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# Prove an inequality relating sums and integrals

Prove that

where is a nonnegative, increasing function defined for all .

Use this to deduce the inequalities

by taking .

Proof. Since is increasing we know (where denotes the greatest integer less than or equal to ) for all . Define step functions and by

Then and are constant on the open subintervals of the partition

So,

Since is integrable we must have

for every pair of step functions . Hence,

Next, if we take (which is nonnegative and increasing on ) we have

From this we then get two inequalities

Therefore,

# Prove some statements about integrals of bounded monotonic increasing functions

Consider a bounded, monotonic, real-valued function on the interval . The define sequences

1. Prove that

and that

2. Prove that the two sequences and converge to .
3. State and prove a generalization of the above to interval .

1. Proof.First, we define two step functions,

where denotes the greatest integer less than or equal to . Then we define a partition of ,

For any we have

So, and are constant on the open subintervals of the partition .
Since is monotonically increasing and for all (by the definition of and ) we have

2. Proof. From part (a) we have

since . Since does not depend on we have

Therefore,

3. Claim: If is a real-valued function that is monotonic increasing and bounded on the interval , then

for and defined as follows:

Proof. Let

be a partition of the interval . Then, define step functions and with and for . By these definitions we have for all (since is monotonic increasing). Since is bounded and monotonic increasing it is integrable, and

And, since , and we have

# Prove inequalities of the logarithm with respect to some series

Consider a partition of the interval for some .

1. Find step functions that are constant on the open subintervals of and integrate to derive the inequalities:

2. Give a geometric interpretation of the inequalities in part (a).
3. Find a particular partition (i.e., choose particular values for ) to establish the following inequalities for ,

1. Proof. We define step function and by

Since is strictly decreasing on , we have

Therefore, using the definition of the integral of a step function as a sum,

2. Geometrically, these inequalities say that the area under the curve lies between the step functions that take on the values and for each .
3. Proof. To establish these inequalities we pick the partition,

Then, applying part (a) we have

The final line follows since so the sum on the left starts with and the sum on the right only runs to . These were the inequalities requested

# Prove the integral of a step function is invariant under translation

For a step function on an interval prove

Proof. Let be a partition of such that is constant on the open k-th subinterval of the partition.
Then is a partition of and for . So,

and,

Thus,

# Prove integrals of step functions are additive wrt interval of integration

For a step function prove that

Proof. Let be a partition of and be a partition of such that is constant on the open subintervals of and . Then, let so is a partition of and is constant on the open subintervals of . Then, let . Then,

# Prove the comparison theorem for integrals of step functions

For step functions for every prove that

Proof. Let be a partition of such that and are constant on the open subintervals of (such a partition exists since we can take the common refinement of partitions on which and are constant on open subintervals). Then, assume if . From the definition of integrals of step functions, we have

Then, (since for every ), which implies,

# Prove the linearity property of integrals of step functions

Prove that for step function defined on an interval and for constants , we have

In Apostol, this is Theorem 1.4.

Proof. By the additive property of the integral of step functions we know

Then, by the homogeneous property (Theorem 1.3 in Apostol) we know

# Prove the additive property of integrals of step functions

For step functions defined on an interval , prove

Proof. Let be a partition of such that is constant on the open subintervals of (we know such a exists since we can take the common refinement of the partition of on which and individually are constant on open subintervals). Let if for . Then, working from the definition of the integral of a step function on an interval, we have

# Properties of another alternate definition of the integral of a step function

Suppose we defined the integral of a step function as:

Which of the following properties of the integral would still hold:

1. .
2. .
3. .
4. .
5. If for all , then .

1. True.
Proof. Let be a partition of and be partition of such that is constant on the open subintervals of and . Then, let , where , so is a partition of and is constant on the open subintervals of . Then,

2. True.
Proof. Let be partition of such that is constant on the open subintervals of . (We know such a partition exists since we can take the common refinement of the partitions of on which and individually are constant.) Assume if for . Then,

3. True.
Proof. Let be a partition of such that is constant on the open subintervals of . Assume if for . Then if so using our alternative definition of the integral we have,

4. False. Since . In particular, a counterexample is given by letting for all and let . Then,

5. False.
A counterexample is given by considering and on the interval . Then, on the interval, but

# Properties of an alternate definition of the integral of a step function

Consider if we chose the following definition for the definite integral of a step function:

Which of the following properties (all valid for the actual definition) would remain valid under this new definition:

1. .
2. .
3. .
4. .
5. If for all , then .

1. True.
Proof. Let be a partition of and be a partition of such that is constant on the open subintervals of and . Then, let , where , so is a partition of and is constant on the open subintervals of . Then,

2. False.
Counterexample: Let for all . Then,

while

Hence, this property does not hold. (More generally, since .)

3. False. Again, this is because . A specific counterexample is given.
Counterexample: Let for all and let . Then,

while,

4. True.
Proof. Let be a partition of such that on the th open subinterval of . Then,

is a partition of and on . So,

Thus, indeed we have

5. True.
Proof. Since implies , the result follows immediately