If is a simple function with canonical form then the Lebesgue integral of is defined as
If is a measurable subset of with finite measure, then is also a simple function, and we define
e.g. Integral of a characteristic function is the measure of the set.
Properties of the Integral of Simple Functions
The integral of simple functions satisfies the following properties:
Independence of the representation: If is any representation of , then
Linearity: If and are simple functions, and , then
Additivity: If and are disjoint subsets of with finite measure, then
Monotonicity: If are simple functions, then
Triangle inequality: If is a simple function, then so is , and
Proposition
If and are a pair of simple functions that agree almost everywhere, then
This identity of integrals for functions that agree almost everywhere will continue to hold for successive definitions of the integral.
Bounded Functions Supported on a Set of Finite Measure
Support
Suppose that is a real valued function whose domain is an arbitrary set . The support of , written , is the set of points in where is non-zero:If for all but a finite number of points in , then is said to have finite support. We shall also say that is supported on a set , if whenever .
is measurable then is a measurable set.
If
Proposition
If is a measurable function bounded by and supported on a set , then there exists a sequence of simple functions, with each bounded by and supported on , and such that
Proof Since is bounded by , is a non-negative function. The theorem states that there exists an increasing sequence of non-negative simple functions converging pointwise to . Define . Then is a sequence of simple functions bounded by and converging pointwise to .
The following lemma allows us to define the integral for bounded functions supported on sets of finite measure:
Lemma
Let be a bounded function supported on a set of finite measure. If is any sequence of simple functions bounded by , supported on , and with for a.e. , then:
The limit exists.
If a.e., then the limit equals .
Lebesgue Integral of a Bounded Function Supported on Sets of Finite Measure
Let be a measurable function bounded by , supported on a set of finite measure. We define the Lebesgue integral of as where is a sequence of simple functions bounded by , , and converging pointwise to a.e.
Bounded Convergence Theorem
Suppose that is a sequence of measurable functions that are all bounded by , are supported on a set of finite measure, and for a.e. as . Then is measurable, bounded, supported on for a.e. , and
Consequently,
Proof Fix any , by Egorov’s theorem, we can find a closed set such that and uniformly on , so there exists such that for all and . Then we have for all . Since is arbitrary, we have .
Remark
The Bounded Convergence Theorem is about interchange of limits and integrals. It says, under certain conditions, we can interchange the limit and the integral:
Proposition
If non-negative is bounded and supported on a set of finite measure and , then a.e.
Proof For each integer , we have this implies by monotonicity. Thus for all , and since , we have .
Relationship with Riemann Integral
Theorem
Suppose is Riemann integrable on the closed interval . Then is measurable, and
where the integral on the left-hand side is the standard Riemann integral, and that on the right-hand side is the Lebesgue integral.
Non-negative Measurable Functions
Extended Lebesgue Integral of a Non-negative Measurable Function
Suppose is a non-negative measurable function, which is not necessarily bounded. We define the extended Lebesgue integral of as where the supremum is taken over all non-negative measurable functions that are bounded and supported on a set of finite measure.
If the supremum is finite, then we say that is Lebesgue integrable. Clearly if is any measurable set, then we define
Properties of the Integral of Non-Negative Measurable Functions
The integral of non-negative measurable functions enjoys the following properties:
Linearity: If , and are positive real numbers, then
Additivity: If and are disjoint subsets of , and , then
Monotonicity: If , then
Triangle inequality: If is a non-negative measurable function, then so is , and
If is integrable and , then is integrable.
If is integrable, then for almost all .
If , then for a.e. .
Note that and for a.e. not necessarily imply . For example, consider Then as for every , but for all . Indeed, the limit of the integrals is greater than the integral of the limit function in general:
Fatou's Lemma
Suppose is a sequence of measurable functions with . If for a.e. , then
Proof Suppose , where is bounded and supported on a set of finite measure. And we let , then is measurable, supported on , and for a.e. . By the bounded convergence theorem, we have By construction, , so that for all , and thereforeby the definition of the limit inferior. Since is arbitrary, taking the supremum over all yields .
Monotone Convergence Theorem
Suppose is a non-negative measurable function, and is a sequence of non-negative measurable functions, and for almost every . Then
Proof Since for a.e. , by monotonicity, we have , so . Hence, it follows that .
Comment
To simplify our notation, we shall write to mean that converges to in a nondecreasing sense, and to mean that converges to in a nonincreasing sense.
Corollary
Consider a series , where is a measurable function for every . Then
If is finite, then the series converges for a.e. .
Lebesgue Integral of a Real-Valued Measurable Function
A real-valued measurable function on is said to be Lebesgue integrable if the non-negative measurable function is integrable, i.e. .
If is Lebesgue integrable, its integral is defined as follows. We first define the positive and negative parts of by
so that both and are non-negative and satisfy
Both functions and are integrable whenever is. We then define the Lebesgue integral of by
Proposition
The integral of Lebesgue integrable functions is independent of the decomposition.
Proposition
The integral of Lebesgue integrable functions is linear, additive, monotonic, and satisfies the triangle inequality.
Lemma
Suppose is integrable on . Then for every :
Existence of a finite measure set: There exists a set of finite measure (a ball, for example) such that
Absolute continuity of the integral: There exists a such that
Dominated Convergence Theorem
Suppose is a sequence of measurable functions such that a.e. as . If , where is integrable, then
and consequently,
Proof For each , let . Fix , by the previous lemma,