Prove that if $P$ and $Q$ are two arbitrary partitions of $[a,b]$, then $L(P)\leq U(Q)$.

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Lesson Parent: 
Integration III: Definite Integrals As Upper and Lower Sums
Problem: 

Assume that $f(x)$ is a bounded function on the closed interval $[a,b]$. Prove that if $P$ and $Q$ are two arbitrary partitions of $[a,b]$, then the upper and lower sums of $f(x)$ are related by $L(P)\leq U(Q)$.

Answer: 

It is true that if $P$ and $Q$ are two arbitrary partitions of $[a,b]$, then $L(P)\leq U(Q)$.

SAMPLE SOLUTION

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Solution: 

Because $f(x)$ is a bounded function on the closed interval $[a,b]$ we can form the upper and lower sums $U(P)$ and $L(P)$.

First, by the definition of upper and lower sums we know that

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and

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Now let's say we combine the partitions $P$ and $Q$. If $P$ and $Q$ are equal then we can say that <Sign in to see all the formulas> and therefore <Sign in to see all the formulas>. But what about the case where $P$ and $Q$ are not different? When we combine them we get a new partition $R$. (Mathematicians call the act of combining two sets intersection and write <Sign in to see all the formulas>.) Now since $P$ and $Q$ are not equal, the partition $R$ differs from both $P$ and $Q$ by at least one point.

Recall from this problem that when we add just one point to a partition we get upper and lower sums that are closer together. This means that

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and

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From our rules of inequalities we see that

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and therefore,

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The proof is now complete.

Note that we could have just as easily shown that <Sign in to see all the formulas> and therefore <Sign in to see all the formulas>. This emphasizes the point that since $P$ and $Q$ are arbitrary that every lower bound is less than or equal to every upper bound. The consequences of this are discussed here.