# The Unapologetic Mathematician

## 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 - Posted by | Analysis, Measure Theory