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Prove that if $f(x)$ is bounded on $[a,b]$ , then $\lim_{||P\,||\to 0}\,L(P)$ is the least upper bound of all lower sums.

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Lesson Parent:

The Logical Framework For Definite Integrals

Problem:

Answer:

It is true that if $f(x)$ is bounded on $[a,b]$ , then $\lim_{||P\,||\to 0}\,L(P)$ is the least upper bound of all lower sums.

Solution:

First, we need to define two variables. Let's denote $\phi$ as the least upper bound of all lower sums. By the definition of $f(x)$ being bounded, we know that there is some constant $K$ where <Sign in to see all the formulas>.

Now we know that there exists a partition <Sign in to see all the formulas> such that <Sign in to see all the formulas>. If this were not the case then $\phi$ would not be a least upper bound of the set of all lower sums. We would then find the actual least upper bound, denote it by $\phi$ and know that there exists a partition $Q$ such that <Sign in to see all the formulas>.

Let's choose <Sign in to see all the formulas> where $N$ comes from $Q$ and $K$ is the bound of $f(x)$ . Now let's assume we have a partition $P$ where <Sign in to see all the formulas>.

Now let's create a new partition made up of the the points in both $P$ and $Q$ and call this new partition $S$ . (Mathematicians call this intersection and write <Sign in to see all the formulas>.) It is important to note that <Sign in to see all the formulas> because it will be used below.

So now let's state some facts to get ourselves ready for the hard part that is coming up:

Every point in $Q$ is an endpoint in $S$ .
Every point in $Q$ may or may not be in an interval of $S$ . In other words, there are two types of intervals in $S$ . Those that contain points that belong to $Q$ and those that don't contain point that belong to $Q$ .

Now we know every element of $Q$ is in $S$ . Recall that <Sign in to see all the formulas>. We say this to show that $Q$ contains $N+1$ elements that are in $S$ . Now the endpoints of $Q$ and $S$ match since they are both partitions of the same definite integral. For each <Sign in to see all the formulas> there are two intervals in $S$ that they are endpoints of. For example, $q_1$ is the right endpoint of the interval before it and the left endpoint of the interval after it. So the maximum number of intervals in $S$ that can have an element of $Q$ is counted by knowing that there are two intervals for each <Sign in to see all the formulas> and one interval for each endpoint. This makes at most <Sign in to see all the formulas> intervals in $S$ that contain an element of $Q$ . (There can be less than this if both the endpoints of an interval are from $Q$ .) So the important thing in this paragraph is to understand that there are at most $2N$ intervals in $S$ that contain an element of $Q$ .

Next let's see if we can get an upper bound on <Sign in to see all the formulas>. We know that <Sign in to see all the formulas> is of the form

<Sign in to see all the formulas>

Since <Sign in to see all the formulas> we know that for every

$M_i$ and

$M_j$ we have <Sign in to see all the formulas> and <Sign in to see all the formulas>. Similarly, for every <Sign in to see all the formulas> and <Sign in to see all the formulas> we know that <Sign in to see all the formulas> and <Sign in to see all the formulas>. (This is where we use <Sign in to see all the formulas> as mentioned above.) Furthermore, the important thing to note is that <Sign in to see all the formulas> is dependent on only those intervals in

$P$ where we have put an element of

$Q$ . This is because those parts of L(P) where where we haven't added an element of

$Q$ are identical in

$L(S)$ and subtract to zero. In the previous paragraph we determined that there are at most

$2N$ intervals in

$S$ that contain an element of

$Q$ . Therefore,

Prove that if f(x)f(x) is bounded on [a,b][a,b], then lim||P||→0L(P)\lim_{||P\,||\to 0}\,L(P) is the least upper bound of all lower sums.

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SAMPLE SOLUTION

<Sign in to see all the formulas>

<Sign in to see all the formulas>

<Sign in to see all the formulas>

<Sign in to see all the formulas>

<Sign in to see all the formulas>

GAIN AN ADVANTAGE

Prove that if $f(x)$ is bounded on $[a,b]$ , then $\lim_{||P\,||\to 0}\,L(P)$ is the least upper bound of all lower sums.