# Lebesgue integration

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File:Integral-area-under-curve.png
The integral can be interpreted as the area under a curve.

In mathematics, the integral of a function can be regarded in the simplest case as the area between the graph of that function and the x-axis. Lebesgue integration is a mathematical theory that extends the integral to a larger class of functions; it also extends the domains on which these functions can be defined. It had long been understood that for functions with a smooth enough graph (such as continuous functions on closed bounded intervals) the area under the curve could be defined as the integral and computed using techniques of approximation of the region by polygons. However, as the need to consider more irregular functions arose (for example, as a result of the limiting processes of mathematical analysis and the mathematical theory of probability) it became clear that more careful approximation techniques would be needed in order to define a suitable integral.

The Lebesgue integral plays an important role in the branch of mathematics called real analysis and in many other fields in the mathematical sciences.

The Lebesgue integral is named for Henri Lebesgue (1875-1941). The pronunciation of his name may be approximated as leh BEG.

## Introduction

The integral of a function f between limits a and b can be interpreted as the area under the graph of f. This is easy to understand for familiar functions such as polynomials, but what does it mean for more exotic functions? In general, what is the class of functions for which "area under the curve" makes sense? The answer to this question has great theoretical and practical importance.

As part of a general movement toward rigour in mathematics in the nineteenth century, attempts were made to put the integral calculus on a firm foundation. The Riemann integral, proposed by Bernhard Riemann (1826-1866), is a broadly successful attempt to provide such a foundation for the integral. Riemann's definition starts with the construction of a sequence of easily-calculated integrals which converge to the integral of a given function. This definition is successful in the sense that it gives the expected answer for many already-solved problems, and gives useful results for many other problems.

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. The Lebesgue definition considers a different class of easily-calculated integrals 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.

In the next section, we discuss the technical definition of the Lebesgue integral. Readers may skip that section and continue on to the Limitations of the Riemann integral section that follows it.

## Construction of the Lebesgue integral

The discussion that follows parallels the most common expository approach to the Lebesgue integral. In this approach the theory of integration has two distinct parts:

1. A theory of measurable sets and measures on these sets.
2. A theory of measurable functions and integrals on these functions.

### Measure theory

Measure theory initially was created to provide a detailed analysis of the notion of length of subsets of the real line and more generally area and volume of subsets of Euclidean spaces. In particular, it provided a systematic answer to the question of which subsets of R have a length. As was shown by later developments in set theory (see non-measurable set), it is actually impossible to assign a length to all subsets of R in a way which preserves some natural additivity and translation invariance properties. This suggests that picking out a suitable class of measurable subsets is an essential prerequisite.

Of course, the Riemann integral uses the notion of length implicitly. Indeed, the element of calculation for the Riemann integral is the rectangle [a, b] × [c, d], whose area is calculated to be (b-a)(d-c). The quantity b-a is the length of the base of the rectangle and d-c 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.

In the development of the theory in most modern textbooks (after 1950), the approach to measure and integration is axiomatic. This means that a measure is any function μ defined on certain subsets X of a set E which satisfies a certain list of properties. These properties can be shown to hold in many different cases.

The theory of measurable sets and measure (including definition and construction of such measures) is discussed in other articles. See measure.

### Integration

We will work in the following abstract setup: μ is a (non-negative) measure on a sigma-algebra X of subsets of E. For example, E can be Euclidean n-space Rn or some Lebesgue measurable subset of it, X will be the sigma-algebra of all Lebesgue measurable subsets of E, and μ will be the Lebesgue measure. In the mathematical theory of probability μ will be a probability measure on a probability space E.

In Lebesgue's theory, integrals are limited to a class of functions called measurable functions. A function f is measurable if the pre-image of any closed interval is in X:

$f^{-1}([a,b])\in X$ 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 from now on. The set of measurable functions are closed under algebraic operations, but more importantly the class is closed under various kinds of pointwise sequential limits:

$\liminf _{k\in \mathbb {N} }f_{k},\quad \limsup _{k\in \mathbb {N} }f_{k}$ are measurable if the original sequence {fk}, where kN, consists of measurable functions.

We build up an integral

$\int _{E}fd\mu \quad$ for measurable complex-valued functions f defined on E in stages:

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

$\int 1_{S}d\mu =\mu (S)$ Simple functions: We extend by linearity to the linear span of indicator functions:

$\int {\bigg (}\sum _{k}a_{k}1_{S_{k}}{\bigg )}d\mu =\sum _{k}a_{k}\int 1_{S_{k}}d\mu$ where the sum is finite and the coefficients ak are real numbers. Such a finite linear combination of indicator functions is called a simple function. Even if a simple function can be written in many ways as a linear combination of indicator functions, the integral will always be the same.

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 values in the extended real number line. We define

$\int _{E}f\,d\mu :=\sup \left\{\,\int _{E}s\,d\mu :s\leq f,\ s\ {\mbox{simple}}\,\right\}$ We need to show this integral coincides with the preceding one, defined on the set of simple functions. There is also the question of whether this corresponds in any way to a Riemann notion of integration. It is not hard to prove that the answer to both questions is yes.

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

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

$f=f^{+}-f^{-},\quad$ where

$f^{+}(x)=\left\{{\begin{matrix}f(x)&{\mbox{if}}\quad f(x)\geq 0\\0&{\mbox{otherwise}}\end{matrix}}\right.$ $f^{-}(x)=\left\{{\begin{matrix}-f(x)&{\mbox{if}}\quad f(x)<0\\0&{\mbox{otherwise}}\end{matrix}}\right.$ Note that both f+ and f are non-negative functions. Also note that

$|f|=f^{+}+f^{-}.\quad$ If

$\int |f|d\mu <\infty ,$ then f is called Lebesgue integrable. In this case, both integrals satisfy

$\int f^{+}d\mu <\infty ,\quad \int f^{-}d\mu <\infty ,$ and it makes sense to define

$\int fd\mu =\int f^{+}d\mu -\int f^{-}d\mu$ It turns out that this definition gives the desirable properties of the integral.

Complex valued functions can be similarly integrated, by considering the real part and the imaginary part separately.

### Intuitive interpretation

To get some intuition about the different approaches to integration, let us imagine that it is desired to find a mountain's volume (above sea level) and that the mountain's boundaries are marked out clearly (they are the boundaries of integration).

The Riemann-Darboux approach: Cut the mountain up into vertical slices, each one having a square base at sea-level. Pick a pair of points inside this square, one where elevation is highest and one where elevation is lowest. Associated to these two elevations are an upper volume and lower volume obtained by multiplying the elevations by the area of the square. The upper Riemann sum is the sum of the upper volumes of all the slices, and similarly for the lower Riemann sum. The Riemann integral exists if the upper and lower Riemann sums converge as the thickness of the slices decreases to 0.

The Lebesgue approach: Draw a contour map of the mountain. For each contour (or set of contours) with the lowest height, find the total area enclosed (within the map) by this set of contours, i.e. find the "measure" of this set of contours. Multiply this measure by the height represented by the set of contours: the product will be one summand of a "Lebesgue sum".

Then find a contour, or set of contours, which are at one step higher in height (lowest height of the remaining contours). Calculate the measure of the area enclosed by them. Multiply the measure by the difference in height (from the previous step), and the product will be another summand of the "Lebesgue sum".

Repeat this process for successive higher levels of contours, until the highest set of contours has been processed. The resulting sum is the linear span: each contour corresponding to an indicator function.

The sum can be refined by adding intermediate contours to the map: halving the difference between successive heights, then recomputing the sum. The Lebesgue integral is the limit of this process.

### Example

Consider the indicator function of the rational numbers, 1Q. It is known 1Q is nowhere continuous.

• 1Q 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, since 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.
• 1Q is Lebesgue-integrable on [0,1]: Indeed it is the indicator function of the rationals so by definition
$\int _{[0,1]}1_{\mathbf {Q} }\,d\mu =\mu (\mathbf {Q} \cap [0,1])=0,$ since Q is countable.

## Limitations of the Riemann integral

Here we discuss the limitations of the Riemann integral and the greater scope offered by the Lebesgue integral. We presume a working understanding of the Riemann integral.

With the advent of Fourier series, many analytical problems involving integrals came up whose satisfactory solution required exchanging infinite summations of functions and integral signs. However, the conditions under which the integrals

$\sum _{k}\int f_{k}(x)dx,\quad \int {\bigg [}\sum _{k}f_{k}(x){\bigg ]}dx$ are equal proved quite elusive in the Riemann framework. There are some other technical difficulties with the Riemann integral. These are linked with the limit taking difficulty discussed above.

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

$g_{k}(x)=\left\{{\begin{matrix}1&{\mbox{if }}x=a_{k}\\0&{\mbox{otherwise}}\end{matrix}}\right.$ Then let

$f_{k}=g_{1}+g_{2}+\ldots +g_{k}.\quad$ The function fk is zero everywhere except on a finite set of points, hence its Riemann integral is zero. The sequence fk is also clearly non-negative and monotonously increasing to 1Q is not Riemann integrable.

Unsuitability for unbounded intervals. The Riemann integral can only integrate functions on a bounded interval. The simplest extension is to define

$\int _{-\infty }^{+\infty }f(x)dx=\lim _{a\rightarrow \infty }\int _{-a}^{+a}f(x)dx$ whenever the limit exists. However, this breaks the desirable property of translation invariance: if f and g are zero outside some interval [a, b] and are Riemann integrable, and if f(x) = g(x + y) for some y, then ∫ f = ∫ g. With this definition of the improper integral (this definition is sometimes called the improper Cauchy principal value about zero), the functions f(x) = (1 if x > 0, −1 otherwise) and g(x) = (1 if x > 1, −1 otherwise) are translations of one another, but their improper integrals are different.

$\int f(x)dx=0,\quad \int g(x)dx=-2.\quad$ ## Basic theorems of the Lebesgue integral

The Lebesgue integral does not distinguish between functions which only differ on a set of μ-measure zero. To make this precise, functions f, g are said to be equal almost everywhere (or equal a.e.) iff

$\mu (\{x\in E:f(x)\neq g(x)\})=0$ • If f, g are non-negative functions (possibly assuming the value +∞) such that f = g almost everywhere, then
$\int fd\mu =\int gd\mu .$ • If f, g are functions such that f = g almost everywhere, then f is integrable iff g is integrable and the integrals of f and g are the same.

The Lebesgue integral has the following properties:

Linearity: If f and g are integrable functions and a and b are real numbers, then af + bg is integrable and

$\int (af+bg)d\mu =a\int fd\mu +b\int gd\mu$ Monotonicity: If fg, then

$\int fd\mu \leq \int gd\mu .$ Monotone convergence theorem: Suppose {fk}kN is a sequence of non-negative measurable functions such that

$f_{k}(x)\leq f_{k+1}(x)\quad \forall k\in \mathbb {N} ,\forall x\in E.$ Then

$\lim _{k}\int f_{k}d\mu =\int \sup _{k}f_{k}d\mu .$ Note: The value of any one the integrals is allowed to be infinite.

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

$\int \liminf _{k}f_{k}d\mu \leq \liminf _{k}\int f_{k}d\mu .$ Again, the value of any one the integrals may be infinite.

Dominated convergence theorem: If {fk}kN is a sequence of measurable functions with pointwise limit f, and if there is an integrable function g such that |fk| ≤ g for all k, then f is integrable and

$\lim _{k}\int f_{k}d\mu =\int fd\mu .$ ## 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

$f=\sup _{k\in \mathbb {N} }f_{k}$ By the monotonicity property of the integral, it is immediate that:

$\int fd\mu \geq \lim _{k}\int f_{k}d\mu$ We now prove the inequality in the other direction, that is

$\int fd\mu \leq \lim _{k}\int f_{k}d\mu .$ It follows from the definition of integral, that there is a non-decreasing sequence gn of non-negative simple functions which converges to f pointwise almost everywhere and such that

$\lim _{k}\int g_{k}d\mu =\int fd\mu .$ Therefore, it suffices to prove that for each kN,

$\int g_{k}d\mu \leq \lim _{j}\int f_{j}d\mu .$ We will show that if g is a simple function and

$\lim _{j}f_{j}(x)\geq g(x)$ almost everywhere, then

$\lim _{j}\int f_{j}d\mu \geq \int gd\mu .$ By breaking up the function g into its constant value parts, this reduces to the case in which g is the indicator function of a set. The result we have to prove is then

Suppose A is a measurable set and {fk}kN is a nondecreasing sequence of measurable functions on E such that
$\lim _{n}f_{n}(x)\geq 1$ for almost all xA. Then
$\lim _{n}\int f_{n}d\mu \geq \mu (A).$ To prove this result, fix ε > 0 and define the sequence of measurable sets

$B_{n}=\{x\in A:f_{n}(x)\geq 1-\epsilon \}.$ By monotonicity of the integral, it follows that for any nN,

$\mu (B_{n})(1-\epsilon )=\int (1-\epsilon )1_{B_{n}}d\mu \leq \int f_{n}d\mu$ By assumption,

$\bigcup _{i}B_{i}=A,$ up to a set of measure 0. Thus by countable additivity of μ

$\mu (A)=\lim _{n}\mu (B_{n})\leq \lim _{n}(1-\epsilon )^{-1}\int f_{n}d\mu .$ As this is true for any positive ε the result follows.

## Alternative formulations

If f is non-negative, then ∫f dμ is precisely the area under the curve as measured by the product measure μ × λ where λ is the Lebesgue measure for R.

One can try to circumvent measure theory entirely. The Riemann integral exists for any continuous function f of compact support. Then we use functional analysis to obtain the integral for more general functions. Let Cc be the space of all real-valued compactly supported continuous functions of R. Define a norm on Cc by

$\|f\|=\int |f(x)|dx$ Then Cc is a normed vector space (and in particular, it is a metric space.) All metric spaces have completions, so let L1 be its completion. This space is isomorphic to the space of Lebesgue integrable functions (modulo sets of measure zero). Furthermore, the Riemann integral ∫ defines a continuous functional on Cc which is dense in L1 hence ∫ has a unique extension to all of L1. This integral is precisely the Lebesgue integral.

The problem with this approach is that integrable functions are represented as elements of an abstractly defined completion and showing how these abstractly defined elements are represented as functions is non-trivial. In particular, the relation between pointwise limits of sequence of functions and the integral is hard to prove.

Another approach is offered by the Daniell integral or the Bourbaki variant of the Daniell integral, often referred to as the Radon measure approach to integration.

## Quote

• "Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane." Richard Hamming