# 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

1. Is this correct: the unit circle in R^2 has measure 0 with respect to the measure of the plane because the circle is the product of sets of at most two points on the vertical lines intersecting the circle (and thus each set having measure 0)?

This made perfect sense if there was a product measure theorem that allowed a product of measure spaces indexed by a continium of real line instead of a product of finite or countably many measure spaces. Have you heard of such a theorem in standard measure theory?

Comment by Hamid | July 20, 2010 | Reply

2. Not offhand, but I’m not sure why you think it would help. In what way is $\mathbb{R}^2$ such a product?

Comment by John Armstrong | July 20, 2010 | Reply

3. Taking only the protion of the x-axis between [-1,1] and considering only the upper half plane, then the intersections of the vertical lines thorough [-1, 1] with the function representing the upper unit circle are just singletons each of measure zero. If this product theorem existed then the measure is zero.

I think what you showed in your upper and lower ordinate sets readily implies that the set is measurable alright.

Actually, this was the remark that our analysis instructor made of the solution to one of the final exam problems right after the exam some years ago. Maybe he was joking, but I still can’t get over it. It seems too easy of a joke.

Comment by Hamid | July 20, 2010 | Reply

4. Of course I am not trying to make a joke here. I am still curious if this makes sense.

Comment by Hamid | July 20, 2010 | Reply

5. I don’t really see how it would help.

Comment by John Armstrong | July 21, 2010 | Reply

6. […] Upper and Lower Ordinate Sets […]

Pingback by Measurable Graphs « The Unapologetic Mathematician | July 21, 2010 | Reply

7. Does it not help to take f(x)=sqrt(1-(sin pi x)^2) for x in [-1, 1] given that we also know somehow that the upper half of unit disk has area pi/2? Don’t ordinate sets apply to show that the graph of (x, f(x)) is measurable?

Comment by Hamid | July 21, 2010 | Reply

8. That’s not what I mean. Where do you see an uncountably-indexed product space?

Comment by John Armstrong | July 21, 2010 | Reply

9. Because first A=[-1, 1] is uncountable; and second, A can be taken as the index set of a product measure (if it exists which is my question) of that many measurable spaces of singletons as subsets of the same real measure spaces as copies of the real vertical lines.

I myself have troubles understanding this and thought afterwards that it must be wrong, but have you not heard of this anywhere either?

Comment by Hamid | July 21, 2010 | Reply

10. Why would the product of all those vertical lines mean anything? The circle lives in the union (with some additional topological information, no less), not their product.

Comment by John Armstrong | July 21, 2010 | Reply

11. True. Let me first again ask you if Fubini’s theorem on the product of measures does not have a generalization to indices larger than omega cardinality; maybe there is a categorical answer. If so, please disregard the rest of the trivialities below.

The thing with this example is that it is a trivial example of more complicated problems, I suppose. My intuition actually fails to see the “right” picture in its most generality. Fibre bundles like sphere bundles where the base is the unit circle, or certain leaves of foliations of a manifold might be good examples. How does one integrate on the manifold globally in local correspondence with integration on each leaf. E.g. The torus as a circle bundle on another circle inherits its topology from those of the circles and as you noted it is a union of circles. How can one build a product topology and a sigma-algebra of Borel sets out of the open sets of torus’ product topology. That again is trivial as the surface of the torus can be envisioned at most as the product of 2 real Lebesgue measures.

Comment by Hamid | July 22, 2010 | Reply

12. I really don’t know about Fubini’s theorem for uncountable products. I also really don’t know what any of the examples you suggest has to do with uncountable products.

Comment by John Armstrong | July 22, 2010 | Reply

13. I was just trying to show with no luck that the vertical lines passing thorough the circle are not just a union of lines. Consider the line bundles on the circle. One gets a cylinder, a Mobius strip, or putting more twists to the lines before gluing the lines, passing thorough 0 and 1 in the interval [0,1], together then one might get line bundles with higher “spins”. In fact, e.g. for a cylinder we have the space of smooth vector bundles on the circle where instead of lines we have smoothly varying vectors on them.

But forget about this for now. Actually, Noncommutative Geometry 1994 of A. Connes chapter 1 section 4 might be of some use here as the pictures of the Kronecker foliation and the Lady in Blue, scrambled and non-scrambled, might also directly apply here.

Take a torus T again colored in black and white stripes or with more number of colors. What is a local or global diffeomorphism D of T so that D(T) has measure 0 of each stripe? It seems like if we do not have an equal partition of the torus into 3 stripes and if we transform the colors according to H: (f(x), g(y)):[0,1]x[0,1]–>[0,1]x[0,1] where f and g are both the adjusted Cantor ternery functions(as in Royden’s Real Analysis) applied to the (x,y) coordinates of the torus, then we whould have been done except that Cantor ternary functions are not measurable. I was almost there if I could kind of forget about measurablility. If the function is applied to 3 stripes in an exact kind of way rotate the torus less than 1/3 before applying. H is also probably not a diffeomorphism but only a homeomorphism.

One might again ask where the uncountable product is. I am afraid that I would read thorough all of noncommutative geometry book and still not get it. Is it a von Neumann algebra of type II_1? I am now more confused due to lack of knowledge of higher category theory 2-Cat and higher. Uncountable product is probably wrong to begin with.

I promise not to say one more extra word to you again unless you say so otherwise. You probably have to stand on your head as John Baez mentions of cocompactness, co-this and co-that, to read the stuff in my website; it has to be forced into becoming strictly mathematical.

Comment by Hamid | July 23, 2010 | Reply

14. Uncountable product is probably wrong to begin with.

That’s pretty much been my point all along.

And what gave you the impression that I’m uncomfortable with category theory (co-this, and co-that)? Please, do go back and read the few-hundred posts I’ve made on that subject. Just because I’m talking about analysis now doesn’t mean I’m always and forever an analyst.

Comment by John Armstrong | July 23, 2010 | Reply

15. I just quoted John Baez from n-category on a homological algebra puzzle. That’s all.

Comment by Hamid | July 23, 2010 | Reply

16. […] Measures of Ordinate Sets If is a -algebra and is a Borel-measurable function we defined the upper and lower ordinate sets and to be measurable subsets of . Now if we have a measure on and Lebesgue measure on the Borel […]

Pingback by The Measures of Ordinate Sets « The Unapologetic Mathematician | July 26, 2010 | Reply

17. Dr. Armstrong,

Please excuse me by bothering you but I think you have a typo in the definition of the “lower ordinate set”.

Comment by Américo Tavares | April 8, 2013 | Reply

18. That being?

Comment by John Armstrong | April 8, 2013 | Reply

19. If I am right it is a very minor typo: “(…) such that $V^{∗}(f)=$” instead of “(…) such that $V_{∗}(f)=$“.

Comment by Américo Tavares | April 9, 2013 | Reply

20. Repetition: If I am right it is a very minor typo: “(…) such that $V_{\ast }(f)$ instead of “(…) such that $V^{\ast }(f)$“.

Comment by Américo Tavares | April 9, 2013 | Reply

21. Ah, yes. Thanks.

Comment by John Armstrong | April 9, 2013 | Reply

22. Maybe I’m wrong, but isn’t supposed to be $V^*(\chi_E) = E\times[0,1] \cup E^{c} \times \{0\}$?

Comment by shamisen | June 29, 2014 | Reply