The Unapologetic Mathematician

Mathematics for the interested outsider

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<f(x)\}

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<f(x) But since \{f_n(x)\} increases to f(x), this means that y<f_n(x) 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<f(x)\}\\&=\{(x,y)\in X\times Y\vert 0\leq y\leq f(x)\}\cap\{(x,y)\in X\times Y\vert 0\leq y<f(x)\}^c\\&=\{(x,y)\in X\times Y\vert 0\leq y\leq f(x)\}\cap\{(x,y)\in X\times Y\vert y<0\textrm{ or }f(x)\leq y\}\\&=\{(x,y)\in X\times Y\vert 0\leq y\leq f(x)\textrm{ and }(y<0\textrm{ or }f(x)\leq y)\}\\&=\{(x,y)\in X\times Y\vert 0\leq y\leq f(x)\leq y)\}\\&=\{(x,y)\in X\times Y\vert y=f(x)\}\end{aligned}

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 | 22 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 389 other followers