this to happen is for them both to be zero, and thus $\nu_s=\bar{\nu}_s$ and $\nu_c-\bar{\nu}_c$.

July 14, 2010

## Corollaries of the Chain Rule

Today we’ll look at a couple corollaries of the Radon-Nikodym chain rule.

First up, we have an analogue of the change of variables formula, which was closely tied to the chain rule in the first place. If $\lambda$ and $\mu$ are totally $\sigma$-finite signed measures with $\mu\ll\lambda$, and if $f$ is a finite-valued $\mu$-integrable function, then

$\displaystyle\int f\,d\mu=\int f\frac{d\mu}{d\lambda}\,d\lambda$

which further justifies the the substitution of one “differential measure” for another.

So, define a signed measure $\nu$ as the indefinite integral of $f$. Immediately we know that $\nu$ is totally $\sigma$-finite and that $\nu\ll\mu$. And, obviously, $f$ is the Radon-Nikodym derivative of $\nu$ with respect to $\mu$. Thus we can invoke the above chain rule to conclude that $\lambda$-a.e. we have

$\displaystyle\frac{d\nu}{d\lambda}=f\frac{d\mu}{d\lambda}$

We then know that for every measurable $E$

$\displaystyle\nu(E)=\int\limits_Ef\frac{d\mu}{d\lambda}\,d\lambda$

and the substitution formula follows by putting $X$ in for $E$.

Secondly, if $\mu$ and $\nu$ are totally $\sigma$-finite signed measures so that $\mu\equiv\nu$ — that is, $\mu\ll\nu$ and $\nu\ll\mu$ — then $\mu$-a.e. we have

$\displaystyle\frac{d\mu}{d\nu}\frac{d\nu}{d\mu}=1$

Indeed, $\mu\ll\mu$, and by definition we have

$\displaystyle\mu(E)=\int\limits_E1\,d\mu$

so $1$ serves as the Radon-Nikodym derivative of $\mu$ with respect to itself. Putting this into the chain rule immediately gives us the desired result.

July 13, 2010 Posted by | Analysis, Measure Theory | 6 Comments

Today we take the Radon-Nikodym derivative and prove that it satisfies an analogue of the chain rule.

If $\lambda$, $\mu$, and $\nu$ are totally $\sigma$-finite signed measures so that $\nu\ll\mu$ and $\mu\ll\lambda$, then $\lambda$-a.e. we have

$\displaystyle\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}$

By the linearity we showed last time, if this holds for the upper and lower variations of $\nu$ then it holds for $\nu$ itself, and so we may assume that $\nu$ is also a measure. We can further simplify by using Hahn decompositions with respect to both $\lambda$ and $\mu$, passing to subspaces on which each of our signed measures has a constant sign. We will from here on assume that $\lambda$ and $\mu$ are (positive) measures, and the case where one (or the other, or both) has a constant negative sign has a similar proof.

Let’s also simplify things by writing

\displaystyle\begin{aligned}f&=\frac{d\nu}{d\mu}\\g&=\frac{d\mu}{d\lambda}\end{aligned}

Since $\mu$ and $\nu$ are both non-negative there is also no loss of generality in assuming that $f$ and $g$ are everywhere non-negative.

So, let $\{f_n\}$ be an increasing sequence of non-negative simple functions converging pointwise to $f$. Then monotone convergence tells us that

\displaystyle\begin{aligned}\lim\limits_{n\to\infty}\int\limits_Ef_n\,d\mu&=\int\limits_Ef\,d\mu\\\lim\limits_{n\to\infty}\int\limits_Ef_ng\,d\lambda&=\int\limits_Efg\,d\lambda\end{aligned}

for every measurable $E$. For every measurable set $F$ we find that

$\displaystyle\int\limits_E\chi_F\,d\mu=\mu(E\cap F)=\int\limits_{E\cap F}\,d\mu=\int\limits_{E\cap F}g\,d\lambda=\int\limits_E\chi_Fg\,d\lambda$

and so for all the simple $f_n$ we conclude that

$\displaystyle\int\limits_Ef_n\,d\mu=\int\limits_Ef_ng\,d\lambda$

Passing to the limit, we find that

$\displaystyle\nu(E)=\int\limits_E\,d\nu=\int\limits_Ef\,d\mu=\int\limits_Efg\,d\lambda$

and so the product $fg$ serves as the Radon-Nikodym derivative of $\nu$ in terms of $\lambda$, and it’s uniquely defined $\lambda$-almost everywhere.

July 12, 2010 Posted by | Analysis, Measure Theory | 10 Comments

Okay, so the Radon-Nikodym theorem and its analogue for signed measures tell us that if we have two $\sigma$-finite signed measures $\mu$ and $\nu$ with $\nu\ll\mu$, then there’s some function $f$ so that

$\displaystyle\nu(E)=\int\limits_Ef\,d\mu$

But we also know that by definition

$\displaystyle\nu(E)=\int\limits_E\,d\nu$

If both of these integrals were taken with respect to the same measure, we would know that the equality

$\displaystyle\int\limits_Ef\,d\mu=\int\limits_Eg\,d\mu$

for all measurable $E$ implies that $f=g$ $\mu$-almost everywhere. The same thing can’t quite be said here, but it motivates us to say that in some sense we have equality of “differential measures” $d\nu=f\,d\mu$. In and of itself this doesn’t really make sense, but we define the symbol

$\displaystyle\frac{d\nu}{d\mu}=f$

and call it the “Radon-Nikodym derivative” of $\nu$ by $\mu$. Now we can write

$\displaystyle\int\limits_E\,d\nu=\int\limits_Ef\,d\mu=\int\limits_E\frac{d\nu}{d\mu}\,d\mu$

The left equality is the Radon-Nikodym theorem, and the right equality is just the substitution of the new symbol for $f$. Of course, this function — and the symbol $\frac{d\nu}{d\mu}$ — is only defined uniquely $\mu$-almost everywhere.

The notation and name is obviously suggestive of differentiation, and indeed the usual laws of derivatives hold. We’ll start today by the easy property of linearity.

That is, if $\nu_1$ and $\nu_2$ are both $\sigma$-finite signed measures, and if $a_1$ and $a_2$, then $a_1\nu_1+a_2\nu_2$ is clearly another $\sigma$-finite signed measure. Further, it’s not hard to see if $\nu_i\ll\mu$ then $a_1\nu_1+a_2\nu_2\ll\mu$ as well. By the Radon-Nikodym theorem we have functions $f_1$ and $f_2$ so that

\displaystyle\begin{aligned}\nu_1(E)&=\int\limits_Ef_1\,d\mu\\\nu_2(E)&=\int\limits_Ef_2\,d\mu\end{aligned}

for all measurable sets $E$. Then it’s clear that

\displaystyle\begin{aligned}\left[a_1\nu_1+a_2\nu_2\right](E)&=a_1\nu_1(E)+a_2\nu_2(E)\\&=a_1\int\limits_Ef_1\,d\mu+a_2\int\limits_Ef_2\,d\mu\\&=\int\limits_Ea_1f_1+a_2f_2\,d\mu\end{aligned}

That is, $a_1f_1+a_2f_2$ can serve as the Radon-Nikodym derivative of $a_1\nu_1+a_2\nu_2$ with respect to $\mu$. We can also write this in our suggestive notation as

$\displaystyle\frac{d(a_1\nu_1+a_2\nu_2)}{d\mu}=a_1\frac{d\nu_1}{d\mu}+a_2\frac{d\nu_2}{d\mu}$

which equation holds $\mu$-almost everywhere.

July 9, 2010 Posted by | Analysis, Measure Theory | 3 Comments

## The Radon-Nikodym Theorem for Signed Measures

Now that we’ve proven the Radon-Nikodym theorem, we can extend it to the case where $\mu$ is a $\sigma$-finite signed measures.

Indeed, let $X=A\uplus B$ be a Hahn decomposition for $\mu$. We find that $\mu^+$ is a $\sigma$-finite measure on $A$, while $\mu^-$ is a $\sigma$-finite measure on $B$.

As it turns out that $\nu\ll\mu^+$ on $A$, while $\nu\ll\mu^-$ on $B$. For the first case, let $E\subseteq A$ be a set for which $\mu^+(E)=0$. Since $E\cap B=\emptyset$, we must have $\mu^-(E)=0$, and so $\lvert\mu\rvert(E)=\mu^+(E)+\mu^-(E)=0$. Then by absolute continuity, we conclude that $\nu(E)=0$, and thus $\nu\ll\mu^+$ on $A$. The proof that $\nu\ll\mu^-$ on $B$ is similar.

So now we can use the Radon-Nikodym theorem to show that there must be functions $f_A$ on $A$ and $f_B$ on $B$ so that

\displaystyle\begin{aligned}\nu(E\cap A)=&\int\limits_{E\cap A}f_A\,d\mu^+\\\nu(E\cap B)=&\int\limits_{E\cap B}f_B\,d\mu^-=-\int\limits_{E\cap B}-f_B\,d\mu^-\end{aligned}

We define a function $f$ on all of $X$ by $f(x)=f^+(x)$ for $x\in A$ and $f(x)=f^-(x)$ for $x\in B$. Then we can calculate

\displaystyle\begin{aligned}\nu(E)&=\nu((E\cap A)\uplus(E\cap B))\\&=\nu(E\cap A)+\nu(E\cap B)\\&=\int\limits_{E\cap A}f_A\,d\mu^+-\int\limits_{E\cap B}-f_B\,d\mu^-\\&=\int\limits_{E\cap A}f\,d\mu^+-\int\limits_{E\cap B}f\,d\mu^-\\&=\int\limits_Ef\,d\mu\end{aligned}

which in exactly the conclusion of the Radon-Nikodym theorem for the signed measure $\mu$.

July 8, 2010 Posted by | Analysis, Measure Theory | 5 Comments

Today we set about the proof of the Radon-Nikodym theorem. We assumed that $(X,\mathcal{S},\mu)$ is a $\sigma$-finite measure space, and that $\nu$ is a $\sigma$-finite signed measure. Thus we can write $X$ as the countable union of subsets on which both $\mu$ and $\nu$ are finite, and so without loss of generality we may as well assume that they’re finite to begin with.

Now, if we assume for the moment that we’re correct and an $f$ does exist so that $\nu$ is its indefinite integral, then the fact that $\nu$ is finite means that $f$ is integrable, and then if $g$ is any other such function we can calculate

$\displaystyle\int\limits_Ef-g\,d\mu=\int\limits_Ef\,d\mu-\int\limits_Ef\,d\mu=\nu(E)-\nu(E)=0$

for every measurable $E\in\mathcal{S}$. Now we know that this implies $f-g=0$ a.e., and thus the uniqueness condition we asserted will hold.

Back to the general case, we know that the absolute continuity $\nu\ll\mu$ is equivalent to the conjunction of $\nu^+\ll\mu$ and $\nu^-\ll\mu$, and so we can reduce to the case where $\nu$ is a finite measure, not just a finite signed measure.

Now we define the collection $\mathcal{K}$ of all nonnegative functions $f$ which are integrable with respect to $\mu$, and for which we have

$\displaystyle\int\limits_Ef\,d\mu\leq\nu(E)$

for every measurable $E$. We define

$\displaystyle\alpha=\sup\limits_{f\in\mathcal{K}}\int f\,d\mu$

Since $\alpha$ is the supremum, we can find a sequence $\{f_n\}$ of functions in $\mathcal{K}$ so that

$\displaystyle\lim\limits_{n\to\infty}\int f\,d\mu=\alpha$

For each $n$ we define

$\displaystyle g_n(x)=\max\limits_{1\leq i\leq n}f_i(x)$

Now if $E$ is some measurable set we can break it into the finite disjoint union of $n$ sets $E_i$ so that $g_n=f_i$ on $E_i$. Thus we have

$\displaystyle\int\limits_Eg_n\,d\mu=\sum\limits_{i=1}^n\int\limits_{E_i}f_i\,d\mu\leq\sum\limits_{i=1}^n\nu(E_i)=\nu(E)$

and so $g_n\in\mathcal{K}$.

We can write $g_n=g_{n-1}\cup f_n$, which tells us that the sequence $\{g_n\}$ is increasing. We define $f_0$ to be the limit of the $g_n$$f_0(x)$ is the maximum of all the $f_i(x)$ — and use the monotone convergence theorem to tell us that

$\displaystyle\int_Ef_0\,d\mu=\lim\limits_{n\to\infty}\int_Eg_n\,d\mu$

Since all of the integrals on the right are bounded above by $\nu(E)$, their limit is as well, and $f_0\in\mathcal{K}$. Further, we can tell that the integral of $f_0$ over all of $X$ must be $\alpha$. Since $f_0$ is integrable, it must be equal $\mu$-a.e. to some finite-valued function $f$. What we must now show is that if we define

$\displaystyle\nu_0(E)=\nu(E)-\int\limits_Ef\,d\mu$

then $\nu_0$ is identically zero.

If it’s not identically zero, then by the lemma from yesterday there is a positive number $\epsilon$ and a set $A$ so that $\mu(A)>0$ and so that

$\displaystyle\epsilon\mu(E\cap A)\leq\nu_0(E\cap A)=\nu(E\cap A)-\int\limits_{E\cap A}f\,d\mu$

for every measurable set $E$. If we define $g=f+\epsilon\chi_A$, then

$\displaystyle\int\limits_Eg\,d\mu=\int\limits_Ef\,d\mu+\epsilon\mu(E\cap A)\leq\int\limits_{E\setminus A}+\nu(E\cap A)\leq\nu(E)$

for every measurable set $E$, which means that $g\in\mathcal{K}$. But

$\displaystyle\int g\,d\mu=\int f\,d\mu+\epsilon\mu(A)>\alpha$

which contradicts the maximality of the integral of $f$. Thus $\nu_0$ must be identically zero, and the proof is complete.

July 7, 2010 Posted by | Analysis, Measure Theory | 3 Comments

Before the main business, a preliminary lemma: if $\mu$ and $\nu$ are totally finite measures so that $\nu$ is absolutely continuous with respect to $\mu$, and $\nu$ is not identically zero, then there is a positive number $\epsilon$ and a measurable set $A$ so that $\mu(A)>0$ and $A$ is a positive set for the signed measure $\nu-\epsilon\mu$. That is, we can subtract off a little bit (but not zero!) of $\mu$ from $\nu$ and still find a non-$\mu$negligible set on which what remains is completely positive.

To show this, let $X=A_n\uplus B_n$ be a Hahn decomposition with respect to the signed measure $\nu-\frac{1}{n}\mu$ for each positive integer $n$. Let $A_0$ be the union of all the $A_n$ and let $B_0$ be the intersection of all the $B_n$. Then since $B_0\subseteq B_n$ and $B_n$ is negative for $\nu-\frac{1}{n}\mu$ we find

$\displaystyle0\leq\nu(B_0)\leq\frac{1}{n}\mu(B_0)$

for every positive integer $n$. This shows that we must have $\nu(B_0)=0$. And then, since $\nu$ is not identically zero we must have $\nu(A_0)=\nu(X\setminus B_0)>0$. By absolute continuity we conclude that $\mu(A_0)>0$, which means that we must have $\mu(A_n)>0$ for at least one value of $n$. So we pick just such a value, set $A=A_n$, and $\epsilon=\frac{1}{n}$, and everything we asserted is true.

Now for the Radon-Nikodym Theorem: we let $(X,\mathcal{S},\mu)$ be a totally $\sigma$-finite measure space and let $\nu$ be a $\sigma$-finite signed measure on $\mathcal{S}$ which is absolutely continuous with respect to $\mu$. Then $\nu$ is an indefinite integral. That is, there is a finite-valued measurable function $f:X\to\mathcal{R}$ so that

$\displaystyle\nu(E)=\int\limits_Ef\,d\mu$

for every measurable set $E$. The function $f$ is unique in the sense that if any other function $g$ has $\nu$ as its indefinite integral, then $f=g$ $\mu$-almost everywhere. It should be noted that we don’t assert that $f$ is integrable, which will only be true of $\nu$ is actually finite. However, either its positive or its negative integral must converge or we wouldn’t use the integral sign for a divergent integral.

Let’s take a moment and consider what this means. We know that if we take an integrable function $f$, or a function whose integral diverges definitely, on a $\sigma$-finite measure space and define its indefinite integral $\nu$, then $\nu$ is a $\sigma$-finite signed measure that is absolutely continuous with respect to the measure against which we integrate. What the Radon-Nikodym theorem tells us that any such signed measure arises as the indefinite integral of some such function $f$. Further, it tells us that such a function is essentially unique, as much as any function is in measure theory land. In particular, we can tell that if we start with a function $f$ and get its indefinite integral, then any other function with the same indefinite integral must be a.e. equal to $f$.

July 6, 2010 Posted by | Analysis, Measure Theory | 12 Comments

## Singularity

Another relation between signed measures besides absolute continuity — indeed, in a sense the opposite of absolute continuity — is singularity. We say that two signed measures $\mu$ and $\nu$ are “mutually singular” and write $\mu\perp\nu$ if there exists a partition of $X$ into two sets $A\uplus B=X$ so that for every measurable set $E$ the intersections $A\cap E$ and $B\cap E$ are measurable, and

$\displaystyle\lvert\mu\rvert(A\cap E)=0=\lvert\nu\lvert(B\cap E)$

We sometimes just say that $\mu$ and $\nu$ are singular, or that (despite the symmetry of the definition) “$\nu$ is singular with respect to $\mu$“, or vice versa.

In a manner of speaking, if $\mu$ and $\nu$ are mutually singular then all of the sets that give $\mu$ a nonzero value are contained in $B$, while all of the sets that give $\nu$ a nonzero value are contained in $A$, and the two never touch. In contradistinction to absolute continuity, not only does the vanishing of $\lvert\mu\rvert$ not imply the vanishing of $\lvert\nu\rvert$, but if we pare away portions of a set for which $\lvert\nu\rvert$ gives zero measure then what remains — essentially the only sets for which $\lvert\nu\rvert$ doesn’t automatically vanish — is necessarily a set for which $\lvert\mu\rvert$ does vanish. Another way to see this is to notice that if $\mu$ and $\nu$ are signed measures with both $\nu\ll\mu$ and $\nu\perp\mu$, then we must necessarily have $\nu=0$; singularity says that $\nu$ must vanish on any set $E$ with $\lvert\mu\rvert(E)\neq0$, and absolute continuity says $\nu$ must vanish on any set $E$ with $\lvert\mu\rvert(E)=0$.

As a quick and easy example, let $\mu^+$ and $\mu^-$ be the Jordan decomposition of a signed measure $\mu$. Then a Hahn decomposition for $\mu$ gives exactly such a partition $X=A\uplus B$ showing that $\mu^+\perp\mu^-$.

One interesting thing is that singular measures can be added. That is, if $\nu_1$ and $\nu_2$ are both singular with respect to $\mu$, then $(\nu_1+\nu_2)\perp\mu$. Indeed, let $X=A_1\uplus B_1$ and $X=A_2\uplus B_2$ be decompositions showing that $\nu_1\perp\mu$ and $\nu_2\perp\mu$, respectively. That is, for any measurable set $E$ we have

\displaystyle\begin{aligned}\nu_1(A_1\cap E)&=0\\\mu(B_1\cap E)&=0\\\nu_2(A_2\cap E)&=0\\\mu(B_2\cap E)&=0\end{aligned}

Then we can write

$\displaystyle X=(A_1\cap A_2)\uplus\left((A_1\cap B_2)\uplus(A_2\cap B_1)\uplus(B_1\cap B_2)\right)$

It’s easy to check that $\nu_1+\nu_2$ must vanish on measurable subsets of $A_1\cap A_2$, and that $\mu$ must vanish on measurable subsets of the remainder of $X$.