In , we defined the Riemann integral on intervals of . We shall now prove some theorems pertaining to it. Below, we will let denote the Riemann sum of associated with tagged partition .
Definition 1 A piecewise function defined on a bounded interval is defined by partitioning into a finite number of sub-intervals and defining on each of the s a function such that for , . We say that is piecewise continuous if each of the s is continuous. One can, here, replace “continuous” with any other qualification of function, such as “smooth”.
Proposition 1 We have the following properties of Riemann integrable functions.
- If is Riemann integrable on two disjoint intervals , with the Riemann integrals equal to , then is Riemann integrable on , with
- Let be an interval containing interval . If is Riemann integrable on , it is Riemann integrable on both and , with
- If on any interval , both and are Riemann integrable, then is also Riemann integrable on , with . Moreover for any , is Riemann integrable on with . In other word, the Riemann integral on operator is a linear operator on the space of Riemann integrable functions on .
Proof: For (1), we take a tagged partition of satisfying and a tagged partition of satisfying . Formally, this says that any refinement of is such that and similar for . For we use the tagged partition . One easily verifies with the triangle inequality that any refinement of it satisfies .
For (2), take any tagged partition of that contains the endpoints of . The equivalence of (3) of Lemma 3 from  with Riemann integrability tells us that for any , there exists a refinement of such that for any which are refinements of ,
Both of can be split into tagged partitions of and , which we denote via and (where the second index represents whether or not the partition is defined on or , such that
If by contradiction were not Riemann integrable on one of , then there would by Lemma 3 of  exist an for which there does not exist a partition that is a refinement of such that the split of any two refinements of it into partitions on , which we denote as and , satisfies both
Setting and using the triangle inequality on the above would thus violate the existence of a refinement of which guarantees , which would imply that is not Riemann integrable on , a contradiction.
With similar logic, one proves Riemann integrability of on . The integral equality of then follows directly from .
The proof of (3) we leave as an exercise to the reader.
Proposition 2 Any continuous function on is Riemann integrable.
Proof: We construct a sequence of tagged partitions of . We let the th tagged partition be given by
in which we have let be the subintervals of corresponding to the partition, where . As for the tags, we let , noting that by the Extreme Value Theorem, the infimum here exists. We notice that under the partial order of partition refinement, this sequence of tagged partitions is a chain. Moreover, the Riemann sums of these partitions form a bounded (by Extreme Value Theorem) monotonically non-decreasing sequence. Thus, by the Monotone Convergence Theorem, this sequence of Riemann sums converges to some . For any , we can find a partition from this sequence such that the distance between its Riemann sum and is less than . Since a refinement of a tagged partition where the tags correspond to infimum of on subintervals cannot decrease the Riemann sum, satisfies the definition of Riemann integral. We emphasize in this proof that the subintervals in the definition of Riemann sum are closed and are not disjoint at endpoints, which allows for two tags (of different subintervals) to have the same value.
Corollary 1 Any function continuous and bounded on an open or half-open interval in is Riemann integrable. Moreover, if we extend such a function to the closure of that interval via the limit and Riemann integrate this on the closure, the result is the same. Formally, . Because of this, we can disregard whether or not the endpoints are included in the interval itself and simply use to denote this Riemann integral.
Proof: We note that and are both closed intervals on which the Riemann integral exists and is zero. The Riemann integral exists on by Proposition 2. Applying (2) of Proposition 1 tells us that the Riemann integral exists and is equal to on each of .
Definition 2 Suppose that . Then, for any Riemann integrable on we define
Proposition 3 Any function piecewise continuous on any interval is Riemann integrable on that interval.
Proof: Corollary 1 tells that we can WLOG only consider closed intervals. Let the interval be , with , where the s are disjoint intervals on each of which is continuous. Proposition 2 tells us that is Riemann integrable on each of the s. Thus, we can apply (1) of Proposition 1 to derive that is Riemann integrable on .
One can also use the equivalence of (2) and (3) in Theorem 1 of , of which this is a very special case.
Proposition 4 On some interval , define Riemann integrable such that for all . Then,
Proof: Trivial and left to the reader.
Theorem 1 (Mean value theorem for definite integrals) Let be a continuous function. Then, there exists such that Proof: The Extreme Value Theorem gives us infimum and supremum values of on , which we denote with and . This gives us
By the Intermediate Value Theorem, for every point in is such that there exists such that . Applying this to the value of yields the desired result.
Theorem 2 (Fundamental theorem of calculus, part I) Let be a continuous real-valued function defined on closed interval . Let be the function defined, for all , by
Then is uniformly continuous on and differentiable on the open interval , and
for all .
Proof: Take arbitrary . We can define for all such that
Applying Mean Value Theorem for Definite Integrals (Theorem 1) to this gives some such that
As , . Because is continuous on , we thus have
By the Extreme Value Theorem, there exists such that for all . Thus, by Proposition 4, for any , take and we have for all . This shows uniform continuity.
Definition 3 We say that is an antiderivative of on if on , .
Lemma 1 Any continuous antiderivative of the zero function on is a constant function on .
Proof: Suppose by contradiction that some that is a continuous anti-derivative of the zero function on is not constant. Then there exists such that . The mean value theorem for differentiable functions tells us that there exists some such that , which means that would then not be a continuous function on with derivative equal to zero on .
Lemma 2 If a function defined on has antiderivatives on , then is a constant function on , where the values of at are simply defined via continuous extension.
Proof: The Fundamental Theorem of Calculus, Part I (Theorem 1) tells us that
is an antiderivative of on that is continuous on . Because the relation defined via constant difference is transitive, it suffices to simply show that is constant on .
That on tells us that is an antiderivative of the zero function on , from which we deduce via Lemma 1 that is a constant function.
Corollary 2 If is a real valued continuous function on and is a continuous antiderivative of on , then
Proof: We define on . Fundamental Theorem of Calculus Part I tells us that it is a continuous antiderivative of on . We have that
By Lemma 2, is a constant function on , we must have that , which completes our proof.
Theorem 2 (Fundamental theorem of calculus part II (Newton-Leibniz axiom)) If is a real valued function on and Riemann integrable on and is an antiderivative of on then
Proof: Take an arbitrary partition . We have that
We can apply the mean value theorem for differentiability on the subintervals to assign tags such that which yields the Riemann sum
By definition of Riemann integrable as given in (4) of Theorem 1 of , for all , there exists such that for any tagged partition of denoted by and for the tags such that its norm or mesh , we have
Another way to put it, per Lemma 2, is that for all , there exists such that any tagged partitions with meshes both less than ,
Thus, we can, setting , take the limit of to derive
which completes our proof.