# College Math Teaching

## January 12, 2011

### The Lebesgue Integral: motivating the mastery of the tools of abstract mathematics

My Ph. D. basic math classes were a long time ago; in the fall of 1985 I enrolled in the beginning topology, algebra, analysis and algebraic topology classes. The algebra, analysis and topology classes were designed to get the student through the preliminary examinations.

The ironic thing is that while I passed analysis on my first try (summer of1986) and algebra on the first try (summer of 1987), it took me three tries to get through topology which I finally passed in the winter of 1988. Of course, that is what I ended up specializing in.

But I can honestly say that most of my analysis was ”studying the problems that might appear on the exam”; I really didn’t know what I was doing nor did I understand the material. I had no perspective whatsoever; I really didn’t know where the course was going. Therefore I remember almost none of it.

What does this have to do with the teaching of college mathematics? I see it this way: often, prior to tackling an abstract concept (say, Lebesgue Integration), one first assembles some tools (say, measure theory, the concept of a sigma algebra of sets, Borel sets, etc.) and then uses these tools as the need arises. But the tools are often developed without any context; I didn’t see why any of this was needed or what it would be used for. I had no perspective; I didn’t know WHY I had to be patient with these esoteric (to me at that time) tools.

So, what I remember now is to constantly remind the students where we are going with a topic; I try to explain WHY we need to be patient with this piece of drudgery or that.

Lebesgue Integration

My goal: I hope to review Lebesgue integration for my own benefit and hopefully, via some notes, perhaps provide some perspective for the student who is going through this process for the first time.

For my references, I am choosing A Primer of Lebesgue Integration by H. S. Bear and Real Analysis by H. L. Royden.

Why Lebesgue Integration?

Ground Rules

For the duration, I’ll be discussing the integration of positive functions over a closed, bounded interval $[a,b]$. In fact, I’ll go ahead and use $[0,1]$ with no loss of generality.

Of course, if we are merely concerned with integrating functions which are piecewise continuous then the Riemann integral (say, as defined as the limit of upper and lower sums) works just fine.

However what happens if we want to, say, extend the class of functions that we can integrate?

Suppose, for example, we wish to compute $\int_{0}^{1}\frac{1}{\sqrt{x}}dx$ or attempt to compute $\int_{0}^{1}\frac{1}{x}dx$?

Ok, strictly speaking, these functions aren’t defined on $[0,1]$, so we can replace the integrand by, say a conditional function:

$f(x)=\left\{ \begin{array}{c}\frac{1}{\sqrt{x}},x>0 \\ 0,x=0\end{array}\right.$ in the first case and similarly in the second case.

Of course, we tell our students that while these functions are NOT Riemann integrable (unbounded functions aren’t as the upper sum will never be finite for any finite partition), we can possibly extend the notion of integrability by using the improper integral technique, which, of course, shows that the first integral converges and the second one does not.

But again, we have a type of piecewise continuity. What happens when we don’t?

Let us examine another couple of functions:

$g(x)=\left\{ \begin{array}{c}1,x\text{ rational} \\ 0,\text{ }x\text{ irrational}\end{array}\right.$

$h(x)=\left\{ \begin{array}{c}q,x=\frac{p}{q},\text{ in lowest terms, }p,q\in \{1,2,3..\} \\ 0\text{ otherwise}\end{array}\right\}$

Notice that $g$ is bounded whereas $h$ is not. Notice that neither is Riemann integrable as there is no hope of a upper and lower sums converging. But notice something else: we know that the rationals compose a minescule portion of the real numbers; hence there might be another theory of integration that would all us to integrate both of these functions and obtain 0 for the integral. But there isn’t a good way to obtain the integrals of these functions as a limit of Riemann integrals; there are no smaller intervals to work with. Hence the need for a more expansive theory of integration.

About the rational numbers: why do they ”take up little space”? I’ll assume that the reader knows that the rationals are countable and the reals are not, hence there are many more irrational numbers than rational ones.

But as far as ”taking up space”: what do we mean by that? We will answer this when we develop ”measure theory”; that is, a way of assigning a generalization of a “length” to more complicated subsets of the real line. For now I’ll explain why the set of rational numbers doesn’t take up much space, even though they are a dense subset of the real line.

I will show that we can cover all of the rationals by a countable set of intervals whose lengths add up to an arbitrarily small number. That is,
for any given $\varepsilon >0,$ I’ll construct a countable set of intervals $[a_{i},b_{i}]$ such that $\cup _{i=1}^{\infty}[a_{i},b_{i}]$ contains ALL of the rational numbers AND $\sum_{i=1}^{\infty }(b_{i}-a_{i})\leq\varepsilon .$ Of course, these intervals are NOT disjoint; far from it. But their union contains all of the rational numbers.

Review: Recall $\sum_{k=0}^{\infty }r^{k}=\frac{1}{1-r}$ for $|r|<1$.

Let $q_{0},q_{1},q_{2}....$ be the set of rational numbers (which are countable). Then for $q_{k},$ consider the interval $[q_{k}-\frac{1}{2}(\frac{\varepsilon +1}{\varepsilon })^{k},q_{k}+\frac{1}{2}(\frac{\varepsilon+1}{\varepsilon })^{k}]$. This interval has length $(\frac{\varepsilon +1}{\varepsilon })^{k}.$

Hence adding the lengths of the intervals together is the same as $\sum_{k=0}^{\infty }(\frac{\varepsilon +1}{\varepsilon})^{k}=\varepsilon$. So the sum of the lengths of closed intervals which covers all of the rationals can be made as small as desired.

So, getting back to the problem of integration: how do we do this? The idea behind the Lebesgue integral is to first define this integral for
“simple functions” that is, functions that take on precisely one value over a set. For example: a step function can be thought of as a kind of
sum of simple functions.

Then $\chi (x)$ is a simple function over a set $I$ and $m(I)$ is the “measure”; (a kind of length) of $I,$ then the new integral would be defined as $\int_{I}\chi (x)=\chi (x)m(I)$ (think of the integral in terms of “area”; and this would be a height (in the “function direction”) times “width of the interval” operation).

What will be new is that we’ll consider $I$‘s that might be much more complicated than a simple disjoint union of intervals. Well also show that if $A=I\cup J$ and the union is disjoint and if $\chi (x)$ is simple over both $I$ and $J,$ then $\int_{A}\chi =$ $\int_{I}\chi +$ $\int_{J}\chi$

So, going back to our example functions $g$ and $h:$ note that the “measure” of the rational numbers is zero as these can be covered by intervals whose lengths sum to arbitrarily small sums. So if we let $Q$ denote the rational numbers in $[0,1]$ and $I$ denote the irrationals, note that $g$ is a simple function over these two sets. Hence,

$\int_{[0,1]}g=$ $\int_{Q}g+$ $\int_{I}g=m(Q)\ast 1+m(I)\ast 0=0\ast 1+1\ast 0=0.$

The second integral is more problematic; we’d have to break $Q$ into a countable union of points whose coordinates have denominators 2, 3, 4,….and to prove a theorem about adding up a countable number of integrals
(e. g., forming a sequence). Getting to that point will take some work, as you can see.

Measure Theory.

Ok, we want to determine a “generalized length” of subsets of the real line, (subsets of $[0,1]$ for now) and these subsets can
be far more complicated than those which are finite disjoint collections of intervals (if you know what the Cantor “middle thirds” set is, you might
review that and if you don’t know what it is, look up “Cantor Set”; that will serve as a nice example of a complicated subset of the real line).
This is a main point of measure theory.

So, in abstract terms, what we are looking for is some sort of a map $m$ from the collection of subsets of $[0,1]$ to the non-negative real numbers that serves as some sort of a length function. It might be a good time to stop and ask yourself: “what properties would we want this map (called a “measure”) to have?

Here are some key properties:

1.$m([a,b])=b-a$ (and the same for open, and half open intervals). After all, this is supposed to be a length function. Of course,
$b > a$.

2. $m(E+r)=m(E):$ that is, measure should be “translation invariant” that is, if you translate a set $E$ by adding the same constant to every element of $E,$ the translated set should have the same measure.

3. $m(\{x\})=m(\emptyset )=0$ (the measure of a single point and the empty set ought to be zero)

4. If $F\subset E$ then $m(F)\leq m(E)$ and we’d like to have some additivity properties:

5. Suppose $E=\cup _{i=1}^{\infty }E_{i}.$ Then we’d want $m(E)\leq\sum_{i=1}^{\infty }m(E_{i})$ provided the left hand sum converges (of course, this convergence is absolute since all terms are positive; hence the order of the terms is of no consequence; this uses our calculus results).

Of course, we are talking about a countable union; sums don’t make sense otherwise.

And, if the sets $E_{i}$ are disjoint, we’d love to have $m(E)=\sum_{i=1}^{\infty }m(E_{i})$

So how would we go about defining such a measure?

Here is one standard way: Every subset of the real line can be covered by a countable union of open intervals. (Note: if you said to yourself
anything about a Lindelof space, you have no business reading this article unless it is to critique it). So given a set $E,$ let $\{(a_{i},b_{i})\}$ be a countable collection of open subintervals whose union contains $E$ (that is, an open cover for $E$). If $\sum_{i}(b_{i}-a_{i})$ is finite, call that the “length” of the cover of $E.$ Then consider the infimum of all lengths of covers of $E$ (note: since $E \subset \lbrack 0,1]$, there exists a cover of length 1 so such an infimum exists). This infimum is
called the measure of $E.$

This definition of measure gives us most of what we want (1-5). Note: proving 1 is a bit trickier than it might first appear; it is immediate that
$m([a,b])\leq b-a.$ To go the other way, one might use the fact that every open cover has a finite subcover (the Heinie-Borel Theorem from your analysis class) and induct on the number of open sets in your finite subcover). Features 2, 3, and 4 are pretty easy to demonstrate.

5 (called “countable subadditivity”) Makes for an interesting exercise. I’ll sketch out a solution here:

Since each $E_{i}$ has a measure, we’ll find a collection of open intervals $I_{ij}$ that cover and, if $l(I_{ij})$ denotes the length of the open interval, we can assume that $m(E_{i})+\frac{\varepsilon }{2^{i}}>\sum_{j}$ $l(I_{ij}).$ Then note that
$\cup _{ij}I_{ij}$ covers $E=\cup _{i=1}^{\infty }E_{i}$ and
$\sum_{i}(m(E_{i})+\frac{\varepsilon }{2^{i}})>\sum_{i}\sum_{j}$ $l(I_{ij})\rightarrow \sum_{i}(m(E_{i}))+\varepsilon >\sum_{i}\sum_{j}$ $l(I_{ij})$ for any $\varepsilon >0.$

Hence $\sum_{i}(m(E_{i}))\geq m(E)$.

So what about the case where the $E_{i}$ are disjoint; do we get equality?

The answer is…well, NO…not for ALL subsets $E.$

It will turn out that we will have to restrict our measure to certain subsets called “measurable sets”. The measurable sets form one large collection of subsets of $[0,1]$ that “work”. These sets will be those sets that have this condition: $m(E)+m(E^{\prime})=1$ (where $E^{\prime }=[0,1]-E$; the set compliment of $E).$

It isn’t obvious that this is the condition that we’ll need; this complement condition is called the Carathedory characterization.

Note: I am being a little bit sloppy here. Strictly speaking, I should use the concept of “outer measure” (that is, the “infimum of the sum of the lengths of the covers”) when talking about the Carathedory characterization and denote that by $m^{*}(E)$ and only use $m(E)$ when I am talking about the “measure” of the “measurable sets”; otherwise I would be using circular logic. But this isn’t a text book so I’ll abuse notation.

Note: it is easy to establish that intervals and one point sets are measurable. To find other sets that are measurable, we’ll need to use
results from the algebra of sets;, sigma-algebras, etc.; the idea is to show that, say, the collection of measurable sets are closed with respect to countable unions, countable intersections, set complementation, etc.

This is one reason why set algebra topics are often covered in real analysis textbooks in the first sections or chapters. We’ll also show an example of a non-measurable set in a future post (2 dimensional versions of these come
into play in the Banach-Tarski “paradox”).

In the next section, we construct a non-measurable set.

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