Sigma-Finite Exhaustions in Tonelli's Theorem

Why Exhaustions Are Needed

On a finite product space, the nonnegative integral identity can be established by first treating simple functions and then passing to general nonnegative measurable functions. For sigma-finite spaces, the whole product need not have finite measure, so the finite-measure identity cannot be applied directly to the entire space. The remedy is to approximate each factor by an increasing sequence of measurable sets of finite measure. Their products are finite-measure rectangles, and their union is the full product space.

Throughout, let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be sigma-finite measure spaces. Sigma-finiteness means that each space is a countable union of measurable sets of finite measure. Write $\mathcal{M}\otimes\mathcal{N}$ for the product sigma-algebra and $\mu\times\nu$ for the product measure. Let $f:X\times Y\to[0,\infty]$ be $\mathcal{M}\otimes\mathcal{N}$-measurable. Integrals below are nonnegative Lebesgue integrals; they may have value $+\infty$. The finite-measure nonnegative integral identity established earlier says that when both factor measures are finite, the integral over the product equals either iterated integral.

Definition: An exhaustion of a measurable space $X$ by finite-measure sets is a sequence $A_1\subseteq A_2\subseteq\cdots$ of measurable sets such that $\mu(A_n)<\infty$ for every $n$ and $\bigcup_{n=1}^{\infty}A_n=X$. An exhaustion of $Y$ is defined in the same way, using $\nu$.

Such exhaustions exist on every sigma-finite measure space. Indeed, if $X=\bigcup_{k=1}^{\infty}E_k$ with $E_k\in\mathcal{M}$ and $\mu(E_k)<\infty$, define $A_n=\bigcup_{k=1}^{n}E_k$. A finite union of finite-measure sets has finite measure, so $\mu(A_n)\le\sum_{k=1}^{n}\mu(E_k)<\infty$. The sets are increasing, and their union is $X$. The same construction gives an exhaustion $(B_n)$ of $Y$.

Finite Rectangles Cover the Product

Proposition: For increasing finite-measure exhaustions $(A_n)$ of $X$ and $(B_n)$ of $Y$, the rectangles $R_n=A_n\times B_n$ increase to $X\times Y$. Each rectangle has finite product measure, with $(\mu\times\nu)(R_n)=\mu(A_n)\nu(B_n)<\infty$.

Proof. If $(x,y)\in R_n$, then $x\in A_n\subseteq A_{n+1}$ and $y\in B_n\subseteq B_{n+1}$, so $(x,y)\in R_{n+1}$. Thus $R_n\subseteq R_{n+1}$. If $(x,y)\in X\times Y$, the exhaustion property gives indices $r,s$ with $x\in A_r$ and $y\in B_s$. For $n=\max(r,s)$, the increasing property gives $x\in A_n$ and $y\in B_n$, hence $(x,y)\in R_n$. Therefore $\bigcup_nR_n=X\times Y$. The rectangle formula for product measure gives $(\mu\times\nu)(R_n)=\mu(A_n)\nu(B_n)$; both factors are finite, so the product is finite. $\square$

This argument uses synchronized indices: even if $x$ and $y$ first enter their respective exhaustions at different stages, both belong to the rectangle with the larger index. For nonnegative $f$, define its restriction to $R_n$ by $f_n=f$ on $R_n$ and $f_n=0$ outside $R_n$. Then $f_n$ is measurable, $0\le f_n\le f_{n+1}$, and $f_n\to f$ pointwise everywhere. The Monotone Convergence Theorem applies because the functions are measurable, nonnegative, and increasing. It gives

\lim_{n\to\infty}\int_{R_n}f\,d(\mu\times\nu) =\lim_{n\to\infty}\int_{X\times Y}f_n\,d(\mu\times\nu) =\int_{X\times Y}f\,d(\mu\times\nu).

No finiteness assumption on the integral of $f$ is needed. If the full integral is infinite, the restricted integrals increase without bound.

Worked Example: Restricted Integrals on the Positive Quadrant

Take $X=Y=[0,\infty)$, each with its Borel sigma-algebra and Lebesgue measure. These spaces are sigma-finite: $A_n=B_n=[0,n]$ are finite-measure exhaustions. Let $f(x,y)=e^{-x-y}$, a nonnegative measurable function. Its restriction to $R_n=[0,n]\times[0,n]$ has integral

\int_{R_n}f\,d(m\times m) =\left(\int_0^n e^{-x}\,dx\right)\left(\int_0^n e^{-y}\,dy\right) =(1-e^{-n})(1-e^{-n}) =(1-e^{-n})^2.

The finite-measure identity justifies the product of the two one-dimensional integrals. Since $1-e^{-n}$ increases to $1$, the restricted integrals increase to $1$. Monotone Convergence also gives $\int_{[0,\infty)^2}f\,d(m\times m)=1$. The full iterated integral has the same value: for each $x\ge0$, $\int_0^\infty e^{-x-y}\,dy=e^{-x}$, and then $\int_0^\infty e^{-x}\,dx=1$. Check: for $n=1$, direct substitution gives the restricted value $(1-e^{-1})^2$; as $n\to\infty$, the same formula tends to $1$, agreeing with the direct iterated calculation.

Increasing Limits for Iterated Integrals

The product integrals are only part of the exhaustion argument. We also need to see that the iterated integrals over the rectangles increase to the full iterated integrals. For $x\in X$, write $f_x(y)=f(x,y)$. Measurability of sections ensures that $f_x$ is $\mathcal{N}$-measurable. The earlier result on nonnegative section integrals ensures that the functions of $x$ obtained by integrating these sections are measurable.

Proposition: With the sigma-finite spaces, exhaustions, and nonnegative measurable $f$ above, set $$ I_n=\int_{A_n}\left(\int_{B_n}f(x,y)\,d\nu(y)\right)d\mu(x). $$ Then $I_n$ increases to $\int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x)$. The corresponding statement holds with the order of integration reversed.

Proof. Define $g_n(x)=\int_{B_n}f_x\,d\nu$ and $g(x)=\int_Y f_x\,d\nu$. For each fixed $x$, the functions $f_x\chi_{B_n}$ are measurable, nonnegative and increasing to $f_x$ everywhere, since $B_n\uparrow Y$. The Monotone Convergence Theorem on $(Y,\mathcal{N},\nu)$ gives $g_n(x)\uparrow g(x)$, allowing infinite values. Now let $h_n(x)=g_n(x)$ for $x\in A_n$ and $h_n(x)=0$ otherwise. The sequence $(h_n)$ is increasing: on $A_n$, $g_{n+1}\ge g_n$, and outside $A_n$, $h_n=0\le h_{n+1}$. For every $x$, there is an index after which $x\in A_n$; from then on $h_n(x)=g_n(x)$, which increases to $g(x)$. Thus $h_n\uparrow g$ everywhere. Applying Monotone Convergence on $(X,\mathcal{M},\mu)$ gives $\lim_n I_n=\int_X g\,d\mu$. Interchanging the roles of $X,\mu,A_n$ and $Y,\nu,B_n$ proves the reversed-order statement. $\square$

On each finite rectangle, the finite-measure nonnegative integral identity applies to the restricted function $f\chi_{R_n}$. Consequently its product integral equals both restricted iterated integrals. The preceding proposition and Monotone Convergence show that all three increasing sequences tend to their respective full integrals. This explains how finite-measure identities pass to sigma-finite spaces without assuming that the full product measure is finite.

Worked Example: A Diagonal Under Counting Measure

Let $X=Y=\mathbb{N}=\{1,2,\ldots\}$, with counting measure on each factor, and let $f(i,j)=1$ when $i=j$, and $f(i,j)=0$ otherwise. Counting measure is sigma-finite because each singleton has measure $1$. Choose $A_n=B_n=\{1,\ldots,n\}$. The rectangle contains exactly the $n$ diagonal points $(1,1),\ldots,(n,n)$, so

\int_{A_n\times B_n}f\,d(\mu\times\nu) =\sum_{i=1}^{n}\sum_{j=1}^{n}\chi_{\{i=j\}} =\sum_{i=1}^{n}1 =n.

For the first iterated integral, each $i\in A_n$ has exactly one matching $j\in B_n$, namely $j=i$. Thus the inner sum is $1$ for each of the $n$ choices of $i$, giving $I_n=n$. The restricted integrals therefore increase to $+\infty$. On the full space, each fixed $i$ still has inner sum $1$, and summing that value over all $i\in\mathbb{N}$ gives $+\infty$. Check: the full product integral also counts the infinitely many diagonal points, so it is $+\infty$, as is the reversed iterated sum.

Worked Example: Making an Exhaustion from Unit Intervals

Let $X=\mathbb{R}$ with Lebesgue measure, $Y=[0,1]$ with Lebesgue measure, and $f(x,y)=1$. The intervals $[k,k+1)$, $k\in\mathbb{Z}$, form a countable finite-measure cover of $\mathbb{R}$. Enumerate their indices as $k_1=0$, $k_2=1$, $k_3=-1$, $k_4=2$, $k_5=-2,\ldots$, and set $A_n=\bigcup_{r=1}^{n}[k_r,k_r+1)$, $B_n=[0,1]$. These are increasing; each $A_n$ is a union of $n$ disjoint intervals of length $1$, so $m(A_n)=n$, while $m(B_n)=1$. The restricted integral is therefore

\int_{A_n\times B_n}1\,d(m\times m) =m(A_n)m(B_n)=n\cdot1=n.

Because the intervals cover $\mathbb{R}$, the rectangles increase to $\mathbb{R}\times[0,1]$. The restricted product and iterated integrals tend to $+\infty$, which agrees with the full product measure $m(\mathbb{R})m([0,1])=+\infty$. Check: for $n=3$, $A_3=[-1,0)\cup[0,1)\cup[1,2)$, a disjoint union of three unit intervals, so the restricted integral is $3$, matching the formula.

A Common Exhaustion Pitfall

A countable cover is not automatically an increasing exhaustion. In the unit-interval example, taking the enumerated intervals themselves as the sequence would give $E_n=[k_n,k_n+1)$, which covers $\mathbb{R}$ but is not increasing. For $f=1$ on $\mathbb{R}\times[0,1]$, the restricted integral over $E_n\times[0,1]$ is $1$ for every $n$. It would be wrong to conclude that these integrals converge to the full integral $+\infty$: the indicators $\chi_{E_n\times[0,1]}$ do not increase pointwise to $1$. Most points leave the restricted sets at every later stage.

The correction is to take finite unions: $A_n=E_1\cup\cdots\cup E_n$. Then $A_n$ increases, still has finite measure, and covers $\mathbb{R}$. The restricted integral becomes $n$, which does increase to the full value $+\infty$. Monotone Convergence requires pointwise increase, not merely that the union of the sets is the whole space. Another necessary detail is to exhaust both factors: rectangles $A_n\times B_n$ cover $X\times Y$ only when each factor sequence covers its own space.

Check Your Understanding

In each question, identify the relevant exhaustion and state why Monotone Convergence applies, or identify the missing condition.

If $X=\bigcup_{k\ge1}E_k$ and every $E_k$ has finite measure, define an increasing finite-measure exhaustion of $X$ and verify both properties.
Why does $A_n\times B_n$ cover $X\times Y$ when both factor sequences are increasing exhaustions, even if a point enters the two factor sequences at different indices?
For counting measure on $\mathbb{N}^2$, use $A_n=B_n=\{1,\ldots,n\}$ and $f(i,j)=\chi_{\{i=j\}}$ to compute the restricted integral and its limit.
Explain why a sequence of sets can have union $X$ and still fail to be an exhaustion suitable for Monotone Convergence.
For a nonnegative measurable $f$, which theorem justifies the limit of the restricted product integrals, and which theorem justifies the limit of the restricted section integrals?

Sigma-Finite Exhaustions in Tonelli's Theorem

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