# Measure Theory/Integration

Let be a -finite measure space. Suppose is a positive simple measurable function, with ; are disjoint.

Define

Let be measurable, and let .

Define

Now let be any measurable function. We say that is integrable if and are integrable and if . Then, we write

The class of measurable functions on is denoted by

For , we define to be the collection of all measurable functions such that

A property is said to hold **almost everywhere** if the set of all points where the property does not hold has measure zero.

## Contents

## Properties[edit]

Let be a measure space and let be measurable on . Then

- If , then
- If , , then
- If and then
- If , , then , even if
- If , , then , even if

**Proof**

## Monotone Convergence Theorem[edit]

Suppose and are measurable for all such that

- for every
- almost everywhere on

Then,

**Proof**

is an increasing sequence in , and hence, (say). We know that is measurable and that . That is,

Hence,

Let

Define ; . Observe that and

Suppose . If then implying that . If , then there exists such that and hence, .

Thus, , therefore . As this is true if , we have that . Thus, .

## Fatou's Lemma[edit]

Let be measurable functions. Then,

**Proof**

For define . Observe that are measurable and increasing for all .

As , . By monotone convergence theorem,

and as , we have the result.

## Dominated convergence theorem[edit]

Let be a complex measure space. Let be a sequence of complex measurable functions that converge pointwise to ; , with

Suppose for some then

and as

**Proof**

We know that and hence , that is,

Therefore, by Fatou's lemma,

As , implying that

## Theorem[edit]

- Suppose is measurable, with such that . Then almost everywhere
- Let and let for every . Then, almost everywhere on
- Let and then there exists constant such that almost everywhere on

**Proof**

- For each define . Observe that

but Thus for all , by continuity, almost everywhere on - Write , where are non-negative real measurable.

Further as are both non-negative, each of them is zero. Thus, by applying part I, we have that vanish almost everywhere on . We can similarly show that vanish almost everywhere on .