7.4. Lebesgue Integral | IRA |
If the limit exists it is called the Riemann integral and the function is called Riemann integrable. Now we will take, in a manner of speaking, the "opposite" approach:
- subdivide the domain of the function (usually a closed, bounded interval) into finitely many subintervals (the partition)
- construct a simple function that has a constant value on each of the subintervals of the partition (the Upper and Lower sums)
- take the limit of these simple functions as you add more and more points to the partition.
If the limit exists it is called the Lebesgue integral and the function is called Lebesgue integrable. To define this new concept we use several steps:
- subdivide the range of the function into finitely many pieces
- construct a simple function by taking a function whose values are those finitely many numbers
- take the limit of these simple functions as you add more and more points in the range of the original function
First, we need to clarify what we mean by "simple function".
- we define the Lebesgue Integral for "simple functions"
- we define the Lebesgue integral for bounded functions over sets of finite measure
- we extend the Lebesgue integral to positive functions (that are not necessarily bounded)
- we define the general Lebesgue integral
A function f defined on a measurable set A that takes no more than finitely many distinct values a1, a2, ... , an can always be written as a simple function
Definition 7.4.1: Characteristic and Simple Function For any set A the function is called the characteristic function of A. A finite linear combination of characteristic functions
s(x) =is called simple function if all sets Ei are measurable.ai XEi(x)
f(x) =wherean XAn(x)
An = { xTherefore simple functions can be thought of as dividing the range of f, where the resulting sets An may or may not be intervals.A: f(x) = an }
For simple functions we define the Lebesgue integral as follows:
Examples 7.4.2: Simple Functions
A step function is a function s(x) such that
s(x) = cj for xj-1 < x < xj
and the { xj } form a partition of [a, b]. Upper, Lower, and Riemann sums are examples of step functions. What is the difference, if any, between step functions and simple functions.Are simple functions uniquely determined? In other words, if s1 and s2 are two simple functions with s1(x) = s2(x), do they have to have the same representation? If different representations are possible, which one is "the best"?
How does the Dirichlet function fit in with this terminology?
Is the function that is equal to 1 if x is part of the Cantor middle-third set and 0 otherwise a simple function?
Are sums, differences, and products of simple functions simple?
Definition 7.4.3: Lebesgue Integral for Simple Function If s(x) = an XAn(x) is a simple function and m(An) is finite for all n, then the Lebesgue Integral of s is defined as
If E is a measurable set, we defines(x) dx =
an m(An)
E s(x) dx =
XE(x) s(x) dx
Just as step functions were used to define the Riemann integral of a bounded function f over an interval [a, b], simple functions are used to define the Lebesgue integral of f over a set of finite measure.
Example 7.4.4: Lebesgue Integral for Simple Functions
Find the Lebesgue integral of the constant function f(x) = c over the interval [a, b].
Find the Lebesgue integral of a step function, i.e. a function s such that s(x) = cj for xj-1 < x < xj and the { xj } form a partition of [a, b].
Find the Lebesgue integral of the Dirichlet function restricted to [0, 1] and of the characteristic function of the Cantor middle-third set.
Define two simple functions
s1(x) = 2 X[0, 2](x) + 4 X[1, 3](x)Show that s1(x) = s2(x) and
s2(x) = 2 X[0, 1)(x) + 6 X[1, 2](x) + 4 X(2, 3](x)s1(x) dx =
s2(x) dx.
We have seen before that the representation of a simple function is not unique. Show that the Lebesgue integral of a simple function is independent of its representation.
Definition 7.4.5: Lebesgue Integral for Bounded Function Suppose f is a bounded function defined on a measurable set E with finite measure. Define the upper and lower Lebesgue integrals, respectively, as I*(f)L = inf{If I*(f)L = I*(f)L the function f is called Lebesgue integrable over E and the Lebesgue integral of f over E is denoted bys(x) dx: s is simple and s
f }
I*(f)L = sup{s(x) dx: s is simple and s
f }
E f(x) dx
Now a function f can be integrated (if it is integrable) using either the Riemann or the Lebesgue integral. Fortunately, for many simple functions the two integrals agree and the Lebesgue integral is indeed a generalization of the Riemann integral.
Examples 7.4.6: Lebesgue Integral for Bounded Functions
Is the function f(x) = x Lebesgue integrable over [0, 1]? If so, find the integral.
Is the function f(x) = x2 Lebesgue integrable over the rational numbers inside [0, 2]? If so, find the integral.
Is the Dirichlet function restricted to [0, 1] Lebesgue integrable? If so, find the integral.
Is every bounded function Lebesgue integrable?
For most practial applications this theorem is all that is necessary: for continuous functions or bounded functions with at most countably many discontinuities over intervals [a, b] there is no need to distinguish between the Lebesgue or Riemann integral. All integration techniques we learned apply equally well, using either integral. But for more complicated situations or more theoretical purposes the Lebesgue integral is more useful, but then techniques such as integration by parts or substitution may no longer apply.
Theorem 7.4.7: Riemann implies Lebesgue Integrable If f is a bounded function defined on [a, b] such that f is Riemann integrable, then f is Lebesgue integrable and f(x) dx =
[a,b] f(x) dx
The Lebesgue integral has properties similar to those of the Riemann integral, but it is "more forgiving": you can change a function on a set of measure zero without changing the integral at all.
Example 7.4.8: Riemann implies Lebesgue Integrable
Find the Lebesgue integral of f(x) = x cos(x) over the interval [-1, 1].
Show that the converse of the above theorem is false, i.e. not every bounded Lebesgue integrable function is Riemann integrable.
If possible, find the Riemann and Lebesgue integrals of the constant function f(x) = 1 over the Cantor middle-third set.
Show that the restriction of a bounded continuous function to a measurable set is Lebesgue integrable.
Proposition 7.4.9: Properties of the Lebesgue Integral Suppose f and g are two bounded, Lebesgue integrable functions defined on a measurable set E with finite measure. Then:
E c f(x) + d g(x) dx = c
E f(x) dx + d
E g(x) dx
- If A and B are disjoint measurable subsets of E then
A
B f(x) dx =
A f(x) dx +
B f(x) dx
- If f(x) = g(x) for all x in E except possibly on a set of measure zero then
E f(x) dx =
E g(x) dx
- If f(x)
g(x) for all x in E except possibly on a set of measure zero then
E f(x) dx
![]()
E g(x) dx
At this point we could stop: we have extended the concept of integration to (bounded) functions defined on general sets (measurable sets with finite measure) without using partitions (subintervals). The new concept, the Lebesgue integral, agrees with the old one, Riemann integral, when both apply and removes some of the oddities mentioned before.
Examples 7.4.10: Properties of the Lebesgue Integral
Is the function f(x) = x ex Lebesgue integrable over the Cantor middle-third set? If so, find the integral.
If f is Lebesgue integrable over E and A
f(x)
B then show that A m(E)
![]()
E f(x) dx
B m(E)
Suppose f is a bounded, non-negative function defined on a measurable set E with finite measure and F
E is measurable with m(F)
m(E). Then show that
F f(x) dx
![]()
E f(x) dx
Suppose f is a bounded, non-negative function defined on a measurable set E with finite measure such that
E f(x) dx = 0. Show that f must then be equal to zero except on a set of measure zero.
But we can continue to extend the Lebesgue integral to functions that are unbounded, including functions that may occasionally be equal to infinity. To do that, we first need to define the concept of a measurable function.
Examples 7.4.11: Lebesgue is more general than Riemann
What happens when you change the value of a Lebesgue integrable function at a single point?
Is it true that a function that is constant except at countably many points is Lebesgue integrable?
What is the difference between Lebesgue integrable functions and bounded continuous functions?
Can you take a Lebesgue integral over anything else but an interval?
Could you define a Lebesgue integral of a function whose domain is not R?
In other words, functions whose values are real numbers or possibly plus or minus infinity are measurable if the inverse image of every interval (-
Definition 7.4.12: Measurable Function Let f be a function from E R into R
{ -
,
} . The function f is called (Lebesgue) measurable if
- the domain E of the function is a measurable set
- for every real number a the set f -1 (-
, a) is a measurable set
, a) is measurable.
That is somewhat comparable to one of the equivalent definitions of continuous functions: a function f is continuous if the inverse image of every open interval is open. However, not every measurable function is continuous, while every continuous function is clearly measurable.
Measurable functions that are bounded are equivalent to Lebesgue integrable functions.
Example 7.4.13: Measurable Functions
Show that every continuous function is measurable
Show that not every measurable function is continuous
If f is measurable and f = g except on a set of measure zero, show that g is also measurable.
Show that a function f is measurable if and only if
- the set { x: f(x) < a } is measurable for all a
- the set { x: f(x)
a } is measurable for all a
- the set { x: f(x) > a } is measurable for all a
- the set { x: f(x)
a } is measurable for all a
Show that if f is measurable, then the set { x: f(x) = a } is measurable for all a
Is it true that if E is open and f is a measurable function, then f -1(E) is also measurable?
Is it true that if E is measurable and f is a measurable function, then f -1(E) is also measurable?
Measurable functions do not have to be continuous, they may be unbounded and they can, in particular, be equal to plus or minus infinity. On the other hand, measurable functions are "almost" continous.
Proposition 7.4.14: Bounded Measurable Functions are Integrable If f is a bounded function defined on a measurable set E with finite measure. Then f is measurable if and only if f is Lebesgue integrable.
Using measurable functions allows us to extend the Lebesgue integral first to non-negative functions that are not necessarily bounded and then to general measurable functions.
Proposition 7.4.15: Measurable Functions are Almost Continuous Suppose f is a measurable function defined on an interval [a, b] such that the set where f is plus or minus infinity has measure zero. Then, for any > 0 we can find a step function g and a continuous function h such that
| f(x) - g(x) | <except on a set of measure less than
| f(x) - h(x) | <![]()
.
Definition 7.4.16: Lebesgue Integral of Non-Negative Functions If f is a measurable function defined on E and h is a bounded measurable function such that m( {x: h(x) # 0} ) is finite, then we define IfE f(x) dx = sup{
E h(x) dx, h
f }
E f(x) dx is finite, then f is called Lebesgue integrable over E.
The final step to define the Lebesgue integral of a general function is now easy.
Examples 7.4.17: Lebesgue Integral of Non-Negative Function
Can a function that is equal to infinity at one or more points be Riemann integrable?
Can a function that is equal to infinity at one or more points be Lebesgue integrable?
What can you say about the set where a function f is equal to infinity, if f is Lebesgue integrable?
Suppose f is a non-negative integrable function such that
E f(x) dx = 0. Show that then f = 0 except on a set of measure zero.
Proposition 7.4.X remains true for general Lebegues integrable functions.
Definition 7.4.18: The General Lebesgue Integral Let f be a measurable function and define the positive and negative parts of f, respectively, as: f +(x) = max(f(x), 0)so that f = f + - f -. Then f is Lebesgue integral if f + and f - are Lebesgue integrable and
f -(x) = max(-f(x), 0)E f(x) dx =
E f +(x) dx -
E f -(x) dx =
Examples 7.4.19: The General Lebesgue Integral Suppose that f is an integrable function over a set E, and take any > 0. Show that