Notes on Chapter 8 (Integration on Product Spaces) of Walter Rudin's Real and Complex Analysis.

This chapter is primarily concerned with Fubini's theorem and some of its applications. We begin with some preliminaries.

Preliminaries

Measurability on Cartesian Products

Let \((X, \mathscr{S})\), \((Y, \mathscr{T})\) be measurable spaces.

A couple of definitions:

  • A measurable rectangle is a set of the form \(A\times B\) with both \(A\) and \(B\) measurable.
  • The class \(\mathscr{E}\) of elementary sets consists of all finite unions of disjoint measurable rectangles.
  • \(\mathscr{S}\times\mathscr{T}\) is the smallest \(\sigma\)-algebra in \(X\times Y\) which contains every measurable rectangle.
  • A monotone class \(\mathfrak{M}\) is a collection of sets such that if \(A_i\in\mathfrak{M}\) and \(A_i\subset A_{i+1}\), then the countable union of the \(A_i\) is in \(\mathfrak{M}\), and where the opposite is true for the intersection of a descending chain of sets.
  • If \(E\subset X\times Y\), define the x- and y-section $$E_x=\{y:(x,y)\in E\},\qquad E^y=\{x:(x,y)\in E\}.$$
  • If \(f: X\times Y\rightarrow Z\), define \(f_x(y) = f(x, y) = f^y(x)\).

We can prove a few basic facts:

Facts about products:

  1. If \(E\in\mathscr{S}\times\mathscr{T}\) then \(E_x\in\mathscr{T}\) and \(E^y\in\mathscr{S}\) for any \(x\in X\) and \(y\in Y\).
  2. \(\mathscr{S}\times\mathscr{T}\) is the smallest monotone class which contains \(\mathscr{E}\).
  3. If \(f\) is \((\mathscr{S}\times\mathscr{T})\)-measurable then \(f_x\) is \(\mathscr{T}\)-measurable and \(f^y\) is \(\mathscr{S}\)-measurable for any \(x\in X\) and \(y\in Y\). □

Product Measures

We use the following theorem to justify our definition of product measures:

Well-definedness of product measures: Let \((X,\mathscr{S},\mu)\) and \((Y,\mathscr{T},\lambda)\) be \(\sigma\)-finite measure spaces. Suppose \(Q\in\mathscr{S}\times\mathscr{T}\). Define $$\varphi(x)=\lambda(Q_x),\qquad \psi(y)=\mu(Q^y)$$ then \(\varphi\) is \(\mathscr{S}\)-measurable, \(\psi\) is \(\mathscr{T}\)-measurable, and $$\int_X\varphi\ d\mu = \int_Y\psi\ d\lambda.$$ □

We show this by setting \(\Omega\) to be the class of all \(Q\in\mathscr{S}\times\mathscr{T}\) that satisfy the conclusion of the theorem. Let \(X = \bigcup\mu(X_n)\) and \(Y = \bigcup\lambda(Y_n)\) be disjoint unions of finite measure sets, using the \(\sigma\)-finiteness property.

Now define $$Q_{mn} = Q\cap(X_n\times Y_m)$$ and let \(\mathfrak{M}\) be the class of all \(Q\in\mathscr{S}\times\mathscr{T}\) such that \(Q_{mn}\in\Omega\) for all \(m\) and \(n\). We show that \(\mathscr{M}\) is monotonic and contains all elementary sets, hence \(\mathfrak{M}=\mathscr{S}\times\mathscr{T}\) (by fact 2 above).

Thus \(Q_{mn}\in\Omega\) for every \(Q\in\mathscr{S}\times\mathscr{T}\) and for all \(m\) and \(n\). We show that \(\Omega\) is closed under countable disjoint unions and we are done.

Note that we went through \(Q_{mn}\) because the proof (in Rudin) used the \(\sigma\)-finite property to guarantee an upper bound on the measure of the largest set in a descending chain of sets in \(\Omega\), which is necessary to use the dominated convergence theorem.

Hence we can define the product measure (on two \(\sigma\)-finite spaces) as $$(\mu\times\lambda)(Q) = \int_X\lambda(Q_x)\ d\mu(x) = \int_Y\mu(Q^y)\ d\lambda(y)\qquad (Q\in\mathscr{S}\times\mathscr{T}).$$

Fubini's Theorem

We can now show the main result of the chapter:

Fubini's Theorem: Let \((X, \mathscr{S}, \mu)\) and \((Y, \mathscr{T}, \lambda)\) be \(\sigma\)-finite measure spaces, and let \(f\) be a \((\mathscr{S}\times\mathscr{T})\)-measurable function on \(X\times Y\).

  • If \(0\leq f\leq \infty\), and if $$\varphi(x)=\int_Y f_x\ d\lambda,\qquad \psi(y)=\int_X f^y\ d\mu\qquad (x\in X, y\in Y),$$ then \(\varphi\) is \(\mathscr{S}\)-measurable, \(\psi\) is \(\mathscr{T}\)-measurable, and $$\int_X\varphi\ d\mu = \int_{X\times Y} f\ d(\mu\times\lambda) = \int_Y\psi\ d\lambda.$$
  • If \(f\) is complex and if $$\varphi^*(x)=\int_Y |f|_x\ d\lambda,\qquad \int_X \varphi^*\ d\mu < \infty,$$ then \(f\in L^1(\mu\times\lambda)\).
  • If \(f\in L^1(\mu\times\lambda)\), then \(f_x\in L^1(\lambda)\) for almost all \(x\in X\), \(f^y\in L^1(\mu)\) for almost all \(y\in Y\); the functions \(\varphi\) and \(\psi\) defined earlier a.e. are in \(L^1(\mu)\) and \(L^1(\lambda)\) respectively, and the equality still holds. □

This is proven using the previous theorem for non-negative simple functions, then extending it to the general case.

Counterexamples

Some counterexamples which occur when the conditions of the theorem are relaxed are instructive:

  • Let \(X=Y=[0,1]\) with the usual Lebesgue measure. Choose \(0 = \delta_1 < \delta_2 < \cdots, \delta_n \rightarrow 1\), and let \(g_n\) be a real continuous function with support in \((\delta_n, \delta_{n+1})\), s.t. \(\int_0^1 g_n(t)\ dt = 1\). Define $$f(x, y) = \sum_{n=1}^\infty [g_n(x) - g_{n+1}(x)] g_n(y).$$ We may observe that $$\int_0^1 \int_0^1 f(x, y)\ dy\ dx = 1 \neq 0 = \int_0^1 \int_0^1 f(x, y)\ dx\ dy,$$ so the conclusion of Fubini's theorem fails. Note that $$\int_0^1 \int_0^1 |f(x, y)|\ dy\ dx = \infty.$$
  • Let \(X=Y=[0,1]\) with the Lebesgue measure on \(X\) and the counting measure on \(Y\), the latter of which is not \(\sigma\)-finite. Put \(f(x, y) = 1\) if \(x = y\) and 0 otherwise. Then $$\int_Y\int_X f(x, y)\ d\mu(x)\ d\lambda(y) = 0 \neq 1 = \int_X\int_Y f(x, y)\ d\lambda(y)\ d\mu(x).$$ Note that our function \(f\) is indeed \((\mathscr{S}\times\mathscr{T})\)-measurable.
  • In both cases above, either the function or the space was 'too big'. We turn our focus to the measurability of \(f\). Once again, let \(X=Y=[0,1]\) with the Lebesgue measure, and assume the continuum hypothesis. There is a well-ordering \(j\) of \([0,1]\) s.t. each initial segment is countable. Let \(Q\) be the set of all \((x, y)\) in the unit square s.t. \(x\) precedes \(y\) in the well-ordering. Thus \(Q_x\) contains all but countably many points of \([0,1]\), and \(Q^y\) contains at most countably many points of \([0,1]\). Hence if \(f=\chi_Q\), both \(f_x\) and \(f^y\) are Borel-measurable and $$\int_0^1 \int_0^1 f(x, y)\ dy\ dx = 1 \neq 0 = \int_0^1 \int_0^1 f(x, y)\ dx\ dy.$$ Here both the function and the space are 'small', and \(f_x\) and \(f^y\) are measurable and the iterated integrals are finite, but \(f\) is not \((\mathscr{S}\times\mathscr{T})\)-measurable. Note that no reference to measurability w.r.t the product measure is needed to define the iterated integrals.

Completions of product measures

It is not necessarily the case that the product of two complete measures is complete. For instance, let \(A\) be any one point in \(\mathbb{R}\), and let \(B\) be any Lebesgue non-measurable set in \(\mathbb{R}\). Then \(A\times B\subset A\times \mathbb{R}\), the latter of which has measure 0. However \(A\times B\) is not \((m_1\times m_1)\)-measurable (because of Fact 1 above). Hence in particular, \(m_1\times m_1\) is not \(m_2\).

However, we do have the following:

Completion of products of Lebesgue measures: Let \(m_k\) denote the Lebesgue measure on \(\mathbb{R}^k\). Then \(m_{r+s}\) is the completion of the product measure \(m_r \times m_s\). □

In view of this, we have an alternative statement of Fubini's theorem:

Fubini's Theorem: Let \((X, \mathscr{S}, \mu)\) and \((Y, \mathscr{T}, \lambda)\) be \(\sigma\)-finite measure spaces, and let \(f\) be a \((\mathscr{S}\times\mathscr{T})^*\)-measurable function on \(X\times Y\), where the \(^*\) denotes the completion. Then all the previous statements of Fubini's theorem hold, except that the measurability of \(f_x\) and \(f^y\) can only be asserted a.e. (instead of everywhere). □

This is done by showing that for any \(\mathfrak{M}^*\)-measurable function there is a \(\mathfrak{M}\)-measurable function that is equal a.e., and if \(h\) is a \((\mathscr{S}\times\mathscr{T})^*\)-measurable function that is 0 a.e. with respect the product measure, then for almost all \(x\in X\), \(h(x, y) = 0\) for almost all \(y\), and \(h_x\) is measurable for almost all \(x\), similarly for \(h^y\).

Applications of Fubini's Theorem

We will quickly state some applications of Fubini's theorem without too much elaboration.

Convolutions: Suppose \(f\in L^1(\mathbb{R})\), \(g\in L^1(\mathbb{R})\). Then $$\int_{-\infty}^\infty |f(x-y)g(y)|\ dy < \infty$$ for almost all \(x\). For this \(x\) define $$h(x)=\int_{-\infty}^\infty |f(x-y)g(y)|\ dy < \infty.$$ Then \(h\in L^1(\mathbb{R})\), and $$\|h\|_1 \leq \|f\|_1\|g\|_1,$$ where $$\|f\|_1 = \int_{-\infty}^\infty |f(x)|\ dx.$$ □

Distribution functions: Suppose that \(f: X\rightarrow [0,\infty]\) is measurable and \(\mu\) is a \(\sigma\)-finite positive measure on \(X\). Let \(\varphi:[0,\infty]\rightarrow [0,\infty]\) is monotonic, absolutely continuous on \([0, T]\) for every \(T<\infty\), and that \(\varphi(0)=0\) and \(\varphi(t)\rightarrow\varphi(\infty)\) as \(t\rightarrow\infty\). Then $$\int_X (\varphi\circ f)\ d\mu = \int_0^\infty \mu\{f > t\}\varphi'(t)\ dt$$ where once again \(\mu\{f>t\} = \mu(\{x\in X: f(x) > t\})\). □

Here uniform continuity was used to use the fundamental theorem of calculus.

Let \(Mf\) denote the maximal function of \(f\). Recall that \(Mf\) lies in weak \(L^1\) when \(f\in L^1(\mathbb{R}^k)\). We also have that \(\|Mf\|_\infty \leq \|f\|_\infty\) for all \(f\in L^\infty(\mathbb{R}^k)\). As for the rest:

Maximal functions: If \(1 < p < \infty\) and \(f\in L^p(\mathbb{R}^k)\) then \(Mf\in L^p(\mathbb{R}^k)\). □

This uses the previous theorem to calculate some tail bounds.