A Brief Discussion of Lebesgue Measure Theory

Introduction

In class, we used the fol lowing function in our development of our definition of area under a
curve:

In particular, if we compute lim R_n for this function over the interval [0, 1], we get the wrong
answer! Namely, we get 1, while the right answer is (I claimed) 0.

This was one of the points that led us to abandon lim R_n as our definition for the area under
a curve.

As a reminder, here is that computation again:

In the second line above, the function evaluates to 1 because it is being evaluated at i/n, which
is rational since i and n are both integers.

As I stated above, I claimed in class that this is actually the wrong answer to the question
of the area under this function. The right answer can be computed with a Lebesgue integral,
which I didn’t have time to discuss in class.

But in this particular instance, this Lebesgue integral actually can be computed without having
to go to particularly gory depths.

The Lebesgue Integral
Roughly speaking, one of the main ideas behind the Lebesgue integral is to slice up the range
instead of the domain. Strangely, this turns out to make a big difference .

Here are the specifics: Suppose we have a function f defined on [a, b]. For the sake of simplicity,
suppose we also have a bounded range given by the interval [c, d].

Similarly to what we have done before, we will slice... but this time we slice up the interval
[c, d] into n equal slices – not the interval [a, b]. We will also eventually consider the limit as n
approaches infinity. Also similarly to before, we define y_i to be the upper endpoint of the ith
such subinterval.

For a given n and a given subinterval [y_i-1, y_i], we consider the set of all points x in the domain
[a, b] for which f(x) ∈ [y_i-1, y_i]. Let’s call this set D_i (sometimes called the “pre-image” of the
interval [y_i-1, y_i].

Effectively, this subdivides the domain [a, b] into n subsets D₁, . . . ,D_n. Accordingly, this sub-
divides the area under the curve into n regions as well. One such region is shown in the figure
above.

The area of one of these regions is computed as the height (y_i) times the width, which is the
width of the set D_i. Note that in the case of the set D_i in the figure above, the “width” of D_i
is the sum of the widths of the two intervals indicated.

At this point, we can see an advantage and a disadvantage of the Lebesgue approach. The
clear advantage is that, because we have sliced horizontally in some sense, the height is easy
to determine – unlike in the case of the Riemann integral, where we have serious problems in
choosing the height. The disadvantage is that, unfortunately, we now have a nontrivial problem
in determining the width – which of course was very easy to compute in the Riemann integral
as just Δx = (b − a)/n.

Computing these widths is in general a hard problem, but can be dealt with effectively by
advanced techniques I won’t describe here.

But, once that has been taken care of, then the Lebesgue integral can be written simply as

which is just the sum of the products of heights times widths for each of the regions the area
was divided into.

Now, for this particular function that we are interested in, there are only two values of y for
which there is any pre-image at all – namely, y = 0 (for which the pre-image is the set of
irrationals between 0 and 1) and y = 1 (for which the pre-image is the set of rationals between
0 and 1). So, the Lebesgue integral just turns into

So, the problem in this case reduces simply to finding the “width” of the set of rational numbers
between 0 and 1.

This is a computation that we will be able to do, but first we have to establish a particular
property of this set of rational numbers between 0 and 1.

Countability
A set A of numbers is said to be “countable” if one can form a list
a₁, a₂, a₃, a₄, . . .

such that every number in A appears somewhere in that list.

For example, the set of positive integers is countable, as we see with the list
1, 2, 3, 4, . . .

The set of all integers is also countable, as shown by the list
0, 1,−1, 2,−2, 3,−3, . . .

The numbers in the list above are not in order , but that is okay – the point is that every number
in the given set (in other words every integer) appears somewhere in the list.
It is tempting to think that all infinite sets are countable, but in fact this is not true. For
example, the set of all real numbers is NOT countable. This can be proved without too much
trouble, but I won’t include that proof here since it isn’t directly relevant.

So, since some infinite sets are countable, and some are not, we cannot simply assume that the
set we are interested in (the set of rational numbers between 0 and 1, which we will refer to
as S throughout the remainder of this discussion) is countable. We need to verify this somehow.

The first step is to observe that we can first make an infinite array that includes all such rational
numbers. This is shown below, where every numerator is the same as the column number, and
every denominator is the same as the row number.

Now, this includes all positive rational numbers, not just the ones in S (the ones between 0
and 1); so, we scratch off the ones that are greater than one... we quickly notice that this is
simply all of the entries above the diagonal.

We then notice that numbers appear multiple times in the array – for example, 1/1 = 2/2 =
3/3 = . . ., and 1/2 = 2/4 = 3/6 = . . ., and so on. Let’s scratch of the duplicates as well, leaving
only the fractions that are in reduced form.
What we are left with at this point is just

Since we have only removed duplicates and numbers that are not in S, we know that every
rational number is on this array somewhere. And we furthermore notice that each row is now
finite. So, we can make a list by just taking one row at a time. We get

This list contains all of the elements of S, and so we can conclude that S is countable.

The “Width” of the Rationals
Now that we have established that the set we are interested in, S, is countable, we can show
that the “width” of S is 0, with the following method.

First we will construct a new set, that we will denote by B_k. This set will be the union of
intervals, each of which has a rational number as its center. We will start with 1 as a center,
and then continue through the list that we formed in the last section.

The radii of these intervals will not be constant though... we will make them smaller as we
proceed through the list. The first one will have radius k, the next one will have radius k/2,
the next one k/4, ... etc., cutting the radius in half each time.

Putting all of this together, here is our new set B_k:

We can make some observations about B_k. First of all, we note that B_k contains S as a subset,
no matter what k is – since in fact every point in S is the center of one of the intervals that
makes up B_k.

And since B_k contains S as a subset, it stands to reason that
width(S) ≤width(B_k)

Second, we can draw some conclusions about the width of B_k, because it is made up of a bunch
of intervals, each of which has a known width. Of course some of these intervals overlap, and
so we can’t conclude that the width of B_k is equal to the sum of the widths of the intervals...
but we can conclude that it will be less than or equal to that sum.

This gives us

Combining these two observations, we get
width(S) ≤ width(B_k) ≤ 4k

Since we know that this must be true for ALL positive values of k, no matter how small, the
only possibility is that
width(S) = 0

Going back to the previous discussion of using the Lebesgue integral to compute area, we finally
conclude that
area under the curve = Lebesgue integral = width(S) = 0