# The Unapologetic Mathematician

## 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

## Infinite Products, Part 2

After all of yesterday’s definitions, we continue examining the countably infinite product of a sequence $\{(X_n,\mathcal{S}_n,\mu_n)\}$ of totally finite measure spaces such that each $\mu_n(X_n)=1$.

If $m$ and $n$ are positive integers with $m, it may happen that a non-empty subset $E\subseteq X$ is both a $\{1,\dots,m\}$-cylinder and a $\{1,\dots,n\}$-cylinder. By yesterday’s result, we find that we can write both $E=A\times X^{(m)}$ and $E=B\times X^{(n)}$ for some subsets $A\subseteq X_1\times\dots\times X_m$ and $B\subseteq X_1\times\dots\times X_n$. But we can rewrite the first of these equations as $E=(A\times X_{m+1}\times\dots\times X_n)\times X^{(n)}$, and thus we conclude that $B=A\times X_{m+1}\times\dots\times X_n$.

If $E$ is measurable, then both $A$ and $B$ are measurable. We calculate

\displaystyle\begin{aligned}\left[\mu_1\times\dots\times\mu_n\right](B)&=\left[\mu_1\times\dots\times\mu_n\right](A\times X_{m+1}\times\dots\times X_n)\\&=\left[\mu_1\times\dots\times\mu_m\right](A)\mu_{m+1}(X_{m+1})\dots\mu_n(X_n)\\&=\left[\mu_1\times\dots\times\mu_m\right](A)\end{aligned}

using the normalization of all the measures $\mu_i(X_i)=1$. It follows that we can define a set function $\mu$ unambiguously on each measurable $\{1,\dots,n\}$-cylinder $A\times X^{(n)}$ by the equation

$\displaystyle\mu\left(A\times X^{(n)}\right)=\left[\mu_1\times\dots\times\mu_n\right](A)$

Since every measurable rectangle $E$ is the product of a sequence $\{A_i\}$ with all but finitely many $A_i=X_i$, we can choose a large enough $n$ so that $A_i=X_i$ for all $i>n$. Then $E$ is a $\{1,\dots,n\}$-cylinder, and $\mu$ is defined on all measurable rectangles. It’s straightforward to see that $\mu$ is finite, non-negative, and finitely additive where it’s defined.

We define the analogue of $\mu$ in $X^{(n)}$ by $\mu^{(n)}$. If $\mu$ is defined on a cylinder $E$ then $\mu^{(n)}$ will be defined on each section $E(x_1,\dots,x_n)$, and we find that

$\displaystyle\mu(E)=\int\cdots\int\mu^{(n)}\left(E(x_1,\dots,x_n\right)\,d\mu_1(x_1)\cdots\,d\mu_n(x_n)$

So now, if $\{(X_n,\mathcal{S}_n,\mu_n)\}$ is a sequence of totally finite measure spaces with $\mu_n(X_n)=1$ for all $n$, then there exists a unique measure $\mu$ defined on the countably infinite product of the measure spaces $\{(X_n,\mathcal{S}_n)\}$ so that for every measurable $E\subseteq X$ of the form $A\times X^{(n)}$ we have

$\displaystyle\mu(E)=\left[\mu_1\times\dots\times\mu_n\right](A)$

This measure $\mu$ is called the product of the measures $\{\mu_n\}$.

From what we know about continuity, we just have to show that $\mu$ is continuous from above at $\emptyset$ to show that it’s a measure. That is, if $\{E_n\}$ is a decreasing sequence of cylinders on which $\mu$ is defined so that $0<\epsilon\leq\mu(E_j)$ for all $j$, then the countable intersection of the $E_j$ is non-empty.

For each $j$ we define the set

$\displaystyle F_j=\left\{x_1\in X_1\bigg\vert\mu^{(1)}(E_j(x_1))>\frac{\epsilon}{2}\right\}$

We then find that

\displaystyle\begin{aligned}\mu(E_j)&=\int\mu^{(1)}(E_j(x_1))\,d\mu_1(x_1)\\&=\int\limits_{F_j}\mu^{(1)}(E_j(x_1))\,d\mu_1(x_1)+\int\limits_{F_j^c}\mu^{(1)}(E_j(x_1))\,d\mu_1(x_1)\end{aligned}

and so $\mu(E_j)\leq\mu(F_j)+\frac{\epsilon}{2}$. Therefore, $\mu_1(F_j)\geq\frac{\epsilon}{2}$.

Now $F_j$ is a decreasing sequence of measurable subsets of $X_1$, which is bounded in measure away from zero. And so by the continuity of the measure $\mu_1$, we conclude that there is at least one point $\bar{x}_1$ in their intersection. That is, $\mu^{(1)}(E_j(\bar{x}_1))\geq\frac{\epsilon}{2}$ for all $j$.

But now everything we’ve said about $X$ is true as well for $X^{(1)}$. We can thus replace the sequence $\{E_j\}$ with the sequence $\{E_j(\bar{x}_1)\}$, and the bound $\epsilon$ with the bound $\frac{\epsilon}{2}$. We then find a point $\bar{x}_2\in X_2$ so that $\mu^{(2)}(E_j(\bar{x}_1,\bar{x}_2))\geq\frac{\epsilon}{4}$ for all $j$. We can repeat this process to find a sequence $(\bar{x}_1,\bar{x}_2,\dots)$ with $\bar{x}_n\in X_n$, such that

$\displaystyle\mu^{(n)}(E_j(\bar{x}_1,\dots,\bar{x}_n))\geq\frac{\epsilon}{2^n}$

for all $j$.

I say that this sequence belongs to all the $E_j$. Indeed, given any of them, choose the $n$ so that $E_j$ is a $\{1,\dots,n\}$-cylinder. We know that $\mu^{(n)}(E_j(\bar{x}_1,\dots,\bar{x}_n))>0$, and so there must be at least one point $(x_1,x_2,\dots)\in E_j$ with $x_i=\bar{x}_i$ for all $i$ from $1$ to $n$. But then, since $E_j$ is a $\{1,\dots,n\}$-cylinder, it must contain the point $(\bar{x}_1,\bar{x}_2,\dots)$.

Thus, as asserted, the intersection of the $E_j$ is nonempty, and so $\mu$ is continuous from above at $\emptyset$, and is thus a measure.

July 30, 2010 Posted by | Analysis, Measure Theory | 2 Comments

## Infinite Products, Part 1

Because we know that product spaces are product objects in the category of measurable spaces — at least for totally measurable spaces — we know that the product functor is monoidal. That is, we can define $n$-ary products unambiguously as iterated binary products. But things start to get more complicated as we pass to infinite products.

If $\{X_n\}_{n=1}^\infty$ is a countably infinite collection of sets, the product is the collection of all sequences $(x_1,x_2,\dots)$ with $x_n\in X_n$ for all $n$. If each $X_n$ is equipped with a $\sigma$-ring $\mathcal{S}_n$ and a measure $\mu_n$, it’s not immediately clear how we should equip the product space with a $\sigma$-ring and a measure. However, we can give meaning to these concepts if we specialize. First we shall insist that each $\mathcal{S}_n$ be a $\sigma$-algebra, and second we shall insist that $\mu_n$ be totally finite. Not just totally finite, though; we will normalize each measure so that $\mu_n(X_n)=1$.

This normalization is, incidentally, always possible for a totally finite measure space. Indeed, if $\mu$ is a totally finite measure on a space $X$, we can define

$\displaystyle\hat{\mu}(E)=\frac{\mu(E)}{\mu(X)}$

It is easily verified that $\hat{\mu}$ is another totally finite measure, and $\hat{\mu}(X)=1$.

Now we will define a rectangle as a product of the form

$\displaystyle\prod\limits_{n=1}^\infty A_n$

where $A_n\subseteq X_n$ for all $n$, and where $A_n=X_n$ for all but finitely many $n$. We define a measurable rectangle to be one for which each $A_n$ is measurable as a subset of $X_n$. We then define a subset of the countably infinite product $X$ to be measurable if it’s in the $\sigma$-algebra $\mathcal{S}$ generated by the measurable rectangles. This defines the product of a countably infinite number of measurable spaces.

Now if $J\subseteq\mathbb{Z}^+$ is any set of positive integers, we say that two sequences $(x_1,x_2,\dots)$ and $(y_1,y_2,\dots)$ agree on $J$ if $x_j=y_j$ for all $j\in J$. A set $E\subseteq X$ is called a $J$-cylinder if any two points which agree on $J$ are either both in or both out of $E$. That is, membership of a sequence $x=(x_n)$ in $E$ is determined only by the coordinates $x_j$ for $j\in J$.

We also define the sets

$\displaystyle X^{(n)}=\prod\limits_{i=n+1}^\infty X_i$

so that we always have $X=(X_1\times\dots\times X_n)\times X^{(n)}$ Each $X^{(n)}$ is itself a countably infinite product space. For every set $E\subseteq X$ and each point $(x_1,\dots,x_n)\in X_1\times\dots\times X_n$, we define the subset $E(x_1,\dots,x_n)\subseteq X^{(n)}$ as the section of $E$ determined by $(x_1,\dots,x_n)$. It should be clear that every section of a measurable rectangle in $X$ is a measurable rectangle in $X^{(n)}$

Now if $J=\{1,\dots,n\}$, and if E\subseteq X\$ is a (measurable) $J$-cylinder, then $E=A\times X^{(n)}$, where $A$ is a (measurable) subset of $X_1\times\dots X_n$. Indeed, let $(\bar{x}_{n+1},\bar{x}_{n+2},\dots)$ be an arbitrary point of $X^{(n)}$, and let $A\subseteq X_1\times\dots\times X_n$ be the $X^{(n)}$-section of $E$ determined by this point. The sets $E$ (by assumption) and $A\times X^{(n)}$ (by construction) are both $J$-cylinders, so if $(x_1,x_2,\dots)$ belongs to either of them, then so does the point $(x_1,\dots,x_n,\bar{x}_{n+1},\bar{x}_{n+2},\dots)$.

It should now be clear that if such a point $(x_1,\dots,x_n,\bar{x}_{n+1},\bar{x}_{n+2},\dots)$ belongs to either $E$ or $A\times X^{(n)}$, then it belongs to the other as well. Again using the fact that both $E$ and $A\times X^{(n)}$ are $J$-cylinders, if $(x_1,\dots,x_n,\bar{x}_{n+1},\bar{x}_{n+2},\dots)$ belongs to either $E$ or $A\times X^{(n)}$ then so does the point $(x_1,\dots,x_n,x_{n+1},x_{n+2},\dots)$. We can conclude that $E$ and $A\times X^{(n)}$ consist of the same points. The parenthetical assertion on measurability follows from the fact that every section of a measurable set is measurable.

July 29, 2010 Posted by | Analysis, Measure Theory | 1 Comment

## Fubini’s Theorem

We continue our assumptions that $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$ are both $\sigma$-finite measure spaces, and we consider the product space $(X\times Y,\mathcal{S}\times\mathcal{T},\mu\times\nu)$.

The first step towards the measure-theoretic version of Fubini’s theorem is a characterization of sets of measure zero. Given a subset $E\subseteq X\times Y$, a necessary and sufficient condition for $E$ to have measure zero is that the $X$-section $E_x$ have $\nu(E_x)=0$ for almost all $x\in X$. Another one is that the $Y$-section $E^y$ have $\mu(E^y)=0$ for almost all $y\in Y$. Indeed, the definition of the product measure tells us that

$\displaystyle\lambda(E)=\int\nu(E_x)\,d\mu(x)=\int\mu(E^y)\,d\nu(y)$

Since the function $x\mapsto\nu(E_x)$ is integrable and nonnegative, our condition for an integral to vanish says that the integral is zero if and only if $\nu(E_x)=0$ $\mu$-almost everywhere. Similarly, we see that the integral of $\mu(E^y)$ is zero if and only if $\mu(E^y)=0$ $\nu$-almost everywhere.

Now if $h$ is a non-negative measurable function on $X\times Y$, then we have the following equalities between the double integral and the two iterated integrals:

$\displaystyle\int h\,d(\mu\times\nu)=\iint h\,d\mu\,d\nu=\iint h\,d\nu\,d\mu$

If $h$ is the characteristic function $\chi_E$ of a measurable set $E$, then we find that

\displaystyle\begin{aligned}\int\chi_E(x,y)\,d\nu(y)&=\int\chi_{E_x}(y)\,d\nu(y)=\nu(E_x)\\\int\chi_E(x,y)\,d\mu(x)&=\int\chi_{E^y}(x)\,d\mu(x)=\mu(E^y)\end{aligned}

and thus

\displaystyle\begin{aligned}\iint\chi_E(x,y)\,d\nu\,d\mu=\int\nu(E_x)\,d\mu&=\left[\mu\times\nu\right](E)=\int\chi_E\,d(\mu\times\nu)\\\iint\chi_E(x,y)\,d\mu\,d\nu=\int\mu(E^y)\,d\nu&=\left[\mu\times\nu\right](E)=\int\chi_E\,d(\mu\times\nu)\end{aligned}

Next we assume that $h$ is a simple function. Then $h$ is a finite linear combination of characteristic functions of measurable sets. But clearly all parts of the asserted equalities are linear in the function $h$, and so since they hold for characteristic functions of measurable sets they must hold for any simple function as well.

Finally, given any non-negative measurable function $h$, we can find an increasing sequence of simple functions $\{h_n\}$ converging pointwise to $h$. The monotone convergence theorem tells us that

$\displaystyle\lim\limits_{n\to\infty}\int h_n\,d(\mu\times\nu)=\int h\,d(\mu\times\nu)$

We define the functions

$\displaystyle f_n(x)=\int h_n(x,y)\,d\nu(y)$

and conclude that since $\{h_n\}$ is an increasing sequence, $\{f_n\}$ must me an increasing sequence of non-negative measurable functions as well. For every $x$ the monotone convergence theorem tells us that

$\displaystyle\lim\limits_{n\to\infty}f_n(x)=f(x)=\int h(x,y)\,d\nu(y)$

As a limit of a sequence of non-negative measurable functions, $f$ must also be a non-negative measurable function. One last invocation of the monotone convergence theorem tells us that

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

which proves the equality of the double integral and one of the iterated integrals. The other equality follows similarly.

And now we come to Fubini’s theorem itself: if $h$ is an integrable function on $X\times Y$, then almost every section of $h$ is integrable. If we define the functions

\displaystyle\begin{aligned}f(x)&=\int h(x,y)\,d\nu(y)\\g(y)&=\int h(x,y)\,d\mu(x)\end{aligned}

wherever these symbols are defined, then $f$ and $g$ are both integrable, and

$\displaystyle\int h\,d(\mu\times\nu)=\int f\,d\mu=\int g\,d\nu$

Since a real-valued function is integrable if and only if both its positive and negative parts are, it suffices to consider non-negative functions $h$. The latter equalities follow, then, from the above discussion. Since the measurable functions $f$ and $g$ have finite integrals, they must be integrable. And since they’re integrable, they must be finite-valued a.e., which implies the assertions about the integrability of sections of $h$.

July 28, 2010 Posted by | Analysis, Measure Theory | 2 Comments

## Double and Iterated Integrals

Let $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$ be two $\sigma$-finite measure spaces, and let $\lambda=\mu\times\nu$ be the product measure on $X\times Y$.

If $h$ is a function on $X\times Y$ so that its integral is defined — either $h$ is integrable, or its integral diverges definitely — then we write it as any of

$\displaystyle\int h\,d\lambda=\int h\,d(\mu\times\nu)=\int h(x,y)\,d\lambda(x,y)=\int h(x,y)\,d(\mu\times\nu)(x,y)$

and call it the “double integral” of $h$ over $X\times Y$. We can also consider the sections $h_x(y)=h(x,y)$ and $h^y(x)=h(x,y)$. For any given $x\in X$, we set

$\displaystyle\int h_x(y)\,d\nu(y)=f(x)$

if the integral exists. If the resulting function $f$ is integrable, then we write

$\displaystyle\int f\,d\mu=\iint h(x,y)\,d\nu(y)\,d\mu(x)=\iint h\,d\nu\,d\mu=\int\,d\mu(x)\int h(x,y)\,d\nu(y)$

The latter notation, with the measure $\mu$ before the integrand is less common, but it can be seen in older texts. I’ll usually stick to the other order.

Similarly, we can define the function $g(y)$ as the integral of the $Y$-section $h^y$ if it exists. If $g$ is integrable, we write

$\displaystyle\int g\,d\nu=\iint h(x,y)\,d\mu(x)\,d\nu(y)=\iint h\,d\mu\,d\nu=\int\,d\nu(y)\int h(x,y)\,d\mu(x)$

where, again, the latter notation is deprecated. These integrals are called the “iterated integrals” of $h$. We can also define double and iterated integrals over a measurable subset $E\subseteq X\times Y$, as usual, and write

\displaystyle\begin{aligned}\int\limits_Eh&\,d\lambda\\\int\limits_Eh&\,d(\mu\times\nu)\\\iint\limits_Eh&\,d\mu\,d\nu\\\iint\limits_Eh&\,d\nu\,d\mu\end{aligned}

July 27, 2010 Posted by | Analysis, Measure Theory | 2 Comments

## The Measures of Ordinate Sets

If $(X,\mathcal{S})$ is a $\sigma$-algebra and $f:X\to\mathbb{R}$ is a Borel-measurable function we defined the upper and lower ordinate sets $V^*(f)$ and $V_*(f)$ to be measurable subsets of $X\times\mathbb{R}$. Now if we have a measure $\mu$ on $X$ and Lebesgue measure on the Borel sets, we can define the product measure $\lambda$ on $X\times\mathbb{R}$. Since we know $V^*(f)$ and $V_*(f)$ are both measurable, we can investigate their measures. I assert that

$\displaystyle\lambda(V^*(f))=\int f\,d\mu=\lambda(V_*(f))$

It will be sufficient to establish this for simple functions, since for either the upper or the lower ordinate set we can approximate any measurable $f$ by a monotone sequence of simple $\{f_n\}$ so that $\lim\limits_{n\to\infty}V^*(f_n)=V^*(f)$ or $\lim\limits_{n\to\infty}V_*(f_n)=V_*(f)$. Then the limit will commute with $\lambda$ (since measures are continuous), and it will commute with the integral as well.

So, we can assume that $f$ is simple, and write

$\displaystyle f=\sum\limits_{i=1}^n\alpha_i\chi_{E_i}$

with the $\{E_i\}$ a pairwise-disjoint collection of measurable sets. But now if the equality holds for each of the summands then it holds for the whole function. That is, we can assume — without loss of generality — that $f=\alpha\chi_E$ for some real number $\alpha$ and some measurable subset $E\subseteq X$.

And now the result should be obvious! Indeed, $V^*(f)$ is the measurable rectangle $E\times[0,\alpha]$, while $V_*(f)$ is the measurable rectangle $E\times[0,\alpha)$. Since the product measure on a measurable rectangle is the product of the measures of the two sides, these both have measure $\alpha\mu(E)$. On the other hand, we calculate the integral as

$\displaystyle\int f\,d\mu=\int\alpha\chi_E\,d\mu=\alpha\mu(E)$

and so the equality holds for such rectangles, and thus for simple functions, and thus for all measurable functions.

Of course, it should now be clear that the graph of $f$ has measure zero. Indeed, we find that

$\displaystyle\lambda(V^*(f)\setminus V_*(f))=\lambda(V^*(f))-\lambda(V_*(f))=\int f\,d\mu-\int f\,d\mu=0$

These results put a precise definition to the concept of the integral as the “area under the graph”, which was the motivation behind our definition of the Riemann integral, way back when we introduced it.

July 26, 2010

## Product Measures

We continue as yesterday, considering the two $\sigma$-finite measure spaces $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$, and the product measure space $(X\times Y,\mathcal{S}\times\mathcal{T})$.

Last time we too a measurable set $E\subseteq X\times Y$ and defined the functions $f(x)=\nu(E_x)$ and $g(y)=\mu(E^y)$. We also showed that

$\displaystyle\int f\,d\mu=\int g\,d\nu$

That is, for every measurable $E$ we can define the real number

$\displaystyle\lambda(E)=\int\nu(E_x)\,d\mu=\int\mu(E^y)\,d\nu$

I say that this function $\lambda$ is itself a $\sigma$-finite measure, and that for any measurable rectangle $A\times B$ we have $\lambda(A\times B)=\mu(A)\nu(B)$. Since measurable rectangles generate the $\sigma$-ring $\mathcal{S}\times\mathcal{T}$, this latter condition specifies $\lambda$ uniquely.

To see that $\lambda$ is a measure, we must show that it is countably additive. If $\{E_n\}$ is a sequence of disjoint sets then we calculate

\displaystyle\begin{aligned}\lambda\left(\biguplus\limits_{n=1}^\infty E_n\right)&=\int\nu\left(\left(\biguplus\limits_{n=1}^\infty E_n\right)_x\right)\,d\mu\\&=\int\nu\left(\biguplus\limits_{n=1}^\infty(E_n)_x\right)\,d\mu\\&=\int\sum\limits_{n=1}^\infty\nu\left((E_n)_x\right)\,d\mu\\&=\sum\limits_{n=1}^\infty\int\nu\left((E_n)_x\right)\,d\mu\\&=\sum\limits_{n=1}^\infty\lambda(E_n)\end{aligned}

where we have used the monotone convergence theorem to exchange the sum and the integral.

We verify the $\sigma$-finiteness of $\lambda$ by covering each measurable set $E$ by countably many measurable rectangles with finite-measure sides. Since the sides’ measures are finite, the measure of the rectangle itself is the product of two finite numbers, and is thus finite.

We call the measure $\lambda$ the “product” of the measures $\mu$ and $\nu$, and we write $\lambda=\mu\times\nu$. We thus have a $\sigma$-finite measure space $(X\times Y,\mathcal{S}\times\mathcal{T},\mu\times\nu)$ that we call the “cartesian product” of the spaces $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$.

July 23, 2010 Posted by | Analysis, Measure Theory | 6 Comments

## Measures on Product Spaces

After considering the product of measurable spaces, let’s consider what happens if our spaces are actually equipped with measures. That is, let $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$ be $\sigma$-finite measure spaces, and consider the product space $(X\times Y,\mathcal{S}\times\mathcal{T})$.

Now, if $E$ is any measurable subset of $X\times Y$, then we can define two functions — $f:X\to\mathbb{R}$ and $g:Y\to\mathbb{R}$ — by the formulæ $f(x)=\nu(E_x)$ and $g(y)=\nu(E^y)$, where $E_x$ and $E^y$ are the $X$section determined by $x$ and $Y$-section determined by $y$, respectively. I say that both $f$ and $g$ are non-negative measurable functions, and that

$\displaystyle\int f\,d\mu=\int g\,d\nu$

where we take the first integral over $X$ and the second over $Y$.

Let $\mathcal{E}$ be the collection of subsets of $X\times Y$ for which the assertions we’ve made are true. It’s straightforward to see that $\mathcal{E}$ is closed under countable disjoint unions. Indeed, if $\{E_n\}$ is a sequence of disjoint sets for which the assertions hold — call the functions $\{f_n\}$ and $\{g_n\}$ respectively — then if $E$ is the disjoint union of the $E_n$ we can calculate

\displaystyle\begin{aligned}f(x)&=\nu(E_x)\\&=\nu\left(\left(\bigcup\limits_{n=1}^\infty E_n\right)_x\right)\\&=\nu\left(\bigcup\limits_{n=1}^\infty(E_n)_x\right)\\&=\sum\limits_{n=1}^\infty\nu\left((E_n)_x\right)\\&=\sum\limits_{n=1}^\infty f_n(x)\end{aligned}

Since each of the $f_n$ are measurable, so is $f$, and $g$ is similarly measurable. The equality of the integrals should be clear.

Now, since $\mu$ and $\nu$ are $\sigma$-finite, every measurable subset of $X\times Y$ can be covered by a countable disjoint union of measurable rectangles, each side of each of which has finite measure. Thus we just have to verify that the result holds for such measurable rectangles with finite-measure sides, and it will hold for all measurable subsets of $X\times Y$, and that $\mathcal{E}$ is a monotone class. That $\mathcal{E}$ is a monotone class follows from the dominated convergence theorem and the monotone convergence theorem, and so we have only to show that the assertions hold for measurable rectangles with finite-measure sides.

Okay, so let $E=A\times B$ be a measurable rectangle with $\mu(A)<\infty$ and $\nu(B)<\infty$. We can simplify our functions to write

\displaystyle\begin{aligned}f(x)&=\nu(B)\chi_A\\g(y)&=\mu(A)\chi_B\end{aligned}

These are clearly measurable, and we find that

$\displaystyle\int f\,d\mu=\mu(A)\nu(B)=\int g\,d\nu$

so the assertions hold for measurable rectangles with finite-measure sides, and thus for all measurable subsets of $X\times Y$.

July 22, 2010 Posted by | Analysis, Measure Theory | 1 Comment

## Measurable Graphs

Yesterday, we showed that the graph of a non-negative measurable function is measurable. Today we’ll explore this further. We continue all the same notation as we used yesterday: $(X,\mathcal{S})$ is a measurable space and $(Y,\mathcal{T})=(\mathbb{R},\mathcal{B})$ is the measurable space of real numbers and Borel sets.

First, if $E\subseteq X\times Y$ is any measurable subset, and if $\alpha\neq0$ and $\beta$ are real numbers, then the set $\{(x,y)\in X\times Y\vert(x,\alpha y+\beta)\in E\}$ is also measurable. The affine transformation sends any measurable set in $Y$ to another such set, so if $E$ is a measurable rectangle, then the transformed set will also be a measurable rectangle. It’s also straightforward to show that the transformation commutes with setwise unions and intersections, and that the assertion is true for $E=\emptyset$. Thus the assertion holds on some $\sigma$-ring $\mathcal{R}$, which contains all measurable rectangles. Since measurable rectangles generate the $\sigma$-algebra of all measurable sets on $X\times Y$, $\mathcal{R}$ must contain all measurable sets, and thus the assertion holds for all measurable $E$.

Now — as a partial converse to yesterday’s final result — if $f$ is a non-negative function so that $V^*(f)$ (or $V_*(f)$) is a measurable set, then $f$ is a measurable function. We will show this using an equivalent definition of measurability — that $f$ will be measurable if we can show that for every real number $c$ the set $\{x\in X\vert f(x)>c\}$ is measurable. This is clearly true for every nonpositive $c$, and so we must show it for the positive $c$.

For each positive integer $n$ we define the set

\displaystyle\begin{aligned}&\left\{(x,y)\in X\times Y\bigg\vert\left(x,\frac{1}{n}y+c\right)\in V^*(f)\textrm{ and }y>0\right\}=\\&\left\{(x,y)\in X\times Y\bigg\vert\left(x,\frac{1}{n}y+c\right)\in V^*(f)\right\}\cap\left\{(x,y)\in X\times Y\vert y>0\right\}\end{aligned}

Our above lemma shows that the first set in the intersection is measurable — as a transformation of $V^*(f)$, which we assumed to be measurable — and the second set is a measurable rectangle. Thus the intersection is measurable. We take the union of these sets for all $n$:

\displaystyle\begin{aligned}&\bigcup\limits_{n=1}^\infty\left\{(x,y)\in X\times Y\bigg\vert\left(x,\frac{1}{n}y+c\right)\in V^*(f)\textrm{ and }y>0\right\}=\\&\bigcup\limits_{n=1}^\infty\left\{(x,y)\in X\times Y\bigg\vert 0\leq\frac{1}{n}y+c0\right\}=\\&\{(x,y)\in X\times Y\vert c0\}\end{aligned}

This is the union of a sequence of measurable sets, and so it is measurable. Taking any $y>0$ we find the $Y$-section determined by $y$ is the set $\{x\in X\vert f(x)>c\}$. And this is thus measurable, since all sections of measurable sets are measurable.

This result is the basis of an alternative characterization of measurable functions. We could have defined a non-negative function $f$ to be measurable if its upper (or lower) ordinate set $V^*(f)$ ($V_*(f)$ is measurable, and extended to general functions by insisting that this hold for both positive and negative parts.

Finally, we can extend our result from last time. If $f$ is any measurable function, then its graph is measurable. Indeed, we can take the positive and negative parts $f^+$ and $f^-$, which are both measurable. Thus all four sets $V^*(f^+)$, $V^*(f^-)$, $V_*(f^+)$, and $V_*(f^-)$ are measurable. Choosing $\alpha=-1$ and $\beta=0$ we reflect $V^*(f^-)$ and $V_*(f^-)$ to $-V^*(f^-)$ and $-V_*(f^-)$. We then form the unions

\displaystyle\begin{aligned}V^*(f+)\cup-V^*(f^-)&=\{(x,y)\in X\times Y\vert0\leq y\leq f^+(x)\textrm{ or }-f^-(x)\leq y\leq0\}\\V_*(f+)\cup-V_*(f^-)&=\{(x,y)\in X\times Y\vert0\leq y

The difference between these two sets is the graph of $f$, which is thus measurable.

July 21, 2010

## Upper and Lower Ordinate Sets

Let $(X,\mathcal{S})$ be a measurable space so that $X$ itself is measurable — that is, so that $\mathcal{S}$ is a $\sigma$-algebra — and let $(Y,\mathcal{T})=(\mathbb{R},\mathcal{B})$ be the real line with the $\sigma$-algebra of Borel sets.

If $f$ is a real-valued, non-negative function on $X$, then we define the “upper ordinate set” to be the subset $V^*(f)\subseteq X\times Y$ such that

$\displaystyle V^*(f)=\{(x,y)\in X\times Y\vert0\leq y\leq f(x)\}$

We also define the “lower ordinate set” to be the subset $V_*(f)\subseteq X\times Y$ such that

$\displaystyle V_*(f)=\{(x,y)\in X\times Y\vert0\leq y

We will explore some basic properties of these sets.

First, if $f$ is the characteristic function $\chi_E$ of a measurable subset $E\subseteq X$, then $V^*(\chi_E)$ is the measurable rectangle $E\times[0,1]$, while $V_*(\chi_E)$ is the measurable rectangle $E\times[0,1)$.

Next, if $f$ is a non-negative simple function, then we can write it as the linear combination of characteristic functions of disjoint subsets. If $f=\sum a_i\chi_{E_i}$, then for each $i$ the upper ordinate set $V^*(a_i\chi_{E_i})$ is the measurable rectangle $E_i\times[0,a_i]$ while the lower ordinate set $V_*(a_i\chi_{E_i})$ is the measurable rectangle $E_i\times[0,a_i)$. Since the $E_i$ are all disjoint, the upper ordinate set $V^*(f)$ is the disjoint union of all the $V^*(a_i\chi_{E_i})$, and similarly for the lower ordinate sets. Thus the upper and lower ordinate sets of a simple function are both measurable.

Next we have some monotonicity properties: if $f$ and $g$ are non-negative functions so that $f(x)\leq g(x)$ for all $x\in X$, then $V^*(f)\subseteq V^*(g)$ and $V_*(f)\subseteq V_*(g)$. Indeed, if $(x,y)\in V^*(f)$ then $0\leq y\leq f(x)\leq g(x)$, so $(x,y)\in V^*(g)$ as well, and similarly for the lower ordinate sets.

If $\{f_n\}$ is an increasing sequence of non-negative functions converging pointwise to a function $f$, then $\{V_*(f_n)\}$ is an increasing sequence of sets whose union is $V_*(f)$. That $\{V_*(f_n)\}$ is increasing is clear from the above monotonicity property, and just as clearly they’re all contained in $V_*(f)$. On the other hand, if $(x,y)\in V_*(f)$, then $0\leq y But since $\{f_n(x)\}$ increases to $f(x)$, this means that $y for some $n$, and so $(x,y)\in V_*(f_n)$. Thus $V_*(f)$ is contained in the union of the $V_*(f_n)$. Similarly, if $\{f_n\}$ is decreasing to $f$, then $\{V^*(f)\}$ decreases to $V^*(f)$.

Finally (for now), if $f$ is a non-negative measurable function, then $V^*(f)$ and $V_*(f)$ are both measurable. The lower ordinate set $V_*(f)$ is easier, since we know that we can pick an increasing sequence of non-negative measurable simple functions converging pointwise to $f$. Their lower ordinate sets are all measurable, and they form an increasing sequence of measurable sets whose union is $V_*(f)$. Since this is a countable union of measurable sets, it must be itself measurable.

For $V^*(f)$ we have to be a little trickier. First, if $g$ is bounded above by $c$, then $c-g$ is non-negative and also bounded above by $c$, and we can find an increasing sequence $\{g_n\}$ of non-negative measurable simple functions converging pointwise to $c-g$. Then $\{c-g_n\}$ is a decreasing sequence of non-negative simple functions converging pointwise to $g$. The catch is that the measurability of a simple function only asks that all the nonzero sets on which it is defined be measurable. That is, in principle the zero set of $g_n$ may be non-measurable. However, the zero set of $g_n$ is the complement of $N(g_n)$, and since this set is measurable we can use the assumed measurability of $X$ to see that $X\setminus N(g_n)$ is measurable as well. And so we see that $c-g_n$ is measurable as well. Thus $\{V^*(g_n)\}$ is a decreasing sequence of measurable sets, converging to $V^*(g)$, which must thus be measurable.

Now, for a general $f$, we can consider the sequence $\{f\cap n\}$ which replaces any value $f(x)>n$ with $n$. Each of these functions is still measurable (again using the measurability of $X$), and is now bounded. Thus $\{V^*(f_n)\}$ is an increasing sequence of measurable sets, and I say that now their union is $V^*(f)$. Indeed, each is contained in $V^*(f)$, so the union must be. On the other hand, if $(x,y)\in V^*(f)$, then $y\leq f(x)$. But since $f(x)\in\mathbb{R}$, there is some $N$ so that $f(x)\leq N$. Thus $y\leq\min(f(x),N)=f_N(x)$, and so $(x,y)\in V^*(f_N)$, and is in the union as well. Since $V^*(f)$ is the union of a countable sequence of measurable sets, it is itself measurable.

Incidentally, this implies that if $f$ is a non-negative measurable function, then the difference $V^*(f)\setminus V_*(f)$ is measurable. But we can calculate this difference as

\displaystyle\begin{aligned}V^*(f)\setminus V_*(f)&=\{(x,y)\in X\times Y\vert 0\leq y\leq f(x)\}\setminus\{(x,y)\in X\times Y\vert 0\leq y

That is, $V^*(f)\setminus V_*(f)$ is exactly the graph of the function $f$, and so we see that the graph of a non-negative measurable function is measurable.

July 20, 2010 Posted by | Analysis, Measure Theory | 22 Comments