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The Lebesgue integral, named after French mathematician Henri Lebesgue, is one way to make this concept rigorous and to extend it to more general functions. The Lebesgue integral is more general than the Riemann integral, which it largely replaced in mathematical analysis since the first half of the 20th century. It can accommodate functions ...
4 Measure theory and the Lebesgue integral. 5 Extensions. 6 Integral equations. 7 Integral transforms. 8 Integral geometry. 9 Other. 10 See also. Toggle the table of ...
An alternative approach (Hewitt & Stromberg 1965) is to define the Lebesgue–Stieltjes integral as the Daniell integral that extends the usual Riemann–Stieltjes integral. Let g be a non-decreasing right-continuous function on [ a , b ] , and define I ( f ) to be the Riemann–Stieltjes integral
Lebesgue measure is both locally finite and inner regular, and so it is a Radon measure. Lebesgue measure is strictly positive on non-empty open sets, and so its support is the whole of R n. If A is a Lebesgue-measurable set with λ(A) = 0 (a null set), then every subset of A is also a null set. A fortiori, every subset of A is measurable.
In Lebesgue integration, this is exactly the requirement for any measurable function f to be considered integrable, with the integral then equaling + (), so that in fact "absolutely integrable" means the same thing as "Lebesgue integrable" for measurable functions.
Lebesgue's dominated convergence theorem is a special case of the Fatou–Lebesgue theorem. Below, however, is a direct proof that uses Fatou’s lemma as the essential tool. Since f is the pointwise limit of the sequence (f n) of measurable functions that are dominated by g, it is also measurable and dominated by g, hence it is integrable ...
However, the Riemann–Lebesgue lemma does not hold for arbitrary distributions. For example, the Dirac delta function distribution formally has a finite integral over the real line, but its Fourier transform is a constant and does not vanish at infinity.
is not locally integrable in x = 0: it is indeed locally integrable near this point since its integral over every compact set not including it is finite. Formally speaking, 1 / x ∈ L 1 , l o c ( R ∖ 0 ) {\displaystyle 1/x\in L_{1,loc}(\mathbb {R} \setminus 0)} : [ 19 ] however, this function can be extended to a distribution on the whole R ...