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Fubini for Signed and Complex Integrands

See how to extend Fubini’s theorem to signed and complex integrands while keeping track of the absolute-integrability condition that makes the iterated integrals well defined.

Intermediate 10 min read

What You'll Learn

  • Split a real integrable function into its positive and negative parts.
  • State and prove Fubini’s theorem for signed functions using Tonelli’s theorem.
  • Identify the almost-everywhere conditions that make section integrals finite.
  • Extend the result to complex functions through their real and imaginary parts.
  • Compute signed and complex iterated integrals on continuous and discrete product spaces.
  • Recognize why absolute integrability prevents undefined differences of infinite quantities.

Why Signed and Complex Integrands Need Care

Tonelli’s theorem applies to nonnegative functions and permits the order of integration to be exchanged, even when the resulting integral is infinite. Fubini’s theorem for an integrable function gives a stronger conclusion: the iterated integrals exist as finite quantities almost everywhere and agree with the product-space integral.

For a signed or complex function, positive and negative contributions can cancel. That cancellation is meaningful only when the total absolute contribution is finite. The key condition is therefore integrability of the absolute value, not merely the existence of a cancellation in one chosen order.

Central principle: If a measurable signed or complex function is absolutely integrable on a product space, Fubini’s theorem applies. For signed functions, separate positive and negative parts; for complex functions, separate real and imaginary parts.

Positive and Negative Parts

Let \(f\) be a real-valued measurable function. Its positive and negative parts are the nonnegative functions

$$ f^+(x,y)=\max\{f(x,y),0\}, \qquad f^-(x,y)=\max\{-f(x,y),0\}. $$

At every point, \(f=f^+-f^-\) and \(|f|=f^++f^-\). Thus, if \(\int |f|\,d(\mu\times\nu)<\infty\), both \(f^+\) and \(f^-\) have finite integrals. Their integrals can be subtracted without encountering an expression of the form infinity minus infinity.

Definition: A real-valued measurable function \(f\) on \(X\times Y\) is integrable if \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). A complex-valued measurable function is integrable if the same condition holds for its complex modulus \(|f|\).

Here \(X\) and \(Y\) are measure spaces with measures \(\mu\) and \(\nu\), and \(\mu\times\nu\) is their product measure. We assume the measures are sigma-finite, as in the product-space form of Fubini’s theorem. The results below rely on Tonelli’s theorem for nonnegative functions, which gives measurability of section integrals and equality of the corresponding nonnegative integrals.

Fubini’s Theorem for Signed Functions

Theorem: Let \((X,\mathcal A,\mu)\) and \((Y,\mathcal B,\nu)\) be sigma-finite measure spaces. Suppose \(f:X\times Y\to\mathbb R\) is measurable and \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Then, for \(\mu\)-almost every \(x\), the section \(y\mapsto f(x,y)\) is integrable, and, for \(\nu\)-almost every \(y\), the section \(x\mapsto f(x,y)\) is integrable. The resulting section integrals are integrable measurable functions, and
$$ \int_{X\times Y} f\,d(\mu\times\nu) = \int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) = \int_Y\left(\int_X f(x,y)\,d\mu(x)\right)d\nu(y). $$
Proof. Apply Tonelli’s theorem to the nonnegative function \(|f|\). It gives
$$ \int_X\left(\int_Y |f(x,y)|\,d\nu(y)\right)d\mu(x) = \int_{X\times Y}|f|\,d(\mu\times\nu) <\infty, $$

and the same identity with the roles of \(X\) and \(Y\) exchanged. A nonnegative measurable function with finite integral is finite almost everywhere. Consequently, \(\int_Y |f(x,y)|\,d\nu(y)<\infty\) for almost every \(x\), and \(\int_X |f(x,y)|\,d\mu(x)<\infty\) for almost every \(y\). This proves the claims about integrable sections.

Tonelli’s theorem applied separately to \(f^+\) and \(f^-\) shows that their section integrals are measurable and that

$$ \int_X\left(\int_Y f^\pm(x,y)\,d\nu(y)\right)d\mu(x) = \int_{X\times Y}f^\pm\,d(\mu\times\nu) <\infty. $$

The finiteness follows from \(0\le f^\pm\le |f|\). Thus, for almost every \(x\), both inner integrals of \(f^+\) and \(f^-\) are finite. Their difference is the section integral \(\int_Y f(x,y)\,d\nu(y)\) wherever the section is integrable. This difference is measurable almost everywhere, and it is integrable because its absolute value is at most \(\int_Y |f(x,y)|\,d\nu(y)\) wherever that quantity is finite. Integrating the difference and using the two Tonelli identities gives

$$ \int_X\left(\int_Y f(x,y)\,d\nu(y)\right)d\mu(x) = \int_{X\times Y}f^+\,d(\mu\times\nu) - \int_{X\times Y}f^-\,d(\mu\times\nu) = \int_{X\times Y}f\,d(\mu\times\nu). $$

The same argument with \(X\) and \(Y\) exchanged proves the other iterated-integral equality. Values assigned to the section integrals on their exceptional null sets do not affect these integrals. \(\square\)

Worked Example: A Signed Function on the Unit Square

Worked Example: A Signed Function on the Unit Square

Let \(f(x,y)=x-y\) on \([0,1]^2\) with Lebesgue measure. Since \(|x-y|\le 1\) and the square has area \(1\), \(\int_{[0,1]^2}|f|\,dx\,dy\le 1\). Fubini’s theorem therefore applies.

For fixed \(x\), integrate first with respect to \(y\):

$$ \int_0^1(x-y)\,dy=x-\frac12, \qquad \int_0^1\left(x-\frac12\right)dx=0. $$

For fixed \(y\), the other order gives

$$ \int_0^1(x-y)\,dx=\frac12-y, \qquad \int_0^1\left(\frac12-y\right)dy=0. $$

The positive part is supported where \(x>y\), and the negative part where \(y>x\). Direct integration confirms that each has integral \(1/6\):

$$ \int_0^1\int_0^x(x-y)\,dy\,dx = \int_0^1\frac{x^2}{2}\,dx = \frac16, \qquad \int_0^1\int_0^y(y-x)\,dx\,dy = \int_0^1\frac{y^2}{2}\,dy = \frac16. $$

Thus the integral of \(f\) is \(1/6-1/6=0\). Here the cancellation is legitimate because both positive and negative parts have finite integrals.

Complex Functions: Real and Imaginary Parts

For a complex-valued measurable function \(f\), write \(f=u+iv\), where \(u=\operatorname{Re}f\) and \(v=\operatorname{Im}f\) are real-valued measurable functions. The inequalities \(|u|\le |f|\) and \(|v|\le |f|\) show that if \(f\) is integrable, then both components are integrable. Define its integral by

$$ \int f\,d(\mu\times\nu) = \int u\,d(\mu\times\nu) + i\int v\,d(\mu\times\nu). $$
Theorem: Under the sigma-finiteness assumptions above, if \(f:X\times Y\to\mathbb C\) is measurable and \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\), then the sections of \(f\) are integrable almost everywhere in either order, their section integrals are measurable and integrable, and both iterated integrals equal \(\int_{X\times Y}f\,d(\mu\times\nu)\).

Proof. Apply the signed-function theorem to \(u=\operatorname{Re}f\) and \(v=\operatorname{Im}f\), which are integrable because their absolute values are bounded by \(|f|\). Outside the union of the two exceptional null sets, both section integrals exist, and the complex section integral is the sum of the real section integral and \(i\) times the imaginary section integral. The signed theorem gives the claimed equality for each component, so combining the two equalities gives it for \(f\). The same reasoning applies with the integration order reversed. \(\square\)

Worked Example: A Complex Integral on the Positive Quadrant

Worked Example: A Complex Integral on the Positive Quadrant

On \([0,\infty)^2\), let \(f(x,y)=e^{-x-y}\bigl(1+i(x-y)\bigr)\). Since

$$ |f(x,y)| = e^{-x-y}\sqrt{1+(x-y)^2} \le e^{-x-y}(1+x+y), $$

and the integral of the bound is \(1+1+1=3\), \(f\) is absolutely integrable. Integrating first in \(y\) gives

$$ \int_0^\infty f(x,y)\,dy = e^{-x}\left(1+i(x-1)\right), \qquad \int_0^\infty e^{-x}\left(1+i(x-1)\right)dx = 1+i(1-1)=1. $$

In the opposite order,

$$ \int_0^\infty f(x,y)\,dx = e^{-y}\left(1+i(1-y)\right), \qquad \int_0^\infty e^{-y}\left(1+i(1-y)\right)dy = 1+i(1-1)=1. $$

The two iterated integrals agree, and the imaginary contribution vanishes because the two terms \(x\) and \(y\) have equal integrals under the symmetric product weight.

Worked Example: An Absolutely Summable Complex Array

Worked Example: An Absolutely Summable Complex Array

Give \(\mathbb N\times\mathbb N\) counting measure, and define \(a_{mn}=i(-1)^m2^{-m-n}\). Its absolute sum is finite:

$$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|a_{mn}| = \left(\sum_{m=1}^{\infty}2^{-m}\right) \left(\sum_{n=1}^{\infty}2^{-n}\right) =1. $$

For fixed \(m\), summing over \(n\) first gives \(\sum_{n=1}^{\infty}a_{mn}=i(-1)^m2^{-m}\), because \(\sum_{n=1}^{\infty}2^{-n}=1\). Therefore

$$ \sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}a_{mn}\right) = i\sum_{m=1}^{\infty}\left(-\frac12\right)^m = -\frac{i}{3}. $$

For fixed \(n\), summing over \(m\) first gives \(\sum_{m=1}^{\infty}a_{mn}=i2^{-n}\sum_{m=1}^{\infty}(-1/2)^m=-i2^{-n}/3\). Hence

$$ \sum_{n=1}^{\infty}\left(\sum_{m=1}^{\infty}a_{mn}\right) = -\frac{i}{3}\sum_{n=1}^{\infty}2^{-n} = -\frac{i}{3}. $$

Both orders give the same finite complex sum. The absolute-sum calculation verifies the hypothesis that justifies exchanging the order.

Why the Absolute-Integrability Condition Matters

The signed theorem is not a rule that permits arbitrary rearrangement of positive and negative contributions. Its proof depends on applying Tonelli’s theorem to \(f^+\) and \(f^-\) and then subtracting two finite quantities. If either part has infinite integral, that subtraction may be undefined, even if a particular order of integration seems to produce a finite answer.

A useful check before using Fubini on a signed or complex integrand is to verify \(\int |f|<\infty\). For a complex function, this means checking the modulus, not merely checking that the real and imaginary parts appear to cancel in an iterated calculation. Absolute integrability guarantees that almost every section is integrable, that the section integrals are measurable, and that the two orders produce the same value.

Check Your Understanding

Use the definitions and theorems in this tutorial to answer the following questions.

  1. How are \(f^+\) and \(f^-\) defined, and how do they express \(f\) and \(|f|\)?
  2. Which application of Tonelli’s theorem shows that almost every section of an integrable signed function is integrable?
  3. Why can the integrals of \(f^+\) and \(f^-\) be subtracted under the theorem’s assumptions?
  4. How does absolute integrability of a complex function imply integrability of its real and imaginary parts?
  5. For the complex array in the final worked example, what is the value of the sum of the absolute values of all entries?

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