11Product Measures, Fubini, Change of Variables

Proof. (a) Good sets: $\{E : E_x \in \mathcal B\ \forall x\}$ is a $\sigma$-algebra (sections commute with complements and countable unions) containing the rectangles. For $f$: $(f_x)^{-1}(B) = (f^{-1}(B))_x$. (b) ($\subseteq$) Rectangles of Borel sets: it suffices that open$\times$open boxes are Borel in $\R^{m+n}$ (they are open) and that general Borel rectangles are limits — good sets again: $\{A : A\times\R^n \in \mathcal B(\R^{m+n})\}$ is a $\sigma$-algebra containing the opens; intersect two such. ($\supseteq$) Every open set of $\R^{m+n}$ is a countable union of rational open boxes $U\times V$: contained in the product $\sigma$-algebra. ∎

Proof. Measurability of $x \mapsto \nu(E_x)$. First let $\nu$ be finite. The class $\mathcal D$ of $E$ for which the map is measurable contains the rectangles ($\nu((A\times B)_x) = \nu(B)\mathbf 1_A(x)$) and is a $\lambda$-system: for $E \subseteq F$ in $\mathcal D$, $\nu((F\setminus E)_x) = \nu(F_x) - \nu(E_x)$ (finiteness); for $E_n \uparrow E$, $\nu((E_n)_x) \uparrow \nu(E_x)$ (continuity from below), and monotone limits of measurable functions are measurable. Rectangles form a $\pi$-system: Dynkin (Theorem 9.4) gives $\mathcal D = \mathcal A\otimes\mathcal B$. If $\nu$ is $\sigma$-finite, write $Y = \bigcup Y_k$, $Y_k \uparrow$, $\nu(Y_k) < \infty$: $\nu(E_x) = \lim_k\nu_k(E_x)$ with $\nu_k = \nu(\cdot\cap Y_k)$ finite.

Measure. $\sigma$-additivity of $E \mapsto \int\nu(E_x)\dd\mu$ follows from Corollary 10.7 (sections of disjoint sets are disjoint). On rectangles it gives $\mu(A)\nu(B)$. Uniqueness: two candidates agree on the $\pi$-system of rectangles; $\sigma$-finiteness provides rectangles $X_k\times Y_k \uparrow X\times Y$ of finite measure: Theorem 9.7. Symmetry: the other-order construction is also a measure agreeing on rectangles — unique, hence the same. ∎

Proof. The standard machine. For $f = \mathbf 1_E$ this is Theorem 11.3 (and its symmetric form). By linearity it holds for simple $f \geq 0$. For general $f \geq 0$: take simple $s_n \nearrow f$ (Theorem 10.4); then $\int_Y(s_n)_x \dd\nu \nearrow \int_Y f_x\dd\nu$ for each $x$ (MCT in $Y$), so the left members converge by MCT in $X$, while $\int s_n\,\dd(\mu\otimes\nu) \nearrow \int f$ by MCT on the product. ∎

Proof. Tonelli applied to $\abs f$ shows $\varphi(x) = \int\abs{f_x} \dd\nu$ has finite integral, hence is finite a.e.: $f_x \in L^1(\nu)$ for a.e. $x$. Split $f = f^+ - f^-$ (real case; complex by components): Tonelli computes each iterated integral of $f^\pm$ as $\int f^\pm\dd(\mu\otimes\nu) < \infty$, and the a.e.-defined difference integrates to the difference. Symmetrically for the other order. ∎

Proof. Apply Tonelli to $\mathbf 1_{\{(x,t) : 0 < t < f(x)\}}$ on $X \times \intoo0{+\infty}$ (measurable: it is $\{(x,t): f(x) - t > 0\}\cap\{t > 0\}$, a Borel-type combination of the measurable $(x,t)\mapsto f(x) - t$): integrating in $t$ first gives $\int f\,\dd\mu$; in $x$ first, $\int_0^\infty\mu(f > t)\dd t$. For $f^p$: substitute $t = s^p$ in $\int\mu(f^p > t) \dd t$, i.e. apply the first formula to $f^p$ and change variables in the one-dimensional integral ($\{f^p > s^p\} = \{f > s\}$). ∎

Proof. $(x, y) \mapsto f(x-y)g(y)$ is measurable ($(x,y)\mapsto x - y$ is continuous; compose and multiply). Tonelli:

(translation invariance of $\lambda_d$ in the inner integral). So the double integral is finite; Fubini gives a.e. absolute convergence and the norm bound $\norm{f*g}_1 \leq \norm f_1\norm g_1$. Commutativity: substitute $y \mapsto x - y$ (translation and reflection invariance — reflection invariance holds on boxes, hence everywhere by uniqueness). Associativity: Tonelli–Fubini on a triple integral. ∎

Proof. The measure $\mu_T(A) = \lambda_d(T(A))$ is a Borel measure (homeomorphisms preserve Borel sets, Problem 9.1), translation-invariant ($T(A + x) = T(A) + Tx$), finite on the unit box: by the characterization of Lebesgue measure (Exercise 9.6, whose proof works verbatim in $\R^d$ with dyadic cubes), $\mu_T = c(T)\lambda_d$ with $c(T) = \lambda_d(T(\intco01^d))$. The map $T \mapsto c(T)$ is multiplicative ($c(ST) = c(S)c(T)$, by composing), so it suffices to compute $c$ on generators of $GL_d$: elementary matrices. Diagonal $\operatorname{diag}(a, 1, \dots, 1)$: maps the unit cube to a box of volume $\abs a$: $c = \abs a = \abs\det$. Transposition of coordinates: permutes the cube: $c = 1 = \abs\det$. Transvection $T(x) = x + \alpha x_2e_1$: the image of the unit cube is a sheared prism; by Tonelli its measure is $\int\lambda_1(\text{section})\dd x_2 \cdots \dd x_d = 1$, each $x_1$-section being an interval of length $1$: $c = 1 = \abs{\det}$. Every invertible matrix is a product of these (Gaussian elimination), and both $c$ and $\abs\det$ are multiplicative: $c(T) = \abs{\det T}$. The integral formula follows by the standard machine (indicators, simple, MCT). ∎

Proof. Write $J(x) = \abs{\det D\Phi(x)}$. The heart of the proof is the inequality

Step 4 below upgrades $(*)$ — applied to both $\Phi$ and $\Phi^{-1}$ — to the equality of the theorem. Note that $\Phi^{-1}$ is itself a $\mathcal C^1$ diffeomorphism with Jacobian $\abs{\det D\Phi^{-1}(y)} = J(\Phi^{-1}y)^{-1}$ (chain rule on $\Phi\circ\Phi^{-1} = \mathrm{id}$).

Step 1: $(*)$ for cubes with a distortion factor. Fix a closed cube $Q \subseteq U$ of center $x_0$ and side $2r$ (sup-norm ball). Claim: for every $\varepsilon > 0$, if $\Phi$ is differentiable on $Q$ with $\norm{D\Phi(x) - D\Phi(x_0)} \leq \varepsilon$ on $Q$ (operator norm for the sup-norm), then

because for $x \in Q$, the mean value inequality applied to $\Phi(x) - \Phi(x_0) - D\Phi(x_0)(x - x_0)$ gives $\norm{\Phi(x) - \Phi(x_0) - D\Phi(x_0)(x-x_0)}_\infty \leq \varepsilon\norm{x - x_0}_\infty \leq \varepsilon r$, and $D\Phi(x_0)^{-1}$ pulls this defect into an $\varepsilon\norm{D\Phi(x_0)^{-1}}\,r$-enlargement of the cube. By Theorem 11.10,

Step 2: $(*)$ for compact cubes, by subdivision. Let $Q \subseteq U$ be a compact cube and $\varepsilon > 0$. On $Q$, $D\Phi$ is uniformly continuous and $\norm{D\Phi^{-1}}$ bounded (compactness); subdivide $Q$ into $2^{kd}$ subcubes $Q_i$ small enough that the oscillation of $D\Phi$ on each is $\leq \varepsilon$. Step 1 on each subcube (center $x_i$):

the last step because $\sum_i\abs{\det D\Phi(x_i)}\mathbf 1_{Q_i} \to J$ uniformly on $Q$ (continuity of $\det D\Phi$) — Riemann-sum comparison. Let $\varepsilon \to 0$: $(*)$ holds for compact cubes.

Step 3: $(*)$ for all Borel $A$. The set function $A \mapsto \lambda_d(\Phi(A))$, on Borel subsets of $U$, is a measure ($\Phi$ is a bijection onto $V$ preserving Borel sets and countable disjointness), and so is $A \mapsto \int_AJ$. Every open subset of $U$ is a countable union of almost-disjoint dyadic compact cubes (standard dyadic decomposition: take maximal dyadic cubes contained in the open set), and both measures are additive across them (boundaries of cubes are $\lambda_d$-null, and their $\Phi$-images are null by Step 2 applied to thin cube-coverings of the faces): $(*)$ passes from cubes to open sets. General Borel $A$: exhaust $U$ by compacts $K_m \uparrow U$ with $K_m \subseteq \mathring K_{m+1}$, and fix $m$; on $\mathring K_{m+1}$, $J$ is bounded by some $M_m$. By outer regularity of $\lambda_d$ (proof as in Theorem 9.13, with boxes), choose open sets $O_n$ with $A \cap K_m \subseteq O_n \subseteq \mathring K_{m+1}$ and $\lambda_d\bigl(O_n \setminus (A\cap K_m)\bigr) \to 0$. Then

Let $m \to \infty$: continuity from below on the left, MCT on the right. This establishes $(*)$.

Step 4: equality and the integral formula. First extend $(*)$ from sets to integrals: for every measurable $g \geq 0$ on $V$,

Indeed, for $g = \mathbf 1_B$ this is $(*)$ with $A = \Phi^{-1}(B)$; linearity extends it to simple $g$, and MCT to all $g \geq 0$ (the standard machine). Now apply $(**)$ twice: first to $g$, then — for the diffeomorphism $\Phi^{-1}$ — to the function $x \mapsto g(\Phi(x))J(x)$:

since $J(\Phi^{-1}y)\abs{\det D\Phi^{-1}(y)} = \abs{\det\bigl( D\Phi(\Phi^{-1}y)\,D\Phi^{-1}(y)\bigr)} = 1$ (chain rule on $\Phi\circ\Phi^{-1} = \mathrm{id}$). All inequalities are equalities: the formula holds for $g \geq 0$, and for $L^1$ functions by decomposition. ∎

Proof. Compute $I = \int_{\R^d}\eu^{-\norm x_2^2}\dd\lambda_d$ twice. By Tonelli it factors: $I = G^d = \pi^{d/2}$. By the layer cake formula (Proposition 11.8) with $f = \eu^{- \norm x^2}$, whose level sets are balls: $\{f > t\} = B\bigl(0, \sqrt{-\ln t}\bigr)$ for $0 < t < 1$, of measure $v_d(-\ln t)^{d/2}$ (dilation by $\rho$ scales $\lambda_d$ by $\rho^d$: Theorem 11.10), so

Equate. (The values: $\Gamma(\frac32) = \frac{\sqrt\pi}2$, $\Gamma(2) = 1$, etc.) Note $v_d \to 0$ as $d \to \infty$ — the weekend problem quantifies how fast, via Stirling. ∎