# The Unapologetic Mathematician

## Upper and Lower Ordinate Sets

Let $(X,\mathcal{S})$ be a measurable space so that $X$ itself is measurable — that is, so that $\mathcal{S}$ is a $\sigma$-algebra — and let $(Y,\mathcal{T})=(\mathbb{R},\mathcal{B})$ be the real line with the $\sigma$-algebra of Borel sets.

If $f$ is a real-valued, non-negative function on $X$, then we define the “upper ordinate set” to be the subset $V^*(f)\subseteq X\times Y$ such that

$\displaystyle V^*(f)=\{(x,y)\in X\times Y\vert0\leq y\leq f(x)\}$

We also define the “lower ordinate set” to be the subset $V_*(f)\subseteq X\times Y$ such that

$\displaystyle V_*(f)=\{(x,y)\in X\times Y\vert0\leq y

We will explore some basic properties of these sets.

First, if $f$ is the characteristic function $\chi_E$ of a measurable subset $E\subseteq X$, then $V^*(\chi_E)$ is the measurable rectangle $E\times[0,1]$, while $V_*(\chi_E)$ is the measurable rectangle $E\times[0,1)$.

Next, if $f$ is a non-negative simple function, then we can write it as the linear combination of characteristic functions of disjoint subsets. If $f=\sum a_i\chi_{E_i}$, then for each $i$ the upper ordinate set $V^*(a_i\chi_{E_i})$ is the measurable rectangle $E_i\times[0,a_i]$ while the lower ordinate set $V_*(a_i\chi_{E_i})$ is the measurable rectangle $E_i\times[0,a_i)$. Since the $E_i$ are all disjoint, the upper ordinate set $V^*(f)$ is the disjoint union of all the $V^*(a_i\chi_{E_i})$, and similarly for the lower ordinate sets. Thus the upper and lower ordinate sets of a simple function are both measurable.

Next we have some monotonicity properties: if $f$ and $g$ are non-negative functions so that $f(x)\leq g(x)$ for all $x\in X$, then $V^*(f)\subseteq V^*(g)$ and $V_*(f)\subseteq V_*(g)$. Indeed, if $(x,y)\in V^*(f)$ then $0\leq y\leq f(x)\leq g(x)$, so $(x,y)\in V^*(g)$ as well, and similarly for the lower ordinate sets.

If $\{f_n\}$ is an increasing sequence of non-negative functions converging pointwise to a function $f$, then $\{V_*(f_n)\}$ is an increasing sequence of sets whose union is $V_*(f)$. That $\{V_*(f_n)\}$ is increasing is clear from the above monotonicity property, and just as clearly they’re all contained in $V_*(f)$. On the other hand, if $(x,y)\in V_*(f)$, then $0\leq y But since $\{f_n(x)\}$ increases to $f(x)$, this means that $y for some $n$, and so $(x,y)\in V_*(f_n)$. Thus $V_*(f)$ is contained in the union of the $V_*(f_n)$. Similarly, if $\{f_n\}$ is decreasing to $f$, then $\{V^*(f)\}$ decreases to $V^*(f)$.

Finally (for now), if $f$ is a non-negative measurable function, then $V^*(f)$ and $V_*(f)$ are both measurable. The lower ordinate set $V_*(f)$ is easier, since we know that we can pick an increasing sequence of non-negative measurable simple functions converging pointwise to $f$. Their lower ordinate sets are all measurable, and they form an increasing sequence of measurable sets whose union is $V_*(f)$. Since this is a countable union of measurable sets, it must be itself measurable.

For $V^*(f)$ we have to be a little trickier. First, if $g$ is bounded above by $c$, then $c-g$ is non-negative and also bounded above by $c$, and we can find an increasing sequence $\{g_n\}$ of non-negative measurable simple functions converging pointwise to $c-g$. Then $\{c-g_n\}$ is a decreasing sequence of non-negative simple functions converging pointwise to $g$. The catch is that the measurability of a simple function only asks that all the nonzero sets on which it is defined be measurable. That is, in principle the zero set of $g_n$ may be non-measurable. However, the zero set of $g_n$ is the complement of $N(g_n)$, and since this set is measurable we can use the assumed measurability of $X$ to see that $X\setminus N(g_n)$ is measurable as well. And so we see that $c-g_n$ is measurable as well. Thus $\{V^*(g_n)\}$ is a decreasing sequence of measurable sets, converging to $V^*(g)$, which must thus be measurable.

Now, for a general $f$, we can consider the sequence $\{f\cap n\}$ which replaces any value $f(x)>n$ with $n$. Each of these functions is still measurable (again using the measurability of $X$), and is now bounded. Thus $\{V^*(f_n)\}$ is an increasing sequence of measurable sets, and I say that now their union is $V^*(f)$. Indeed, each is contained in $V^*(f)$, so the union must be. On the other hand, if $(x,y)\in V^*(f)$, then $y\leq f(x)$. But since $f(x)\in\mathbb{R}$, there is some $N$ so that $f(x)\leq N$. Thus $y\leq\min(f(x),N)=f_N(x)$, and so $(x,y)\in V^*(f_N)$, and is in the union as well. Since $V^*(f)$ is the union of a countable sequence of measurable sets, it is itself measurable.

Incidentally, this implies that if $f$ is a non-negative measurable function, then the difference $V^*(f)\setminus V_*(f)$ is measurable. But we can calculate this difference as

\displaystyle\begin{aligned}V^*(f)\setminus V_*(f)&=\{(x,y)\in X\times Y\vert 0\leq y\leq f(x)\}\setminus\{(x,y)\in X\times Y\vert 0\leq y

That is, $V^*(f)\setminus V_*(f)$ is exactly the graph of the function $f$, and so we see that the graph of a non-negative measurable function is measurable.

July 20, 2010 Posted by | Analysis, Measure Theory | 21 Comments