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Moreover, a function f defined on a bounded interval is Riemann-integrable if and only if it is bounded and the set of points where f is discontinuous has Lebesgue measure zero. An integral which is in fact a direct generalization of the Riemann integral is the Henstock–Kurzweil integral .
The Riemann–Stieltjes integral appears in the original formulation of F. Riesz's theorem which represents the dual space of the Banach space C[a,b] of continuous functions in an interval [a,b] as Riemann–Stieltjes integrals against functions of bounded variation. Later, that theorem was reformulated in terms of measures.
of a Riemann integrable function defined on a closed and bounded interval are the real numbers and , in which is called the lower limit and the upper limit. The region that is bounded can be seen as the area inside a {\displaystyle a} and b {\displaystyle b} .
The Lebesgue criterion for integrability states that a bounded function is Riemann integrable if and only if the set of all discontinuities has measure zero. [5] Every countable subset of the real numbers - such as the rational numbers - has measure zero, so the above discussion shows that Thomae's function is Riemann integrable on any interval.
Although all bounded piecewise continuous functions are Riemann-integrable on a bounded interval, subsequently more general functions were considered—particularly in the context of Fourier analysis—to which Riemann's definition does not apply, and Lebesgue formulated a different definition of integral, founded in measure theory (a subfield ...
A bounded function, , is Riemann integrable on [,] if and only if the correspondent set of all essential discontinuities of first kind of has Lebesgue's measure zero. The case where E 1 = ∅ {\displaystyle E_{1}=\varnothing } correspond to the following well-known classical complementary situations of Riemann integrability of a bounded ...
A version holds for Fourier series as well: if is an integrable function on a bounded interval, then the Fourier coefficients ^ of tend to 0 as . This follows by extending f {\displaystyle f} by zero outside the interval, and then applying the version of the Riemann–Lebesgue lemma on the entire real line.
The narrow definition of the Riemann integral also does not cover the function / on the interval [0, 1]. The problem here is that the integrand is unbounded in the domain of integration. In other words, the definition of the Riemann integral requires that both the domain of integration and the integrand be bounded. However, the improper ...