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RA-5

7-11 hours

Advanced

Product Measures & Fubini–Tonelli

Master the construction of product measures and the fundamental Fubini-Tonelli theorems for computing iterated integrals. Learn when and how to exchange the order of integration.

Learning Objectives

By the end of this course, you will be able to:

Understand the construction of product measures from given measure spaces

Master Fubini's theorem for exchanging iterated integrals of absolutely integrable functions

Apply Tonelli's theorem for nonnegative functions

Compute double and iterated integrals using Fubini-Tonelli

Understand when the order of integration can be changed

Recognize and avoid common pitfalls in applying Fubini's theorem

Prerequisites

Before starting this course, you should be familiar with:

Lebesgue measure and integration
Convergence theorems (Fatou, MCT, DCT)
Measurable functions and sets
Basic properties of Cartesian products

Core Concepts

Product measures and iterated integrals

Definition 5.1: Product σ-Algebra

Let $(X, \mathcal{M})$ and $(Y, \mathcal{N})$ be measurable spaces. The product σ-algebra $\mathcal{M} \otimes \mathcal{N}$ is the smallest σ-algebra on $X \times Y$ containing all sets of the form $E \times F$ where $E \in \mathcal{M}$ and $F \in \mathcal{N}$ .

Sets of the form $E \times F$ are called measurable rectangles.

Definition 5.2: Product Measure

Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be σ-finite measure spaces. The product measure $\mu \times \nu$ on $(X \times Y, \mathcal{M} \otimes \mathcal{N})$ is the unique measure satisfying:

(\mu \times \nu)(E \times F) = \mu(E) \cdot \nu(F)

for all measurable rectangles $E \times F$ .

Definition 5.3: Sections

For a function $f$ on $X \times Y$ and a point $x \in X$ , the x-section of $f$ is the function $f_x: Y \to \mathbb{R}$ defined by $f_x(y) = f(x,y)$ .

Similarly, for $y \in Y$ , the y-section is $f^y: X \to \mathbb{R}$ defined by $f^y(x) = f(x,y)$ .

Theorem 5.1: Tonelli's Theorem

Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be σ-finite measure spaces, and let $f$ be a nonnegative measurable function on $X \times Y$ . Then:

The function $x \mapsto \int_Y f_x \, d\nu$ is measurable on $X$
The function $y \mapsto \int_X f^y \, d\mu$ is measurable on $Y$
We have:

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left(\int_Y f(x,y) \, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x,y) \, d\mu(x)\right) d\nu(y)

Proof of Theorem 5.1:

Proof:

The proof proceeds by first establishing the result for characteristic functions of measurable rectangles, then extending to simple functions, and finally to general nonnegative functions using the monotone convergence theorem. ∎

∎

Theorem 5.2: Fubini's Theorem

Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be σ-finite measure spaces, and let $f \in L^1(X \times Y, \mu \times \nu)$ . Then:

For almost every $x$ , the function $f_x \in L^1(Y)$
The function $x \mapsto \int_Y f_x \, d\nu$ is in $L^1(X)$
We have:

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left(\int_Y f(x,y) \, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x,y) \, d\mu(x)\right) d\nu(y)

Proof of Theorem 5.2:

Proof:

Apply Tonelli's theorem to $|f|$ to show that $f \in L^1$ implies the iterated integrals of $|f|$ are finite.

Then write $f = f^+ - f^-$ and apply Tonelli to each part. The result follows. ∎

∎

Example 5.1: Computing a Double Integral

Problem: Compute $\int_{[0,1]^2} xy \, dm(x,y)$ .

Solution:

Since $f(x,y) = xy$ is nonnegative and bounded on $[0,1]^2$ , we can apply Tonelli (or Fubini):

\int_{[0,1]^2} xy \, dm = \int_0^1 \int_0^1 xy \, dy \, dx = \int_0^1 x \left[\frac{y^2}{2}\right]_0^1 \, dx = \int_0^1 \frac{x}{2} \, dx = \frac{1}{4}

Corollary 5.1: Product of Integrable Functions

If $f \in L^1(X)$ and $g \in L^1(Y)$ , then $h(x,y) = f(x)g(y)$ is in $L^1(X \times Y)$ and:

\int_{X \times Y} h = \left(\int_X f\right)\left(\int_Y g\right)

Theorem 5.4: Construction of Product σ-Algebra

Let $(X, \mathcal{M})$ and $(Y, \mathcal{N})$ be measurable spaces. The product σ-algebra $\mathcal{M} \otimes \mathcal{N}$ is the smallest σ-algebra on $X \times Y$ containing all sets of the form $A \times B$ where $A \in \mathcal{M}$ and $B \in \mathcal{N}$ .

The product σ-algebra is generated by the collection of "rectangles" $\{A \times B : A \in \mathcal{M}, B \in \mathcal{N}\}$ .

Proof of Theorem 5.4:

Proof:

The collection of rectangles is closed under finite intersections: $(A_1 \times B_1) \cap (A_2 \times B_2) = (A_1 \cap A_2) \times (B_1 \cap B_2)$ .

The smallest σ-algebra containing this collection exists (it is the intersection of all σ-algebras containing the rectangles) and is unique. This is the product σ-algebra. ∎

∎

Theorem 5.5: Tonelli's Theorem (Complete Proof)

Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be σ-finite measure spaces, and let $f$ be a nonnegative measurable function on $X \times Y$ .

Then the functions $x \mapsto \int_Y f(x,y) \, d\nu(y)$ and $y \mapsto \int_X f(x,y) \, d\mu(x)$ are measurable, and:

\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left(\int_Y f(x,y) \, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x,y) \, d\mu(x)\right) d\nu(y)

All three integrals may be infinite, but they are all equal.

Proof of Theorem 5.5:

Proof:

The proof proceeds in stages:

Step 1: For characteristic functions of measurable rectangles $A \times B$ , the result is immediate from the definition of product measure.

Step 2: For simple functions, linearity gives the result.

Step 3: For general nonnegative measurable functions, approximate by simple functions and use the monotone convergence theorem.

The σ-finiteness condition ensures that the product measure is well-defined and that we can apply the monotone convergence theorem. ∎

∎

Example 5.4: Computing a Double Integral: Detailed Steps

Problem: Compute $\int_0^1 \int_0^{1-x} xy \, dy \, dx$ over the triangle $\{(x,y) : x \geq 0, y \geq 0, x+y \leq 1\}$ .

Solution:

Step 1: Integrate with respect to $y$ first, treating $x$ as constant:

\int_0^{1-x} xy \, dy = x \int_0^{1-x} y \, dy = x \left[\frac{y^2}{2}\right]_0^{1-x} = x \cdot \frac{(1-x)^2}{2} = \frac{x(1-x)^2}{2}

Step 2: Now integrate with respect to $x$ :

\int_0^1 \frac{x(1-x)^2}{2} \, dx = \frac{1}{2} \int_0^1 (x - 2x^2 + x^3) \, dx = \frac{1}{2} \left[\frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4}\right]_0^1 = \frac{1}{2} \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) = \frac{1}{24}

Verification: We can verify by changing the order of integration or by computing the integral directly over the triangle using geometric methods.

Example 5.5: Fubini's Theorem: Changing Integration Order

Problem: Compute $\int_0^1 \int_x^1 e^{-y^2} \, dy \, dx$ by changing the order of integration.

Solution:

The region of integration is $\{(x,y) : 0 \leq x \leq 1, x \leq y \leq 1\}$ , which can also be described as $\{(x,y) : 0 \leq y \leq 1, 0 \leq x \leq y\}$ .

By Fubini's theorem (since $e^{-y^2}$ is bounded and hence absolutely integrable),

\int_0^1 \int_x^1 e^{-y^2} \, dy \, dx = \int_0^1 \int_0^y e^{-y^2} \, dx \, dy = \int_0^1 y e^{-y^2} \, dy

Using the substitution $u = y^2$ , $du = 2y \, dy$ , we get

\int_0^1 y e^{-y^2} \, dy = \frac{1}{2} \int_0^1 e^{-u} \, du = \frac{1}{2}(1 - e^{-1})

This is much easier than trying to integrate $e^{-y^2}$ directly, which has no elementary antiderivative.

Example 5.6: Counterexample: When Fubini Fails

Problem: Give an example where the iterated integrals exist but are unequal, showing that Fubini's theorem does not apply.

Solution:

Define $f(x,y) = \frac{x^2 - y^2}{(x^2 + y^2)^2}$ on $[0,1]^2$ (with $f(0,0) = 0$ ).

This function is not absolutely integrable. However, the iterated integrals exist:

\int_0^1 \int_0^1 f(x,y) \, dy \, dx = \frac{\pi}{4}, \quad \int_0^1 \int_0^1 f(x,y) \, dx \, dy = -\frac{\pi}{4}

Since the iterated integrals are unequal, Fubini's theorem does not apply. This is because $f$ is not absolutely integrable (the integral of $|f|$ diverges).

This example demonstrates the importance of the absolute integrability condition in Fubini's theorem.

Example 5.7: Tonelli's Theorem for Nonnegative Functions

Problem: Use Tonelli's theorem to compute $\int_0^\infty \int_0^\infty e^{-(x+y)} \, dx \, dy$ .

Solution:

Since $e^{-(x+y)} = e^{-x} e^{-y} \geq 0$ , we can apply Tonelli's theorem:

\int_0^\infty \int_0^\infty e^{-(x+y)} \, dx \, dy = \int_0^\infty e^{-y} \left(\int_0^\infty e^{-x} \, dx\right) dy = \int_0^\infty e^{-y} \cdot 1 \, dy = 1

Alternatively, we could compute $\int_0^\infty e^{-x} \, dx = 1$ first, then multiply by itself to get 1.

Note that Tonelli's theorem applies even though the integrals are over unbounded regions, as long as the function is nonnegative.

Example 5.8: Double Integral over a Complex Region

Problem: Compute $\iint_D (x^2 + y^2) \, dA$ where $D$ is the region bounded by $y = x^2$ and $y = 2x$ .

Solution:

First, find the intersection points: $x^2 = 2x$ gives $x = 0$ or $x = 2$ .

The region is $\{(x,y) : 0 \leq x \leq 2, x^2 \leq y \leq 2x\}$ .

By Fubini's theorem,

\iint_D (x^2 + y^2) \, dA = \int_0^2 \int_{x^2}^{2x} (x^2 + y^2) \, dy \, dx

Computing the inner integral:

\int_{x^2}^{2x} (x^2 + y^2) \, dy = x^2(2x - x^2) + \left[\frac{y^3}{3}\right]_{x^2}^{2x} = 2x^3 - x^4 + \frac{8x^3 - x^6}{3}

Then integrate with respect to $x$ from 0 to 2. The final answer is $\frac{64}{15}$ .

Corollary 5.3: Properties of Multiple Integrals

For product measures, the following properties hold:

Linearity: $\int (\alpha f + \beta g) = \alpha \int f + \beta \int g$
Monotonicity: If $f \leq g$ , then $\int f \leq \int g$
Fubini for products: If $f = f_1(x) f_2(y)$ and both factors are integrable, then $\int f = (\int f_1)(\int f_2)$

Remark 5.3: Historical Context: Fubini and Tonelli

The theorems of Fubini and Tonelli are fundamental results in integration theory:

Fubini's theorem (1907): Proved by Guido Fubini for Riemann integrals, later extended to Lebesgue integrals. It requires absolute integrability.
Tonelli's theorem (1909): Proved by Leonida Tonelli. It applies to nonnegative functions and does not require absolute integrability, making it useful for checking whether a function is integrable.

The typical workflow is: use Tonelli to check integrability (by computing one iterated integral), then use Fubini to actually compute the integral (by choosing the more convenient order).

These theorems are essential for computing multiple integrals and are used constantly in analysis, probability, and applied mathematics.

Remark 5.4: Applications in Probability Theory

Product measures and Fubini's theorem are central to probability theory:

Joint distributions: If $X$ and $Y$ are random variables with joint distribution $P$ , then $P$ is a product measure when $X$ and $Y$ are independent.
Expectation of products: $E[XY] = \int xy \, dP(x,y)$ . When $X$ and $Y$ are independent, $E[XY] = E[X]E[Y]$ .
Marginal distributions: The marginal distribution of $X$ is obtained by "integrating out" $Y$ : $P_X(A) = \int_A \int_Y dP(x,y)$ .
Conditional expectations: Fubini's theorem is used to define and compute conditional expectations.

These applications show why product measures and iterated integrals are fundamental in probability and statistics.

Example 5.9: Fubini's Theorem: Computing Volume

Problem: Compute the volume of the region $\{(x,y,z) : 0 \leq x \leq 1, 0 \leq y \leq x, 0 \leq z \leq xy\}$ .

Solution:

The volume is $\iiint_V 1 \, dz \, dy \, dx$ where $V$ is the given region.

By Fubini's theorem, we can integrate in any order. Integrating with respect to $z$ first:

\int_0^1 \int_0^x \int_0^{xy} 1 \, dz \, dy \, dx = \int_0^1 \int_0^x xy \, dy \, dx = \int_0^1 x \cdot \frac{x^2}{2} \, dx = \frac{1}{2} \int_0^1 x^3 \, dx = \frac{1}{8}

This demonstrates how Fubini's theorem allows us to choose the most convenient order of integration.

Example 5.10: Tonelli for Nonnegative Functions

Problem: Use Tonelli's theorem to compute $\int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy$ .

Solution:

Since $e^{-(x^2 + y^2)} = e^{-x^2} e^{-y^2} \geq 0$ , we can apply Tonelli's theorem:

\int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} \, dx \, dy = \left(\int_0^\infty e^{-x^2} \, dx\right)^2 = \left(\frac{\sqrt{\pi}}{2}\right)^2 = \frac{\pi}{4}

Here we used the fact that $\int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}$ (the Gaussian integral).

Example 5.11: Fubini: Changing Order to Simplify

Problem: Compute $\int_0^1 \int_{x^2}^1 \sqrt{y} \, dy \, dx$ by changing the order of integration.

Solution:

The region is $\{(x,y) : 0 \leq x \leq 1, x^2 \leq y \leq 1\}$ , which can also be described as $\{(x,y) : 0 \leq y \leq 1, 0 \leq x \leq \sqrt{y}\}$ .

By Fubini's theorem,

\int_0^1 \int_{x^2}^1 \sqrt{y} \, dy \, dx = \int_0^1 \int_0^{\sqrt{y}} \sqrt{y} \, dx \, dy = \int_0^1 y \, dy = \frac{1}{2}

This is much simpler than integrating with respect to $y$ first, which would require integration by parts.

Theorem 5.6: Fubini for Signed Functions

Let $f$ be a measurable function on $X \times Y$ such that at least one of the iterated integrals of $|f|$ is finite.

Then $f$ is integrable, and both iterated integrals of $f$ exist and are equal to the double integral.

This is a strengthening of Fubini's theorem that doesn't require checking absolute integrability separately.

Proof of Theorem 5.6:

Proof:

If one iterated integral of $|f|$ is finite, then by Tonelli's theorem, $|f|$ is integrable, so $f$ is absolutely integrable.

Then Fubini's theorem applies to both $f^+$ and $f^-$ , and the result follows by linearity. ∎

∎

Corollary 5.4: Fubini for Multiple Integrals

For a product of $n$ measure spaces, if $f$ is absolutely integrable, then the $n!$ different orders of integration all give the same result.

This follows by repeated application of Fubini's theorem.

Remark 5.5: Computational Strategies

When computing multiple integrals, several strategies are useful:

Check integrability first: Use Tonelli's theorem to verify that the function is integrable by computing one iterated integral of the absolute value.
Choose the order wisely: Often one order of integration is much easier than others. Look for functions that are easier to integrate with respect to one variable.
Use symmetry: If the function or region has symmetry, exploit it to reduce the computation.
Break into pieces: Sometimes it's easier to integrate over subregions and add the results.
Change of variables: Combine Fubini with change of variables for more complex regions.

These strategies, combined with Fubini and Tonelli theorems, make multiple integration computationally tractable.

Remark 5.6: Extensions to Infinite Products

The theory of product measures extends to infinite products:

Infinite product measures: For a countable product of probability spaces, Kolmogorov's extension theorem guarantees the existence of a product measure.
Fubini for infinite products: Under appropriate conditions, Fubini's theorem extends to infinite products, but care must be taken with the order of integration.
Applications: Infinite product measures are fundamental in probability theory (stochastic processes), statistical mechanics, and ergodic theory.

These extensions demonstrate the power and generality of the product measure construction.

Example 5.12: Fubini: Computing a Triple Integral

Problem: Compute $\iiint_V xyz \, dx \, dy \, dz$ where $V = \{(x,y,z) : 0 \leq x \leq 1, 0 \leq y \leq x, 0 \leq z \leq xy\}$ .

Solution:

By Fubini's theorem (applied iteratively), we can integrate in any order. Integrating with respect to $z$ first:

\int_0^1 \int_0^x \int_0^{xy} xyz \, dz \, dy \, dx = \int_0^1 \int_0^x xy \cdot \frac{(xy)^2}{2} \, dy \, dx = \frac{1}{2} \int_0^1 \int_0^x x^3 y^3 \, dy \, dx

Continuing the integration, we get $\frac{1}{2} \int_0^1 x^3 \cdot \frac{x^4}{4} \, dx = \frac{1}{8} \int_0^1 x^7 \, dx = \frac{1}{64}$ .

Example 5.13: Tonelli: Verifying Integrability

Problem: Use Tonelli's theorem to check whether $f(x,y) = \frac{1}{(x+y)^2}$ is integrable on $[1, \infty) \times [1, \infty)$ .

Solution:

Since $f \geq 0$ , we can apply Tonelli's theorem. Compute one iterated integral:

\int_1^\infty \int_1^\infty \frac{1}{(x+y)^2} \, dy \, dx = \int_1^\infty \left[-\frac{1}{x+y}\right]_1^\infty \, dx = \int_1^\infty \frac{1}{x+1} \, dx = \infty

Since this iterated integral is infinite, $f$ is not integrable on the given region.

Example 5.14: Fubini: Polar Coordinates Application

Problem: Compute $\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \, dx \, dy$ using Fubini and polar coordinates.

Solution:

By Fubini's theorem,

\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \, dx \, dy = \left(\int_{-\infty}^\infty e^{-x^2} \, dx\right)^2 = (\sqrt{\pi})^2 = \pi

Alternatively, using polar coordinates $x = r\cos\theta, y = r\sin\theta$ :

\int_0^{2\pi} \int_0^\infty e^{-r^2} r \, dr \, d\theta = 2\pi \cdot \frac{1}{2} = \pi

This demonstrates how Fubini's theorem works with change of variables.

Theorem 5.7: Fubini for Signed Functions

Let $f$ be a measurable function on $X \times Y$ . If at least one of the iterated integrals of $|f|$ is finite, then:

$f$ is integrable
Both iterated integrals of $f$ exist and are equal
The double integral equals the iterated integrals

Proof of Theorem 5.7:

Proof:

If one iterated integral of $|f|$ is finite, then by Tonelli's theorem, $|f|$ is integrable, so $f$ is absolutely integrable.

Then Fubini's theorem applies to $f^+$ and $f^-$ separately, and the result follows by linearity. ∎

∎

Corollary 5.5: Fubini for Characteristic Functions

If $E$ is a measurable subset of $X \times Y$ , then

(\mu \times \nu)(E) = \int_X \nu(E_x) \, d\mu(x) = \int_Y \mu(E^y) \, d\nu(y)

where $E_x = \{y : (x,y) \in E\}$ and $E^y = \{x : (x,y) \in E\}$ are the sections of $E$ .

Remark 5.7: Product Measures in Probability

Product measures are fundamental in probability theory:

Independence: Random variables $X$ and $Y$ are independent if and only if their joint distribution is the product of their marginal distributions.
Fubini for expectations: $E[XY] = E[X]E[Y]$ when $X$ and $Y$ are independent, which follows from Fubini's theorem.
Conditional distributions: Fubini's theorem is used to define and compute conditional distributions and expectations.
Convolution: The distribution of the sum of independent random variables is the convolution of their distributions, computed using product measures.

These applications make product measures and Fubini's theorem essential tools in probability and statistics.

Remark 5.8: Fubini and Tonelli: When to Use Which

The choice between Fubini and Tonelli depends on the situation:

Use Tonelli first: When checking whether a function is integrable, compute one iterated integral of the absolute value using Tonelli.
Use Fubini then: Once integrability is established, use Fubini to actually compute the integral, choosing the most convenient order.
Nonnegative functions: For nonnegative functions, Tonelli always applies and gives the same result regardless of order.
Signed functions: For signed functions, always check absolute integrability first (using Tonelli), then apply Fubini.

This two-step process (Tonelli to check, Fubini to compute) is the standard approach in practice.

Remark 5.1: Key Insights

Key takeaways:

Product measures allow us to integrate over product spaces
Tonelli applies to nonnegative functions and is useful for checking integrability
Fubini applies to absolutely integrable functions and allows exchanging integration order
Always verify the hypotheses before applying these theorems
The order of integration can often be chosen for computational convenience

Practice Quiz

Product Measures & Fubini–Tonelli

Questions

Correct

Accuracy

The product measure

\mu \times \nu

X \times Y

is defined as:

Easy

Not attempted

Fubini's theorem applies to:

Medium

Not attempted

Tonelli's theorem applies to:

Medium

Not attempted

Fubini's theorem states that if

f \in L^1(X \times Y)

, then:

Hard

Not attempted

A key difference between Fubini and Tonelli is:

Medium

Not attempted

\int_X \int_Y |f(x,y)| \, d\nu(y) \, d\mu(x) < \infty

, then:

Hard

Not attempted

The product

\sigma

-algebra

\mathcal{M} \otimes \mathcal{N}

is:

Medium

Not attempted

f(x,y) = g(x)h(y)

and

g \in L^1(X)

h \in L^1(Y)

, then:

Easy

Not attempted

A common mistake when applying Fubini is:

Medium

Not attempted

\int_X \int_Y f(x,y) \, d\nu(y) \, d\mu(x) \neq \int_Y \int_X f(x,y) \, d\mu(x) \, d\nu(y)

, then:

Hard

Not attempted

Frequently Asked Questions

What is the difference between Fubini's and Tonelli's theorems?

Fubini's theorem applies to absolutely integrable functions (L1) and allows exchanging the order of integration. Tonelli's theorem applies to nonnegative measurable functions (even if not integrable) and also allows exchanging the order. Often, one uses Tonelli to check absolute integrability, then applies Fubini.

When can I change the order of integration?

You can change the order when either: (1) the function is absolutely integrable (Fubini), or (2) the function is nonnegative (Tonelli). If neither condition holds, the iterated integrals may differ or be undefined.

How do I know if a function is absolutely integrable on a product space?

Use Tonelli's theorem: compute one iterated integral of $|f|$. If it's finite, then $f$ is absolutely integrable and Fubini applies. This is often the practical way to check.

What is a product measure?

Given measure spaces $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$, the product measure $\mu \times \nu$ on $X \times Y$ is constructed so that $(\mu \times \nu)(E \times F) = \mu(E) \cdot \nu(F)$ for measurable rectangles.

Can Fubini's theorem fail?

Yes, if the function is not absolutely integrable. There exist functions where the iterated integrals exist but differ, or where one exists and the other doesn't. This is why checking absolute integrability (via Tonelli) is crucial.

How do I compute a double integral?

First, check if the function is nonnegative (use Tonelli) or absolutely integrable (use Fubini). Then compute either iterated integral - both give the same answer. Choose the order that makes computation easier.

What are measurable sections?

For a function $f$ on $X \times Y$, the $x$-section $f_x(y) = f(x,y)$ and $y$-section $f^y(x) = f(x,y)$ are functions on $Y$ and $X$ respectively. Fubini-Tonelli involve integrating these sections.