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An indicator function of a bounded set is Riemann-integrable if and only if the set is Jordan measurable. The Riemann integral can be interpreted measure-theoretically as the integral with respect to the Jordan measure.
The region that is bounded can be seen as the area inside and . For example, the function f ( x ) = x 3 {\displaystyle f(x)=x^{3}} is defined on the interval [ 2 , 4 ] {\displaystyle [2,4]} ∫ 2 4 x 3 d x {\displaystyle \int _{2}^{4}x^{3}\,dx} with the limits of integration being 2 {\displaystyle 2} and 4 {\displaystyle 4} .
The Riemann–Stieltjes integral admits integration by parts in the form () = () () ()and the existence of either integral implies the existence of the other. [2]On the other hand, a classical result [3] shows that the integral is well-defined if f is α-Hölder continuous and g is β-Hölder continuous with α + β > 1 .
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.
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.
If the function space of locally integrable functions, i.e. functions belonging to (), is considered in the preceding definitions 1.2, 2.1 and 2.2 instead of the one of globally integrable functions, then the function space defined is that of functions of locally bounded variation.
Sometimes integrals may have two singularities where they are improper. Consider, for example, the function 1/((x + 1) √ x) integrated from 0 to ∞ (shown right). At the lower bound of the integration domain, as x goes to 0 the function goes to ∞, and the upper bound is itself ∞, though the function goes to 0. Thus this is a doubly ...
The Dirichlet function is not Riemann-integrable on any segment of despite being bounded because the set of its discontinuity points is not negligible (for the Lebesgue measure). The Dirichlet function provides a counterexample showing that the monotone convergence theorem is not true in the context of the Riemann integral.