Riemann integral
This area can be described as the set of all points (x, y) on the graph that follow these rules: a ≤ x ≤ b (the x-coordinate is between a and b) and 0 < y < f(x) (the y-coordinate is between 0 and the height of the curve f(x)).
This notation means “the integral of f(x) from a to b,” and it represents the exact area under the curve f(x) and above the x-axis, between x = a and x = b.
In the end, when the rectangles are infinitely small, the sum gives the exact area, which is what the integral represents.
Each term in the sum is the product of the value of the function at a given point and the length of an interval.
These are similar to Riemann sums, but the tags are replaced by the infimum and supremum (respectively) of f on each sub-interval:
Our new definition says that the Riemann integral of f exists and equals s if the following condition holds: For all ε > 0, there exists a tagged partition y0, ..., ym and r0, ..., rm − 1 such that for any tagged partition x0, ..., xn and t0, ..., tn − 1 which is a refinement of y0, ..., ym and r0, ..., rm − 1, we have
To show that the first definition implies the second, start with an ε, and choose a δ that satisfies the condition.
Its Riemann sum is within ε of s, and any refinement of this partition will also have mesh less than δ, so the Riemann sum of the refinement will also be within ε of s. To show that the second definition implies the first, it is easiest to use the Darboux integral.
Choose a tagged partition x0, ..., xn and t0, ..., tn − 1 with mesh smaller than δ.
(We may assume that all the inequalities are strict because otherwise we are in the previous case by our assumption on the length of δ.)
To prove this, we will show how to construct tagged partitions whose Riemann sums get arbitrarily close to both zero and one.
Then, by carefully choosing the new tags, we can make the value of the Riemann sum turn out to be within ε of either zero or one.
The first way is to always choose a rational point, so that the Riemann sum is as large as possible.
The second way is to always choose an irrational point, so that the Riemann sum is as small as possible.
Since we started from an arbitrary partition and ended up as close as we wanted to either zero or one, it is false to say that we are eventually trapped near some number s, so this function is not Riemann integrable.
Alone this restriction does not impose a problem: we can refine any partition in a way that makes it a left-hand or right-hand sum by subdividing it at each ti.
Again, alone this restriction does not impose a problem, but the reasoning required to see this fact is more difficult than in the case of left-hand and right-hand Riemann sums.
However, combining these restrictions, so that one uses only left-hand or right-hand Riemann sums on regularly divided intervals, is dangerous.
The Riemann integral is a linear transformation; that is, if f and g are Riemann-integrable on [a, b] and α and β are constants, then
This subcover is a finite collection of open intervals, which are subintervals of J(ε1)i (except for those that include an edge point, for which we only take their intersection with J(ε1)i).
Another sufficient criterion to Riemann integrability over [a, b], but which does not involve the concept of measure, is the existence of a right-hand (or left-hand) limit at every point in [a, b) (or (a, b]).
If a real-valued function is monotone on the interval [a, b] it is Riemann integrable, since its set of discontinuities is at most countable, and therefore of Lebesgue measure zero.
The Lebesgue–Vitali theorem does not imply that all type of discontinuities have the same weight on the obstruction that a real-valued bounded function be Riemann integrable on [a, b].
[11] It is easy to extend the Riemann integral to functions with values in the Euclidean vector space
This definition carries with it some subtleties, such as the fact that it is not always equivalent to compute the Cauchy principal value
Even standardizing a way for the interval to approach the real line does not work because it leads to disturbingly counterintuitive results.
then the integral of the translation f(x − 1) is −2, so this definition is not invariant under shifts, a highly undesirable property.
The most severe problem is that there are no widely applicable theorems for commuting improper Riemann integrals with limits of functions.
For proper Riemann integrals, a standard theorem states that if fn is a sequence of functions that converge uniformly to f on a compact set [a, b], then
Another way of generalizing the Riemann integral is to replace the factors xk + 1 − xk in the definition of a Riemann sum by something else; roughly speaking, this gives the interval of integration a different notion of length.
