Improper Iterated Integrals Without Absolute Convergence

Separate Limits Can Depend on Order

For an absolutely integrable function on a product measure space, Fubini's theorem guarantees that the two iterated integrals agree. Without absolute integrability, a different issue arises: even if every one-dimensional section has a finite integral, taking the limits in the two possible orders can give different answers. Rectangular cutoffs can depend on their shape as well. The example below makes these distinctions exact using only unit rectangles and values \(1\), \(-1\), and \(0\).

Let \(X=Y=[0,\infty)\), with the Borel sigma-algebras and Lebesgue measure \(m\) on each factor. We use the product Lebesgue measure \(m\times m\) on \([0,\infty)^2\), a sigma-finite measure space. For \(i,j\in\mathbb{N}=\{1,2,\ldots\}\), let \(Q_{ij}=[i-1,i)\times[j-1,j)\). These half-open rectangles partition \([0,\infty)^2\). Define the real-valued measurable function \(f\) by

f(x,y)= \begin{cases} 1,&(x,y)\in Q_{ii}\text{ for some }i,\\ -1,&(x,y)\in Q_{i,i+1}\text{ for some }i,\\ 0,&\text{otherwise}. \end{cases}

The half-open convention assigns every point to exactly one cell, including points on integer grid lines. Each nonzero cell has area \(1\). There are infinitely many cells where \(f=1\), and infinitely many where \(f=-1\). Thus \(\int |f|\,d(m\times m)=\infty\); in particular, \(f\) is not integrable on the product space. We will instead calculate its section integrals and its finite rectangular truncations.

Computing the Two Lebesgue Iterated Integrals

Definition: For a measurable real-valued function on \([0,\infty)^2\), its Lebesgue iterated integral in the \(y\)-then-\(x\) order means \(\int_0^\infty\bigl(\int_0^\infty f(x,y)\,dy\bigr)\,dx\), when the inner integrals exist as extended Lebesgue integrals and the resulting outer integral is defined. The other order is defined by integrating in \(x\) first. Separately truncated improper integrals are limits of integrals over finite rectangles; their limits need not be the same as each other or as a joint limit over all expanding rectangles.

Worked Example: Integrating Each Horizontal Section

Fix \(x\in[i-1,i)\). As a function of \(y\), \(f(x,y)\) equals \(1\) on \([i-1,i)\), equals \(-1\) on \([i,i+1)\), and is zero elsewhere. Each of these intervals has length \(1\). Therefore the section is absolutely integrable and

\int_0^\infty f(x,y)\,dy =1\cdot m([i-1,i))-1\cdot m([i,i+1)) =1-1=0.

This holds for every \(x\in[0,\infty)\), including integer \(x\), because the half-open intervals specify the section there. The resulting function of \(x\) is identically zero, so its Lebesgue integral is \(0\). Hence the \(y\)-then-\(x\) iterated integral exists and equals \(0\).

Worked Example: Integrating Each Vertical Section

Fix \(y\in[j-1,j)\). If \(j=1\), then \(f(x,y)=1\) for \(x\in[0,1)\) and is zero elsewhere. Its integral over \(x\) is \(1\). If \(j\ge2\), then \(f(x,y)=1\) on \([j-1,j)\), \(f(x,y)=-1\) on \([j-2,j-1)\), and is zero elsewhere. These two intervals each have length \(1\), so the section integral is \(1-1=0\). In every case the section is absolutely integrable.

Consequently, the function of \(y\) obtained by integrating in \(x\) is \(\chi_{[0,1)}(y)\). Its integral is

\int_0^\infty \chi_{[0,1)}(y)\,dy =m([0,1))=1.

The \(x\)-then-\(y\) iterated integral therefore exists and equals \(1\). The two Lebesgue iterated integrals are finite, but they disagree: they are \(0\) and \(1\).

Separate Limits and Rectangular Cutoffs

To compare these calculations with finite truncations, define \(I(M,N)=\int_{[0,M)\times[0,N)}f\,d(m\times m)\) for positive integers \(M,N\). The rectangle consists of \(M N\) unit cells. Among them, the diagonal cells \(Q_{ii}\) contribute \(+1\) when \(i\le \min(M,N)\), and the cells \(Q_{i,i+1}\) contribute \(-1\) when \(i\le M\) and \(i+1\le N\). Thus

I(M,N)=\min(M,N)-\min(M,N-1) = \begin{cases} 1,&M\ge N,\\ 0,&M<N. \end{cases}

This formula also verifies the separate limits. Fix a positive integer \(M\). For every \(N>M\), \(I(M,N)=0\), so \(\lim_{N\to\infty}I(M,N)=0\), and taking \(M\to\infty\) afterwards still gives \(0\). In the reverse order, fix \(N\). For every \(M\ge N\), \(I(M,N)=1\), so \(\lim_{M\to\infty}I(M,N)=1\); taking \(N\to\infty\) afterwards gives \(1\). These integer truncations reproduce the two iterated values.

Worked Example: Square Cutoffs and a Different Expanding Path

For a square cutoff \(M=N=k\), where \(k\) is any positive integer, the formula gives \(I(k,k)=1\). For the rectangular cutoff \(M=k-1,\ N=k\), where \(k\ge2\), it gives \(I(k-1,k)=0\). Both sequences of rectangles expand without bound as \(k\to\infty\), but the integrals have different limits. Therefore there is no joint limit of \(I(M,N)\) as \(M,N\to\infty\) without a restriction on how the two cutoffs approach infinity.

Check: at \(k=3\), the square \((M,N)=(3,3)\) contains three \(+1\) cells and two \(-1\) cells, so its integral is \(3-2=1\). The rectangle \((M,N)=(2,3)\) contains two \(+1\) cells and two \(-1\) cells, so its integral is \(2-2=0\). These counts agree with the formula in both cases.

Why Fubini Does Not Force Agreement

Theorem (Fubini, absolute-integrability form): On a product of sigma-finite measure spaces, if a measurable real-valued function \(g\) satisfies \(\int |g|\,d(\mu\times\nu)<\infty\), then its section integrals exist almost everywhere, are integrable, and the two iterated integrals agree with the product integral.

Our spaces are sigma-finite and \(f\) is measurable, but the required absolute-integrability hypothesis fails. Indeed, the \(k\) diagonal cells \(Q_{ii}\), \(1\le i\le k\), each contribute \(1\) to the integral of \(f^+\), so \(\int f^+\,d(m\times m)\ge k\) for every \(k\). The \(k\) cells \(Q_{i,i+1}\), \(1\le i\le k\), each contribute \(1\) to the integral of \(f^-\), so \(\int f^-\,d(m\times m)\ge k\) for every \(k\). Both integrals are infinite, as is \(\int|f|\,d(m\times m)\). Consequently, the product Lebesgue integral of the signed function is not defined as a finite signed integral. The finite iterated values arise from integrating sections, not from a finite product integral to which Fubini could be applied.

This distinction is important: “each section is integrable” does not imply “the function is integrable on the product space.” Nor does existence of both iterated integrals imply they agree. The absolute-integrability hypothesis in Fubini is what rules out order-dependent cancellation. In this example, each section has only one or two nonzero unit intervals, so section integrability is immediate; across the full plane, however, there are infinitely many positive and negative cells.

A Common Cancellation Error

A tempting but invalid calculation is to pair each positive diagonal cell with the negative cell immediately above it and conclude that the whole integral is \(0\). That pairing describes one truncation scheme: if the \(x\)-cutoff is held fixed and the \(y\)-cutoff is then taken sufficiently large, every included row contributes \(1-1=0\). It does not justify cancellation over the entire plane. For square cutoffs, the last diagonal cell has no matching negative cell inside the square, and the integral is \(1\). For \(M=k-1,\ N=k\), every positive cell in the rectangle is matched, and the integral is \(0\).

The correction is to specify the domain of integration before pairing terms. On each finite rectangle the integral is an ordinary finite sum and can be computed exactly. Only after that computation should a limit be taken, with the order or path of the limits stated. A cancellation argument over infinitely many cells without a fixed truncation has silently changed the summation or integration procedure.

The example gives a useful checklist for improper double integrals:

Check whether the sections have finite integrals, and compute the resulting section-integral functions.
Distinguish each iterated integral from a joint limit over expanding rectangles.
For finite truncations, state the shape of the cutoff and calculate before passing to a limit.
Before invoking Fubini, verify absolute integrability on the product space.

Check Your Understanding

Use the cell definition of \(f\) and the truncation formula to answer the following.

For \(x\in[4,5)\), identify the two intervals on which \(f(x,\cdot)\) is nonzero and compute its section integral.
For \(y\in[2,3)\), identify the two intervals on which \(f(\cdot,y)\) is nonzero and compute its section integral.
Compute \(I(4,3)\) and \(I(3,4)\). What do these values show about rectangular truncations?
Explain why the existence of both finite Lebesgue iterated integrals does not allow Fubini's theorem to be applied here.
For each positive integer \(k\), compute the number of positive and negative cells in \([0,k)\times[0,k)\), and use them to recover its integral.