Let be a real valued measurable function. Let

.

These are the positive and negative parts of . An easy observation is

Definition. Let be a real valued measurable function. If or is finite, we denote

We say is integrable if both terms are finite.

Proposition 2.21. The set of integrable real valued functions on is a real vector space, and the integral is a linear functional on it.

Proof. We need to prove two claims.

(1). for any This is easy: we distinguish three cases,

(2). It is to observe that, if $h = f+g,$

then . We rearrange it to obtain

. Taking integrals on both sides yields (2).

Definition. The complex valued function is integrable if and are both integrable.

Remark. It is easy to see that the space of complex valued integrable functions is a complex vector space and the integral is a linear functional over it. It is denoted by .

Remark. We will regard as a set of equivalence classes of a.e. defined integrable functions on , where and are considered to be equivalent if and only if .

Proposition 2.22. If , then .

Proof. When is real valued, the proof is easy. For complex valued , there exists ,

by linearity of integrals. It further equals

Proposition 2.24. (The dominated convergence theorem. ) Let be a sequence in such that

(a).

(b). There exists s.t. for a.e. for all . Then

and

Proof. Without loss of generality, we assume that is real valued.

Step 1. The claim that is easy.

(2). By Fatou’s theorem,

This yields,

The same process can be applied to $\int g+f $ to show

Remark. Examples. (1). and

(2). and .

(3). and

Theorem 2.26. If and , there is an integrable simple function such that If is the Lebesgue-Stieljes measure on , the sets in the definition of can be taken to finite unions of open intervals; moreover there is a continuous function that vanishes outside a bounded interval such that

Proof. Step 1. For an integrable function , there exists a sequence of simple functions such that

and pointwise. By DCT, ,

Step 2. From Step 1, That is to say

for .

Thus for each . $ For each , by Theorem 1.20, there exists a set that is a finite unions of open intervals such that for some small constant . By taking small, we can take the sets in the definition of to be finite unions of open intervals.

Step 3. Since we can approximate each , where is an open interval of finite length, by continuous functions with the following property

for some constant . By taking small, we can replace by a continuous function that vanishes outside a bounded interval such that

Theorem 2.27. Suppose that and that is integrable for each . Let .

(a). Suppose that there exists such that for all . If for every , then

. In particular, if is continuous for each $latrex x$, then is continuous.

(b). Suppose that exists and there is a such that for all . Then is differentiable and

Proof. (a). We are proving the claim in (a) by using sequential characterization of continuity. Let , the sequence of measurable functions is uniformly by a function and it converges to $f(x,t_0)$ pointwise. Therefore by the dominated convergence theorem,

(b). The proof is similar. We need to prove that, for each ,

By applying the mean function for in , we see that the sequence $ \{ \frac {f(x,t+h_n) -f(x,t)}{h_n}\}$ is uniformly bounded by a function ; moreover it converges to . Again by the dominated convergence theorem,

We next discuss the relation between Riemann integrable functions and Lebesgue measurable function.

Theorem. Let be a bounded real-valued function on . If is Riemann integrable, then it is Lebesgue integrable. Moreover the two integrals are equal.

Remark. Without loss of generality, we assume that is a nonnegative function. As in Theorem 2.10, there is a sequence of simple functions that satisfies,

and uniformly on . Although is a pointwise limit of , the Lebesgue measurability of is not immediately clear. We should seek to prove the Lebesgue measurability in other ways.

Proof. Let by a partition of . Let

where are supremum and infimum of on . Since $f$ is Riemann integrable, we can choose a sequence of partitions whose mesh size and

.

These are understood that the two sequences and has the same limit.

This implies

On the other hand, by the monotone convergence theorem for functions,

Therefore (the following integrals are understood as in the Lebesgue sense. )

Since pointwise, , which yields that . Since $G$ is Lebesgue measurable as simple functions and is Lebesgue measurable.

That the two integrals are equal is immediate.