The Unapologetic Mathematician

Mathematics for the interested outsider

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 | 4 Comments

   

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