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

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