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

First, recall that $\phi$ is the least upper bound of all lower sums and that $\Phi$ is the greatest lower bound of all upper sums. In general we know that for every partition

<Sign in to see all the formulas>

To prove that there exits a unique number $I$ such that for all partitions $P$, <Sign in to see all the formulas> we need to do two things. First, we need to prove that there exists at least one $I$ such that all partitions $P$, <Sign in to see all the formulas>. This is easy to do. We could pick either $\phi$ or $\Phi$ or anything in between them.

The second thing we need to do is to prove that $I$ is unique. Let's pick two arbitrary numbers $I_1$ and $I_2$ such that

<Sign in to see all the formulas>

<Sign in to see all the formulas>

Because we assume that <Sign in to see all the formulas> (we will refer to them both as just $\phi$ now) we know that <Sign in to see all the formulas> and therefore <Sign in to see all the formulas> and that <Sign in to see all the formulas> and therefore <Sign in to see all the formulas>. We have now shown that <Sign in to see all the formulas>.

The proof is now complete. In fact we have shown more than what we set out to prove. We have actually shown that when we say "if <Sign in to see all the formulas>, then there exits a unique number $I$ such that for all partitions $P$, <Sign in to see all the formulas>" we know that <Sign in to see all the formulas>. So we have actually determined what $I$ is!

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