The Unapologetic Mathematician

Mathematics for the interested outsider

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+c<f(x)\textrm{ and }y>0\right\}=\\&\{(x,y)\in X\times Y\vert c<f(x)\textrm{ and }y>0\}\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<f^+(x)\textrm{ or }-f^-(x)<y\leq0\}\end{aligned}

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

July 21, 2010 Posted by | Analysis, Measure Theory | Leave a comment