MATH 409 Lecture 19
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Darboux vs. Riemann Sums
Riemann Sums
Recall upper and lower Darboux sums and upper and lower integrals
A Riemann sum of a function with respect to a partition of generated by samples is a sum
The riemann sums converge to a limit as the norm if for every there exists such that implies for any partition and choice of samples .
Theorem. The Riemann sums converge to a limit as if and only if the function is integrable on and .
Proof of (⇒). Assume the Riemann sums converge to a limit as . Given , we choose so that for every partition with , we have for any choice of samples . Let be a different set of samples for the same partition . Then . We can choose the samples so that is arbitrarily close to while is arbitrarily close to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \inf{f([x_{j-1},x_])}} . That way gets arbitrarily close to , while gets arbitarily close to .
Hence it follows from the above inequalities that and . As a consequence, the distance between and is . In particular, the function is bounded. We conclude that is integrable.
Let . The number lies between and . The inequalities and imply that as can be arbitrarily small, .
Properties of Integrals
Integration as a Linear Operation
Theorem. If are integrable on an interval , then the sum is also integrable on and
Theorem. If is integrable on an interval and is a real constant, then the scalar product is integrable on and
Proof of both. Let denote the integral of and denote the integral of over . The key observation is that the Riemann sums depend linearly on a function. Namely, and for any partition of and choice of samples . It follows that
As , the Riemann sums and get arbitrarily close to and , respectively. Then will be getting arbitrarily close to while will be getting arbitrarily close to . Thus is the integral of and is the integral of over .
Integrability of Subintervals
Theorem. If a function is integrable on , then it is integrable on each subinterval .
Proof. Since is integrable on , for any there is a partition of such that . Given a subinterval , let and . Then is a partition of that refines . Hence
Since is a partition of contained in , it follows that
We conclude that is integrable on .
Integration over Subintervals
Theorem. If a function is integrable on , then for any ,
Proof. Since is integrable on , it is also integrable on the subintervals and . Let be a partition of and be some samples for that partition. Further, let be a partition of and be some samples for that partition. Then is a partition of and are samples for it. The key observation is that
If and , then as well. Therefore the Riemann sums in the latter equality will converge to the integrals , , and , respectively.
Theorem. If a function is integrable on and , then for each continuous function , the composition is also integrable on .
(proof omitted)
Corollary If function and are integrable on , then so is .
Proof. We have . Since and are integrable on , so is . Since is a continuous function on , the compositions , , and are integrable on . Then is integrable on .
Comparison Theorem for Integrals
Theorem. [Comparison theorem]. If functions are integrable on and for all , then
Proof. Since on the interval it follows that for any partition of and choice of samples . As , the sum gets arbitrarily close to the integral of . The theorem follows.
Corollary. If is integrable on and , for , then .
Proof. Comparison theorem with .
Corollary. If is integrable on and for , then
Corollary. If is integrable on , then the function is also integrable on and
Proof. The function is the composition of with a continuous function . Therefore is integrable on . Since for , the Comparison Theorem for integrals implies that
Integral with Variable Limit
Suppose is an integrable function. For any , let (we assume ).
Theorem. The function is well-defined and continuous on .
Proof. Since is integrable on , it is also integrable on each subinterval of . Hence the function is well-defined on . Besides, is bounded: for some and all .
For any , where , we have . Equivalently, . It follows that
Thus is a Lipschitz function on , which implies is uniformly continuous on .
Sets of Measure Zero
A subset of the real line is said to have measure zero if for any , the set can be covered by countably many open intervals such that .
Examples
Theorem. Any countable set has measure zero.
Proof. Indeed, suppose is a countable set and let be a list of all elements of . Given , let
Then and for all , so that .
Theorem. A nondegenerate interval is not a set of measure zero.
Theorem. There exist sets of measure zero that are of the same cardinality as .
Lebesgue's Criterion for Riemann Integrability
Suppose is a property depending on , where . We say that holds for almost all (or almost everywhere on ) if the set has measure zero.
Theorem. A function is Riemann integrable on the interval if and only if is bounded on and continuous almost everywhere on .