A detailed proof of Tonelli's and Fubini's theorems

Preliminaries

Product Measure Spaces

Let and be two -finite measure spaces. Let

Sets in are called rectangles.

Note that is a semi-algebra but not a -algebra: - is a semi-algebra. - Closed under binary intersections: - Complements are finite disjoint unions: - is not an algebra (thus not a -algebra). - Not closed under finite unions: Let , where is the Borel -algebra on . Consider the rectangles and . Their union is not a rectangle and hence not in .

Denote as the smallest -algebra containing . Our discussion below will focus on the measurable space .

Product Measure

Let be a measure on such that for all rectangles , we have

It can be shown that when are -finite, such a measure is unique and is called the product measure of and , denoted by . To rigorously prove this, we will present several lemmas and then apply Carathéodory's extension theorem.

Lemma 1. The algebra generated by the semi-algebra consists of all finite disjoint unions of sets in .

Proof. First, by the definition of algebra, no other algebra can be smaller than . Thus it suffices to show that is an algebra. - Closed under finite unions: Let . Then for each we can write where . Hence is a finite union of sets in and thus in . - Closed under complements: Let where . By the definition of semi-algebra, we can write where . Then . Since by definition, is a finite union of sets in and thus in .

Lemma 2. Let be a semi-algebra and let defined on have . Suppose: - (i) If is a finite disjoint union of sets , then . - (ii) If is a countable disjoint union of sets , then .

Then has a unique extension that is a pre-measure (the version of measure defined on an algebra rather than a -algebra) on , the algebra generated by .

Proof. By Lemma 1, every can be written as a finite disjoint union of sets in , i.e., where . By the definition of pre-measure, we must have . Thus the extension is unique if it exists.

Let where are disjoint sets in , and where are disjoint. Then we have

$\tag not allowed in aligned environment \begin{aligned} \bar{\mu}(A) &= \sum_{i=1}^n \mu(A_i) \\ &\le \sum_{i=1}^n \sum_{j,k} \mu(A_i \cap S_{jk}) \\ &= \sum_{j,k} \sum_{i=1}^n \mu(A_i \cap S_{jk}) \\ &= \sum_{j \ge 1} \sum_{k=1}^{m_j} \mu(S_{jk}) \\ &= \sum_{j \ge 1} \bar{\mu}(S_j) \tag{1} \end{aligned}$

where: - The first equality follows from the construction of . - The inequality follows from (ii). Note that by the definition of semi-algebra. - The second equality follows from the non-negativity of . - The third equality follows from (i). - The last equality follows from the construction of .

When the number of is finite, the above inequality becomes an equality. Thus when and are disjoint, we have

To obtain the opposite inequality, we write

where - is the part of not covered by - is an arbitrary positive integer. - The second equality follows from .

Taking , we obtain

Combining (1) and (3), we conclude that

Hence is a pre-measure on .

Theorem 1 (Carathéodory's Extension Theorem). Let be a pre-measure on an algebra . If is -finite, then can be extended to a unique measure on :

Proof. Omitted. See e.g. Appendix A.1 in Probability: Theory and Examples by Rick Durrett.

Now we can prove the existence and uniqueness of the product measure: - is a semi-algebra. - is -finite, since both and are -finite. - Let and where are disjoint. Then we have

where:
- The first equality follows from the definition of product measure.
- The second equality follows from the definition of Lebesgue integral.
- The third equality follows from the the property of measure (countable additivity) and the fact that are disjoint.
- The fourth equality follows from the linearity of integral.
- The last equality follows from the definition of Lebesgue integral.

Therefore, we can apply Lemma 2 to show that can be uniquely extended to a pre-measure on the algebra generated by . By further applying Carathéodory's extension theorem, we conclude that can be uniquely extended to a measure on .

Fubini and Tonelli Theorems

Theorem 2 (Tonelli's Theorem). Let and be two -finite measure spaces. Let be a non-negative -measurable function. Then

Theorem 3 (Fubini's Theorem). Let and be two -finite measure spaces. Let be an integrable function with respect to . i.e.

Then

Our proof will proceed in six steps: - Step 1: Prove both theorems for indicator functions of rectangles. - Step 2: Prove both theorems for indicator functions of countable disjoint unions of rectangles. - Step 3: Prove both theorems for indicator functions of sets with finite measure. ※ - Step 4: Prove both theorems for indicator functions of measurable sets. - Step 5: Prove both theorems for simple functions. - Step 6: Prove both theorems for non-negative measurable functions. - Step 7: Prove Fubini's theorem for integrable functions.

For simplicity, we will only prove the first equality in both theorems. The second equality can be proved symmetrically.

Proof

Step 1: Indicator Functions of Rectangles

Let where . Then we need to show that

We have

where the last equality follows from the definition of product measure.

Step 2: Indicator Functions of Countable Disjoint Unions of Rectangles

Let where and are disjoint. We have

where: - The first equality follows from the definition of . - The second and third equalities follow from the Monotone Convergence Theorem (MCT). - The fourth equality follows from Step 1. - The last equality follows from the property of measure (countable additivity).

When the union is finite, the above argument still holds except that we do not need to apply MCT.

Step 3: Indicator Functions of Sets with Finite Measure

Let where and . We need to show that

where .

To prove this, we first introduce a useful theorem about approximating measurable sets using sets in an algebra.

Theorem 4 (Approximation of Measurable Sets). Let be an algebra and be a pre-measure on that is -finite. Let be the unique extension of on by Carathéodory's extension theorem. Then for any and , there exists a finite disjoint union of sets , i.e. , such that .

Proof. By Theorem 1, when is -finite, the only possible definition of is

By the definition of infimum, for any , there exists a sequence of sets such that and . Without loss of generality, we can assume are disjoint, since we can always replace by without increasing the sum .

Let , then we have

Thus

Let

By the continuity of measure from below, we have

There exists an integer such that

Therefore

For simplicity, we denote

for any .

Now we can prove the claim. For any , by Theorem 4, there exists a finite disjoint union of rectangles (since every set in the extended algebra is a finite disjoint union of rectangles) such that . Then

where: - The first equality follows from the property of measure (finite additivity). - The first inequality follows from the triangle inequality. - The second equality follows from Step 2. - The second inequality follows from the triangle inequality and the non-negativity of measure. - The third equality follows from the fact that for any sets . - The third inequality follows from the construction of .

To complete the proof, the only thing left is to show that can be made arbitrarily small.

Similar to the argument above, there exists a sequence of rectangles such that

Then we have

where: - The first inequality follows from the monotonicity of measure and integral. - The first equality follows from Step 2. - The second equality follows from the property of measure (countable additivity). - The last inequality follows from the construction of .

Step 4: Indicator Functions of Measurable Sets

Let where .

Since is -finite, there exist sets such that and . By Step 4, we have

Also, we have

By the continuity of measure from below, we have

To complete the proof, we only need to show that

Expanding the right-hand side, we have

where: - The first equality follows from the definition of . - The second equality follows from the Monotone Convergence Theorem (MCT). - The third equality follows from the continuity of measure from below. - The last equality follows from the definition of .

Therefore,

Step 5: Simple Functions

Let where and . By linearity of integral, we have

Step 6: Non-negative Measurable Functions

Let be a sequence of simple functions such that . By the Monotone Convergence Theorem (MCT), we have

Step 5 shows that

Thus it suffices to show that

We have

where: - The first equality follows from . - The second equality follows from for all , monotonicity of integral, and the MCT. - The third equality follows from (by the MCT) for all , monotonicity of integral, and the MCT.

Thus we complete the proof for Tonelli's theorem. Importantly, we did not assume is integrable in this step, hence Tonelli's theorem holds for all non-negative measurable functions.

Step 7: Integrable Functions

Let . We have

where: - The first equality follows from the definition of Lebesgue integral. This step requires to be integrable, since is undefined. - The second equality follows from Tonelli's theorem (Step 6). - The third equality follows from the linearity of integral. - The last equality follows from the definition of Lebesgue integral.

Thus we complete the proof for Fubini's theorem.