Prove for a bounded function on $[a,b]$ that if there exists a unique number $I$ such that for all partitions $P$, $L(P)\leq I\leq U(P)$, then $\phi=\Phi$.

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Lesson Parent: 
The Logical Framework For Definite Integrals
Problem: 

Prove for a bounded function on $[a,b]$ that if there exists a unique number $I$ such that for all partitions $P$, $L(P)\leq I\leq U(P)$, then $\phi=\Phi$.

Answer: 

It is true that if there exists a unique number $I$ such that for all partitions $P$, $L(P)\leq I\leq U(P)$, then $\phi=\Phi$.

SAMPLE SOLUTION

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Solution: 

Recall from this lesson that $\phi$ and $\Phi$ are the least upper bound and greatest lower bound for the lower and upper sums, respectively.

This proof is actually very easy if you are comfortable with the statements that have ". . . there exists a unique . . ." This means two things. First, it means that something exists. Second, it means that it is unique. Unique means that there is only one. (You are unique, aren't you?) Let's examine what both "there exists" and "there is only one" means for this problem.

So we assume that there exists an $I$ such that for every $P$, <Sign in to see all the formulas>. First this means that $I$ is an upper bound on $L(P)$. Well know that $\phi$ is the least upper bound on all $L(P)$'s. Therefore, by the definition of least upper bounds we know that <Sign in to see all the formulas>. Similarly, because $\Phi$ is the greatest lower bound on all $U(P)$'s and by the definition of greatest lower bounds we know that <Sign in to see all the formulas>. So now we know that

<Sign in to see all the formulas>


It's important to note that this does not tell us how many there are. Just that there is at least one. In fact, there could be an infinite number of $I$'s.

Let's denote the statement "for every partition $P$, <Sign in to see all the formulas>" by <Sign in to see all the formulas>. Since we can have more than one $I$ we could denote the statements "for every partition $P$, <Sign in to see all the formulas>" and "for every partition $P$, <Sign in to see all the formulas>" by <Sign in to see all the formulas> and <Sign in to see all the formulas>, respectively.

One of logical forms of completeness is to say if <Sign in to see all the formulas> and <Sign in to see all the formulas> are true, then <Sign in to see all the formulas>. (Math aficionados should recognize this as <Sign in to see all the formulas> where $P$ is a general statement and not a partition.) In other words, if we can find an $I_1$ such that for every partition $P$, <Sign in to see all the formulas> is true and an $I_2$ such that for every partition $P$, <Sign in to see all the formulas> is true, then we know that <Sign in to see all the formulas>.

We are told to assume that $I_1$ and/or $I_2$ exists, but we can actually do much better. We know that for every partition $P$ that <Sign in to see all the formulas> and <Sign in to see all the formulas>. In other words, we have an $I_1$ and an $I_2$. Let's call <Sign in to see all the formulas> and <Sign in to see all the formulas>. Because we are told to assume uniqueness then we know that <Sign in to see all the formulas> and therefore <Sign in to see all the formulas>.

The proof is now complete. In fact, we have shown more than what we were asked in this proof. We know that if there is a unique number then it has to be equal to $\phi$ (and $\Phi$ since they are equal).

Finally, it is important to note that this relationship is even stronger and we can also show that if <Sign in to see all the formulas>, then there exists a unique number $I$ such that for all partitions $P$, <Sign in to see all the formulas>.