# College Math Teaching

## January 17, 2011

### Integration: Riemann Integration, Limitations and Lebesgue’s Idea

What about integration? Here we will see what Lebesgue integration is about, how it differs from Riemann integration and why we need to learn about the algebra of measurable sets.

Brief review of Riemann Integration

Remember that the idea was as follows: we limit ourselves to bounded functions. suppose we want to compute $\int_{a}^{b}f(x)dx$. We partitioned the interval $[a,b]$ into several subintervals:

$a=x_{0} Let $m_{i}=\inf f(\xi ),\xi \in \lbrack x_{i-1},x_{i}]$ and let $M_{i}=\sup f(\omega ),\omega \in \lbrack x_{i-1},x_{i}]$. Let $\Delta x_{i}=x_{i}-x_{i-1}$. Call this partition $P$.

Then $L_{P}=\sum_{j=1}^{n}m_{j}\Delta x_{j}$ and $U_{P}=\sum_{j=1}^{n}M_{j}\Delta x_{j}$ are called the lower sums and upper sums for $f$ with respect
to the partition $P$.

One proves theorems such as if $Q$ is a refinement of partition $P$ then $L_{P}\leq L_{Q}$ and $U_{Q}\leq U_{P}$ (that is, as you make the refinement finer…with smaller intervals, the lower sums go up (or stay the same) and the upper sums go down (or stay the same) and then one can define $U$ to the
be infimum (greatest lower bound) of all of the possible upper sums and $L$ to the the supremum (least upper bound) of all of the possible lower sums. If $U=L$ we then declare that to be the (Riemann) integral of $f$ over $[a,b].$

Note that this puts some restrictions on functions that can be integrated; for example $f$ being unbounded, say from above, on a finite interval will prevent upper sums from being finite. Or, if there is some dense subset of $[a,b]$ for which $f$ obtains values that are a set distance away from the the values that $f$ attains on the compliment of that subset, the upper and lower sums will never converge to a single value. So this not only puts restrictions on which functions have a Riemann integral, but it also precludes some “reasonable sounding” convergence theorems from being true.

For example, suppose we enumerate the rational numbers by $q_{1},q_{2},...q_{k}...$ and define $f_{1}(x)=\left\{\begin{array}{c}1,x\neq q_{1} \\ 0,x=q_{1}\end{array}\right.$ and then inductively define $f_{k}(x)=\left\{\begin{array}{c}1,x\notin \{q_{1},q_{2},..q_{k}\} \\ 0,x\in \{q_{1},q_{2},..q_{k}\}\end{array}\right.$. Then $f_{k}\rightarrow f=\left\{\begin{array}{c}1,x\notin \{q_{1},q_{2},..q_{k}....\} \\ 0,x\in \{q_{1},q_{2},..q_{k},...\}\end{array}\right.$ and for each $k$, $\int_{0}^{1}f_{k}(x)dx=1$ but $f$, the limit function, is not Riemann integrable.

So, there are a couple of things to note here:

1. The Riemann integral involves partitioning the interval to be integrated over without regards to the function being integrated at all; that is, if you were doing $\int_{0}^{1}e^{\sqrt{x}}dx$ or $\int_{0}^{1}\sin (x^{2})dx$ you wouldn’t partition $[0,1]$ any differently.

2. The elements of any partition of the Riemann integral are intervals of finite length.

The Lebesgue integral changes these two features;

1. We’ll use information about the function being integrated to help us select partitions and

2. The elements of our partition need not be intervals of finite length; they just need to be measurable sets.

For example, suppose we wish to compute $\int_{0}^{1}4x-x^{2}dx$ by using a Lebesgue integral.

Partition the range of $f$ into 4 subintervals:

$Y_{1}=0\leq y<.25,$

$Y_{2}=.25\leq y<.5,$

$Y_{3}=.5\leq y<.75,$

$Y_{4}=.75\leq y\leq 1.$

Now consider the inverse image of these subintervals and label these:

$E_{1}=f^{-1}(Y_{1})=[0,\frac{1}{2}-\frac{1}{4}\sqrt{3})\cup (\frac{1}{2}+\frac{1}{4}\sqrt{3},1]$

$E_{2}=f^{-1}(Y_{2})=[\frac{1}{2}-\frac{1}{4}\sqrt{3},\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}})\cup \lbrack \frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}},\frac{1}{2}+\frac{1}{4}\sqrt{3})$

$E_{3}=f^{-1}(Y_{3})=[\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}},\frac{1}{4})\cup \lbrack \frac{3}{4},\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}})$

$E_{4}=f^{-1}(Y_{4})=[\frac{1}{4},\frac{3}{4})$

Then we form something similar to upper and lower sums. Recall the measure of an interval is just its length.

So we obtain something like an upper sum:

$U=\frac{1}{4}m(E_{1})+\frac{1}{2}m(E_{2})+\frac{3}{4}m(E_{3})+1m(E_{4})$

and a lower sum as well:

$L=0m(E_{1})+\frac{1}{4}m(E_{2})+\frac{1}{2}m(E_{3})+\frac{3}{4}m(E_{4})$

See the above figure for an illustration of an upper sum.

Then we proceed by refining the partitions of our range; for this to work we need for the inverse image of the partitions of the range to be measurable sets; this is why we need theorems about what constitues a measureable set.

A measurable function is one whose inverse images (or partitions of its range into intervals) are measurable sets.

The Lebesgue integral can be defined as either the infimum of all the upper sums or the supremum of all of the lower sums.

If one wants to see how this works, try doing this for $\int_{0}^{1}g(x)dx$ where
$g(x)=\left\{ \begin{array}{c}1,x\notin Q \\ \frac{1}{p},x=\frac{p}{q}\end{array}\right.$ where $Q$ is the rationals and $\frac{p}{q}$ is in lowest terms.

Then the subinterval which includes 1 in any partition of the range will have an inverse image with measure 1 whereas all subintervals whose upper bounds are strictly less than 1 will have measure zero. Hence it follows that $\int_{0}^{1}g(x)dx=1$ though $g$ is not Riemann integrable.

Of course, this has been sketchy and we haven’t covered which types of sets are measurable. We’ll also discuss some convergence theorems as well.