# The Unapologetic Mathematician

## Sections of Sets and Functions

Our definitions today are purely set-theoretical. If $E$ is a subset of $X\times Y$, then given a point $x\in X$ we define the “section” of $E$ determined by $x$ to be the set

$\displaystyle E_x=\{y\in Y\vert(x,y)\in E\}$

Similarly, we define the section of $E$ determined by a point $y\in Y$ to be the set

$\displaystyle E^y=\{x\in X\vert(x,y)\in E\}$

Of course, the two concepts are essentially equivalent, and neither really depends on the fact that we have only two factors, but we choose this notation for now. If we don’t so much care about the particular point $x$ or $y$, we refer to “an $X$-section” or “a $Y$-section”. It should be stressed that these sections are not (as might be supposed) subsets of $E$, but rather an $X$-section $E_x$ is a subset of $Y$, while a $Y$-section $E^y$ is a subset of $X$.

It should be clear that taking sections commutes with most common set theoretic operations. For example, we compute

\displaystyle\begin{aligned}(E\cup F)_x&=\{y\in Y\vert(x,y)\in E\cup F\}\\&=\{y\in Y\vert(x,y)\in E\textrm{ or }(x,y)\in F\}\\&=\{y\in Y\vert(x,y)\in E\}\cup\{y\in Y\vert(x,y)\in F\}\\&=E_x\cup F_x\end{aligned}

Similarly, $(E\setminus F)_x=E_x\setminus F_x$ and $(E\cap F)_x=E_x\cap F_x$; and similarly for $Y$-sections.

Now if $f$ is any function defined on $E$ and $x\in X$ is any point, we define the section of $f$ determined by $x$ to be the function $f_x$ on $E_x$ defined by $f_x(y)=f(x,y)$. Similarly, the section of $f$ determined by a point $y\in Y$ is the function $f^y$ on $E^y$ defined by $f^y(x)=f(x,y)$. Again, we say that $f_x$ is an $X$-section, and $f^y$ is a $Y$-section.

With these definitions down, we turn to measure theory. Let $(X,\mathcal{S})$ and $(Y,\mathcal{T})$ be measurable spaces, and let $(X\times Y,\mathcal{S}\times\mathcal{T})$ be the product space.

If $E=A\times B$ is a measurable rectangle, then every $X$-section $E_x$ is either $B$ or $\emptyset\subseteq Y$, according as $x\in A$ or not. Similarly, every $Y$-section is either $A$ or $\emptyset\subseteq X$. That is, every section of a measurable rectangle is measurable. Now we let $\mathcal{E}$ be the collection of all subsets of $X\times Y$ for which this is true — $E\in\mathcal{E}$ if and only if every section of $E$ is measurable. Clearly $\mathcal{E}$ contains all measurable rectangles. It’s also closed under unions and setwise differences — making it a ring — and under monotone limits — making it a $\sigma$-ring. Since $\mathcal{E}$ is a $\sigma$-ring containing all measurable rectangles, it must contain $\mathcal{S}\times\mathcal{T}$. Therefore, every section of every measurable set is measurable.

Now if $f:X\times Y\to Z$ and $M\subseteq Z$ is any measurable set, then we calculate

\displaystyle\begin{aligned}f_x^{-1}(M)&=\{y\in Y\vert f_x(y)\in M\}\\&=\{y\in Y\vert f(x,y)\in M\}\\&=\{y\in Y\vert (x,y)\in f^{-1}(M)\}\\&=f^{-1}(M)_x\end{aligned}

Since $f$ is measurable, $f^{-1}(M)$ must be measurable, and thus all of its sections are measurable. In particular, $f_x^{-1}(M)$ is measurable for any measurable $M$, and thus $f_x$ is a measurable function. Similarly we can show that the $Y$-section $f^y$ of a measurable function $f$ is measurable.

The one caveat is that we treated measurable real-valued functions differently from other ones. Just to be sure, let $f:X\times Y\to\mathbb{R}$ be a measurable real-valued function, and let $M$ be a Borel set. Then we need to ask that $N(f_x)\cap f_x^{-1}(M)$ be measurable. We can use the above fact that $f_x^{-1}(M)=f^{-1}(M)_x$, and the result will follow if we can show that $N(f_x)=N(f)_x$. But we easily calculate

\displaystyle\begin{aligned}N(f_x)&=\{y\in Y\vert f_x(y)\neq0\}\\&=\{y\in Y\vert f(x,y)\neq0\}\\&=\{y\in Y\vert(x,y)\in N(f)\}\\&=N(f)_x\end{aligned}

and thus the result follows. The proof that the $Y$-section $f^y$ is measurable is similar.

July 19, 2010 - Posted by | Analysis, Measure Theory

1. […] This is the union of a sequence of measurable sets, and so it is measurable. Taking any we find the -section determined by is the set . And this is thus measurable, since all sections of measurable sets are measurable. […]

Pingback by Measurable Graphs « The Unapologetic Mathematician | July 21, 2010 | Reply

2. […] then we can define two functions — and — by the formulæ and , where and are the -section determined by and -section determined by , respectively. I say that both and are non-negative […]

Pingback by Measures on Product Spaces « The Unapologetic Mathematician | July 22, 2010 | Reply

3. […] call it the “double integral” of over . We can also consider the sections and . For any given , we […]

Pingback by Double and Iterated Integrals « The Unapologetic Mathematician | July 27, 2010 | Reply

4. […] a countably infinite product space. For every set and each point , we define the subset as the section of determined by . It should be clear that every section of a measurable rectangle in is a […]

Pingback by Infinite Products, Part 1 « The Unapologetic Mathematician | July 29, 2010 | Reply

5. Just to think about it. there exists a relation between $(f^{-1}(E))_{x}$ and $E_{x}$? Under what conditions one of them contains the other?

Comment by Camilo | March 4, 2016 | Reply

• Well, just write out the definitions. One set is $E_x=\{y\in Y\vert(x,y)\in E\}$ and the other is $f^{-1}(E)_x=\{y\in Y\vert(f(x),y)\in E\}$. What can you conclude from that?

Comment by John Armstrong | March 12, 2016 | Reply

However, I think it is not correct. Look:

E_x=\{y\in Y\vert(x,y)\in E\}

\left(f^{-1}(E)\right)_x=\{y\in Y\vert(x,y)\in f^{-1}(E)\}=\{y\in Y\vert f(x,y)\in E\}

I don’t see any relation between those two sets.

Maybe if we impose some condition on the function f, we can compare the sets.

I hope you read this and leave some comments. Best regards. Camilo Chaparro.

Comment by Camilo Chaparro | July 28, 2016 | Reply

• You asked if there exists a relation between them. I did the tiniest bit of work, just writing out the definitions that you should have been able to do yourself. From that point you were able to say “I don’t see any relation between those two sets” and “maybe if we impose some condition on the function f, we can compare the sets.” Which certainly seems like an answer to your original question.

I don’t really see how my answer is “not correct”. Maybe you’re upset that I didn’t do all of the work for you? Seems a funny way to show the gratitude you start off by expressing.

Comment by John Armstrong | July 28, 2016 | Reply

• John. My native language is not English, and so I apologize if I don’t express myself correctly.

When I said “It is not correct” I mean that the second definition that you wrote is not correct. If you compare your definition of $(f^{-1}(E))_x$ with mine, you will realize that they are different.

Comment by Camilo Chaparro | July 28, 2016 | Reply

• ohn. My native language is not English, and so I apologize if I don’t express myself correctly.

When I said “It is not correct” I mean that the second definition that you wrote is not correct. If you compare your definition of $(f^{-1}(E))_x$ with mine, you will realize that they are different.

When I said “It is not correct” I mean that the second definition that you wrote is not correct. If you compare your definition of $(f^{-1}(E))_x$ with mine, you will realize that they are different.