Lebesgue integration

From formulasearchengine
Jump to navigation Jump to search

{{#invoke: Sidebar | collapsible }}


The Lebesgue integral plays an important role in the branch of mathematics called real analysis and in many other fields in the mathematical sciences, and is named after Henri Lebesgue (1875–1941) who introduced the integral in Template:Harv. It is also a pivotal portion of the axiomatic theory of probability.





However, Riemann integration does not interact well with taking limits of sequences of functions, making such limiting processes difficult to analyze. This is of prime importance, for instance, in the study of Fourier series, Fourier transforms and other topics. The Lebesgue integral is better able to describe how and when it is possible to take limits under the integral sign (via the powerful monotone convergence theorem and dominated convergence theorem). The Lebesgue definition considers a different class of easily calculated areas than the Riemann definition, which is the main reason the Lebesgue integral is better behaved. The Lebesgue definition also makes it possible to calculate integrals for a broader class of functions. For example, the Dirichlet function, which is 0 where its argument is irrational and 1 otherwise, has a Lebesgue integral, but it does not have a Riemann integral.



Source: Template:Harv


Intuitive interpretation



The Riemann-Darboux approach
Divide the base of the mountain into a grid of 1 meter squares. Measure the altitude of the mountain at the center of each square. The volume on a single grid square is approximately 1 m2 × (that square's altitude), so the total volume is 1 m2 times the sum of the altitudes.
The Lebesgue approach
Draw a contour map of the mountain, where adjacent contours are 1 meter of altitude apart. The volume of earth contained in a single contour is approximately 1 m × (that contour's area), so the total volume is the sum of these areas times 1 m.

Folland[1] summarizes the difference between the Riemann and Lebesgue approaches thus: "to compute the Riemann integral of f, one partitions the domain [a, b] into subintervals", while in the Lebesgue integral, "one is in effect partitioning the range of f ".

Towards a formal definition

To define the Lebesgue integral formally requires the notion of a measure which, roughly, associates to each set A of real numbers a nonnegative number μ(A) representing the "size" of A. This notion of "size" should agree with the usual length of an interval or disjoint union of intervals. Suppose that f : RR+ is a non-negative real-valued function. Using the "partitioning the range of f " philosophy, the integral of f should be the sum over t of the area of the thin horizontal strip between y = t and y = t + dt. This area is just


The Lebesgue integral of f is then defined by[2]

where the integral on the right is an ordinary improper Riemann integral (note that f* is a non-negative decreasing function, and therefore has a well-defined improper Riemann integral). For a suitable class of functions (the measurable functions) this defines the Lebesgue integral.





Approximating a function by simple functions.




Measure theory


The Riemann integral uses the notion of length explicitly. Indeed, the element of calculation for the Riemann integral is the rectangle [a, b] × [c, d], whose area is calculated to be (ba)(dc). The quantity ba is the length of the base of the rectangle and dc is the height of the rectangle. Riemann could only use planar rectangles to approximate the area under the curve because there was no adequate theory for measuring more general sets.



We start with a measure space (E, X, μ) where E is a set, X is a σ-algebra of subsets of E, and μ is a (non-negative) measure on E defined on the sets of X.


In Lebesgue's theory, integrals are defined for a class of functions called measurable functions. A real-valued function f on E is measurable if the pre-image of every interval of the form (t, ∞) is in X:

It can be shown that this is equivalent to requiring that the pre-image of any Borel subset of R be in X. We will make this assumption henceforth. The set of measurable functions is closed under algebraic operations, but more importantly it is closed under various kinds of pointwise sequential limits:

are measurable if the original sequence (fk)k, where kN, consists of measurable functions.

We build up an integral

for measurable real-valued functions f defined on E in stages:

Indicator functions: To assign a value to the integral of the indicator function 1S of a measurable set S consistent with the given measure μ, the only reasonable choice is to set:


Simple functions: A finite linear combination of indicator functions



Some care is needed when defining the integral of a real-valued simple function, to avoid the undefined expression ∞ − ∞: one assumes that the representation

is such that μ(Sk) < ∞ whenever ak ≠ 0. Then the above formula for the integral of f makes sense, and the result does not depend upon the particular representation of f satisfying the assumptions.


Non-negative functions: Let f be a non-negative measurable function on E which we allow to attain the value +∞, in other words, f takes non-negative values in the extended real number line. We define


We have defined the integral of f for any non-negative extended real-valued measurable function on E. For some functions, this integral  ∫E f dμ  will be infinite.

Signed functions: To handle signed functions, we need a few more definitions. If f is a measurable function of the set E to the reals (including ±∞), then we can write


Note that both f+ and f are non-negative measurable functions. Also note that

We say that the Lebesgue integral of the measurable function f exists, or is defined if at least one of and is finite:

In this case we define


we say that f is Lebesgue integrable.



If h=f+ig for real-valued integrable functions f, g, then the integral of h is defined by



  • is not Riemann-integrable on [0, 1]: No matter how the set [0, 1] is partitioned into subintervals, each partition will contain at least one rational and at least one irrational number, because rationals and irrationals are both dense in the reals. Thus the upper Darboux sums will all be one, and the lower Darboux sums will all be zero.
because Q is countable.

Domain of integration




Limitations of the Riemann integral




Failure of monotone convergence. As shown above, the indicator function 1Q on the rationals is not Riemann integrable. In particular, the Monotone convergence theorem fails. To see why, let {ak} be an enumeration of all the rational numbers in [0, 1] (they are countable so this can be done.) Then let

The function gk is zero everywhere except on a finite set of points, hence its Riemann integral is zero. Each gk is non-negative, and this sequence of functions is monotonically increasing, but its limit as k → ∞ is 1Q, which is not Riemann integrable.


Integrating on structures other than Euclidean space. The Riemann integral is inextricably linked to the order structure of the real line.

Basic theorems of the Lebesgue integral




  • If f, g are functions such that f = g almost everywhere, then f is Lebesgue integrable if and only if g is Lebesgue integrable, and the integrals of f and g are the same if they exist.

The Lebesgue integral has the following properties:


Monotonicity: If fg, then

Monotone convergence theorem: Suppose { fk}kN is a sequence of non-negative measurable functions such that

Then, the pointwise limit f of fk is Lebesgue integrable and


Fatou's lemma: If { fk}kN is a sequence of non-negative measurable functions, then


Dominated convergence theorem: Suppose { fk}kN is a sequence of complex measurable functions with pointwise limit f, and there is a Lebesgue integrable function g (i.e., g belongs to the space L1) such that | fk | ≤ g for all k.

Then, f is Lebesgue integrable and

Proof techniques

To illustrate some of the proof techniques used in Lebesgue integration theory, we sketch a proof of the above mentioned Lebesgue monotone convergence theorem. Let { fk}kN be a non-decreasing sequence of non-negative measurable functions and put


and the limit on the right exists, because the sequence is monotonic. We now prove the inequality in the other direction. It follows from the definition of integral that there is a non-decreasing sequence (gn) of non-negative simple functions such that gnf and

Therefore, it suffices to prove that for each nN,

We will show that if g is a simple function and

almost everywhere, then


Suppose A is a measurable set and { fk}kN is a nondecreasing sequence of non-negative measurable functions on E such that

for almost all xA. Then


By monotonicity of the integral, it follows that for any kN,


up to a set of measure 0. Thus by countable additivity of μ, and because Bk increases with k,

As this is true for any positive ε the result follows.

For another Proof of the Monotone Convergence Theorem, we follow:[3]

Let (X, M, μ) be a measure space.

{ fn} is an increasing sequence of numbers, therefore its limit exists, even if it's equal to . We know that

for all n, so that


Now we need to establish the reverse inequality. Fix α ∈ (0, 1), let φ be a simple function with 0 ≤ φf and let


Then {En} is an increasing sequence of measurable sets with . We know that


This is true for all n, including the limit:




This was true for all α ∈ (0, 1), so it remains true for α = 1, and taking the supremum over simple φf by the definition of integration in L+,


Now we have both inequalities, so we've shown the Monotone Convergence theorem:

for f{n+1}fn, and fnf pointwise, {fn} ∈ L+, the set of positive measurable functions from X → [0, ∞].

Alternative formulations




Then Cc is a normed vector space (and in particular, it is a metric space.) All metric spaces have Hausdorff completions, so let L1 be its completion. This space is isomorphic to the space of Lebesgue integrable functions modulo the subspace of functions with integral zero. Furthermore, the Riemann integral is a uniformly continuous functional with respect to the norm on Cc, which is dense in L1. Hence has a unique extension to all of L1. This integral is precisely the Lebesgue integral.


Limitations of Lebesgue integral

The main purpose of Lebesgue integral is to provide an integral notion where limits of integrals hold under mild assumptions. There is no guarantee that every function is Lebesgue integrable. But it may happen that improper integrals exist for functions that are not Lebesgue integrable. One example would be

over the entire real line. This function is not Lebesgue integrable, as

On the other hand, exists as an improper integral and can be computed to be finite.

See also


  1. Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 1984, p. 56.
  2. Template:Harvnb
  3. Folland


  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} Very thorough treatment, particularly for probabilists with good notes and historical references.

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} A classic, though somewhat dated presentation.

  • {{#invoke:citation/CS1|citation

|CitationClass=citation }}

  • {{#invoke:Citation/CS1|citation

|CitationClass=journal }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} Includes a presentation of the Daniell integral.

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} Good treatment of the theory of outer measures.

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} Known as Little Rudin, contains the basics of the Lebesgue theory, but does not treat material such as Fubini's theorem.

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} Known as Big Rudin. A complete and careful presentation of the theory. Good presentation of the Riesz extension theorems. However, there is a minor flaw (in the first edition) in the proof of one of the extension theorems, the discovery of which constitutes exercise 21 of Chapter 2.

  • {{#invoke:Citation/CS1|citation

|CitationClass=journal }}. English translation by Laurence Chisholm Young, with two additional notes by Stefan Banach.

  • {{#invoke:citation/CS1|citation

|CitationClass=book }} Emphasizes the Daniell integral.

  • {{#invoke:citation/CS1|citation

|CitationClass=citation }}.

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

  • {{#invoke:citation/CS1|citation

|CitationClass=book }}

{{#invoke: Navbox | navbox }}