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.
Primary tabs
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.
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.
SAMPLE SOLUTION
All of our problems have fully worked out, logical solutions. We connect every solution to its underlying lesson so you can understand the concepts needed to solve any problem and reinforce your learning by practicing with similar problems. All of the formulas are available to our subscribers, but here is a sample of the problem solution you will see when you log in:
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>.
From our definition of partition limits we need to show that for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that for every partition where <Sign in to see all the formulas>, then <Sign in to see all the formulas>. Since <Sign in to see all the formulas> this reduces to <Sign in to see all the formulas>. So we need to show that for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that for every partition where <Sign in to see all the formulas>, then <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,
<Sign in to see all the formulas>
Now since
<Sign in to see all the formulas>
and <Sign in to see all the formulas>, by our rules of inequalities we know that
<Sign in to see all the formulas>
The proof is now complete and we can say that
