Prove that if $f(x)$ is bounded on $[a,b]$, then \[\lim_{||P\,||\to 0}\,U(P)\] is the greatest lower bound of all upper sums.
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Prove that if $f(x)$ is bounded on $[a,b]$, then \[\lim_{||P\,||\to 0}\,U(P)\] is the greatest lower bound of all upper sums.
It is true that if $f(x)$ is bounded on $[a,b]$, then \[\lim_{||P\,||\to 0}\,U(P)\] is the greatest lower bound of all upper sums.
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First, we need to define two variables. Let's denote $\Phi$ as the greatest lower bound of all upper 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 greatest lower bound of the set of all upper sums. We would then find the actual greatest lower 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 U(P) where where we haven't added an element of $Q$ are identical in $U(S)$ 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
