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Prove that $L(P)\leq R(P)\leq U(P)$.

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Problem:

Prove that $L(P)\leq R(P)\leq U(P)$ where $L(P)$ and $U(P)$ are upper and lower sums, respectively, and $R(P)$ is an arbitrary Riemann sum.

Answer:

It is true that $L(P)\leq R(P)\leq U(P)$.

Solution:

Recall our lessons on upper and lower sums here, Riemann sums here, and partitions here.

For each closed interval <Sign in to see all the formulas> in the partition $P$, $f(x)$ will have a greatest lower bound, $m_i$ and a least upper bound, $M_i$. By definition of the least upper bound and greatest lower bound, any arbitrary point <Sign in to see all the formulas> will result in <Sign in to see all the formulas>.

Furthermore, because <Sign in to see all the formulas> is positive we know that

for every closed interval <Sign in to see all the formulas> in the partition. This means that we have

where $N$ is the number of closed intervals in the partition.

Because of this we can write

Now since

and

we have proven that