Product Measures

Product

σ

-Algebra

Suppose that

(X, 𝒮)

and

(Y, 𝒯)

are measurable spaces. The product $σ$ -algebra

𝒮 \otimes 𝒯

is the smallest

σ

-algebra on

X \times Y

containing every rectangle

A \times B

with

A \in 𝒮

and

B \in 𝒯

Measurable Rectangle

Suppose that

(X, 𝒮)

and

(Y, 𝒯)

are measurable spaces. A measurable rectangle in

X \times Y

is a set of the form

A \times B

, where

A \in 𝒮

and

B \in 𝒯

Section of a Set

Suppose that

E \subseteq X \times Y

. For

x \in X

, define

E_{x} : = {y \in Y : (x, y) \in E} .

For

y \in Y

, define

E^{y} : = {x \in X : (x, y) \in E} .

Sections of Measurable Sets are Measurable

Suppose that

E \in 𝒮 \otimes 𝒯

. Then

E_{x} \in 𝒯

for every

x \in X

, and

E^{y} \in 𝒮

for every

y \in Y

Let $𝒜$ be the class of sets $E \subseteq X \times Y$ whose $x$ -sections are in $𝒯$ and whose $y$ -sections are in $𝒮$ . Rectangles $A \times B$ with $A \in 𝒮$ and $B \in 𝒯$ belong to $𝒜$ . The class $𝒜$ is a $σ$ -algebra, because sections commute with complements and countable unions. Therefore $𝒮 \otimes 𝒯 \subseteq 𝒜$ , proving the claim.

Product Measure

Suppose that

(X, 𝒮, μ)

and

(Y, 𝒯, ν)

are

σ

-finite measure spaces. The product measure

μ \times ν

is the measure on

𝒮 \otimes 𝒯

satisfying

(μ \times ν) (A \times B) = μ (A) ν (B)

for every

A \in 𝒮

and

B \in 𝒯

Tonelli's Theorem

Suppose that

(X, 𝒮, μ)

and

(Y, 𝒯, ν)

are

σ

-finite measure spaces and

f : X \times Y \to [0, \infty]

𝒮 \otimes 𝒯

-measurable. Then the section functions are measurable and

\int_{X \times Y} f d (μ \times ν) = \int_{X} (\int_{Y} f (x, y) d ν (y)) d μ (x) = \int_{Y} (\int_{X} f (x, y) d μ (x)) d ν (y) .

First prove the formula for characteristic functions of measurable rectangles, where both sides equal $μ (A) ν (B)$ . By finite linearity, it holds for nonnegative simple functions. For a nonnegative measurable $f$ , choose increasing simple functions converging to $f$ by approximation by simple functions. The section functions are measurable by the same simple-function approximation and measurable sections. Passing to the limit with the monotone convergence theorem gives both iterated-integral identities.

Fubini's Theorem

Suppose that

(X, 𝒮, μ)

and

(Y, 𝒯, ν)

are

σ

-finite measure spaces and

f \in L^{1} (μ \times ν)

. Then

f (x, \cdot) \in L^{1} (ν)

for almost every

x

f (\cdot, y) \in L^{1} (μ)

for almost every

y

, and the iterated integrals exist and are equal to

\int f d (μ \times ν)

Apply Tonelli's theorem to $| f |$ . Since $f \in L^{1} (μ \times ν)$ , the iterated integrals of $| f |$ are finite, so the sections are integrable for almost every parameter. Decompose $f = f^{+} - f^{-}$ , apply Tonelli to $f^{+}$ and $f^{-}$ , and subtract the two finite identities to obtain the Fubini formula.

Lebesgue Measure on Products

Lebesgue measure on

ℝ^{m + n}

agrees with the product of Lebesgue measure on

ℝ^{m}

and Lebesgue measure on

ℝ^{n}

on product-measurable sets.

Both sides are measures on the product $σ$ -algebra and agree on rectangles, since the Lebesgue measure of a rectangle is the product of the side measures. Rectangles form a generating $π$ -system for the product $σ$ -algebra, and the measures are $σ$ -finite on countable unions of bounded rectangles. The uniqueness part of the product-measure construction therefore implies that the two measures agree on all product-measurable sets.

Measure Transformation under a Diffeomorphism

Suppose that

U, V \subseteq ℝ^{n}

are open,

Φ : U \to V

is a continuously differentiable bijection with continuously differentiable inverse, and

λ

is Lebesgue measure on

ℝ^{n}

. If

E \subseteq U

is measurable, then

λ (Φ (E)) = \int_{E} | \det {[D Φ (x)]}_{E} | d λ (x) .

Start with an invertible linear map $T : ℝ^{n} \to ℝ^{n}$ . For elementary linear maps the claim is direct: coordinate permutations and orthogonal maps preserve Euclidean volume, a shear preserves the volume of boxes, and scaling one coordinate by $c \neq 0$ multiplies volume by $| c |$ . Every invertible matrix is a product of these elementary maps, so for every measurable $E \subseteq ℝ^{n}$ ,

λ (T (E)) = | \det {[T]}_{E} | λ (E) .

Now let $Φ : U \to V$ be a $C^{1}$ -diffeomorphism. First work on a compact rectangle $K \subseteq U$ . Since $x \mapsto {[D Φ (x)]}_{E}$ is continuous, it is uniformly continuous on $K$ . Therefore, after subdividing $K$ into sufficiently small cubes, ${[D Φ (x)]}_{E}$ changes very little on each cube.

Fix one such cube $C$ and choose $a \in C$ . By differentiability,

Φ (x) = Φ (a) + D Φ (a) (x - a) + R_{a} (x), \frac{| R_{a} (x) |}{| x - a |} \to 0.

On a small cube this says that $Φ (C)$ is squeezed between small enlargements of the affine image

Φ (a) + D Φ (a) (C - a) .

By the linear case, that affine image has Lebesgue measure

| \det {[D Φ (a)]}_{E} | λ (C) .

Because the determinant is continuous and ${[D Φ (x)]}_{E}$ varies little on $C$ , this differs only slightly from

\int_{C} | \det {[D Φ (x)]}_{E} | d λ (x) .

Summing over the cubes in the subdivision and refining the subdivision gives

λ (Φ (K)) = \int_{K} | \det {[D Φ (x)]}_{E} | d λ (x) .

For a general measurable $E \subseteq U$ , write $E$ as an increasing union of bounded measurable pieces whose closures lie in $U$ , up to a null remainder. Applying the compact-rectangle result to these pieces and passing to the limit with the monotone convergence theorem gives the displayed measure transformation formula for $E$ .

Change of Variables for Lebesgue Integrals

Suppose that

U, V \subseteq ℝ^{n}

are open,

Φ : U \to V

is a continuously differentiable bijection with continuously differentiable inverse, and

λ

is Lebesgue measure on

ℝ^{n}

. If

f : V \to [0, \infty]

is measurable, then

\int_{V} f d λ = \int_{U} f (Φ (x)) | \det {[D Φ (x)]}_{E} | d λ (x) .

The integral formula follows from measure transformation under a diffeomorphism and the construction of the Lebesgue integral.

First suppose that $f = χ_{B}$ is the characteristic function of a measurable set $B \subseteq V$ . Apply measure transformation under a diffeomorphism to $E = Φ^{- 1} (B)$ . Then

\int_{V} χ_{B} d λ = λ (B) = λ (Φ (Φ^{- 1} (B))) = \int_{U} χ_{B} (Φ (x)) | \det {[D Φ (x)]}_{E} | d λ (x) .

By finite additivity and linearity, the same formula holds for every nonnegative simple function on $V$ . Finally, if $f : V \to [0, \infty]$ is measurable, choose an increasing sequence of nonnegative simple functions $s_{k}$ with $s_{k} \to f$ pointwise by approximation by simple functions. Applying the simple-function case to each $s_{k}$ and passing to the limit on both sides by the monotone convergence theorem proves the change of variables formula for $f$ .