From Real to Complex Integration
For a real-valued integrable function, Fubini’s theorem identifies the integral over a product space with either order of iterated integration. A complex-valued function has real and imaginary parts, so the same conclusion should hold by applying the real theorem twice. The condition that makes this argument valid is absolute integrability: the integral of the modulus must be finite. It guarantees that almost every section is integrable and that the section integrals themselves are integrable.
Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces. Their product measure is denoted by \(\mu\times\nu\), on \(\mathcal{M}\otimes\mathcal{N}\). A complex-valued measurable function \(f:X\times Y\to\mathbb{C}\) is integrable if \(\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Writing \(f=u+iv\), where \(u\) and \(v\) are real-valued, its integral is defined by \(\int f\,d(\mu\times\nu)=\int u\,d(\mu\times\nu)+i\int v\,d(\mu\times\nu)\). Both real integrals are finite because \(|u|\le |f|\) and \(|v|\le |f|\).
We use Tonelli’s theorem for nonnegative measurable functions and the real-valued Fubini theorem established earlier. In the latter, an exceptional measurable null set of fibers may be assigned the value zero when forming the measurable section-integral function. That convention matters on measure spaces that are not complete: arbitrary changes on a null set need not preserve measurability.
Complex-Valued Fubini
Proof. Put \(g(x)=\int_Y|f_x|\,d\nu\), allowing \(+\infty\). Tonelli’s theorem says that \(g\) is measurable and \(\int_Xg\,d\mu=\int_{X\times Y}|f|\,d(\mu\times\nu)<\infty\). Therefore \(A=\{x:g(x)=\infty\}\) is measurable and \(\mu(A)=0\). Indeed, a nonnegative measurable function with finite integral is finite almost everywhere. Applying Tonelli in the other order gives a measurable null set \(B=\{y:\int_X|f^y|\,d\mu=\infty\}\) with \(\nu(B)=0\). Product-measurable functions have measurable sections, so each section of \(f\) is measurable; outside the respective exceptional set, its absolute integral is finite and hence the complex section integral exists.
We verify measurability without making arbitrary assignments on null sets. Write \(u=\operatorname{Re}f\) and \(v=\operatorname{Im}f\). By Tonelli, the functions \(P_u(x)=\int_Yu_x^+\,d\nu\) and \(N_u(x)=\int_Yu_x^-\,d\nu\) are measurable extended-real-valued functions. On \(A^c\), both are finite, since \(u_x^+,u_x^-\le |f_x|\). Thus the function equal to \(P_u-N_u\) on \(A^c\) and zero on \(A\) is measurable: it is a difference of finite measurable functions on the measurable set \(A^c\), extended by a fixed value on \(A\). It is precisely \(\operatorname{Re}F\). The same argument using \(v^+\) and \(v^-\) proves that \(\operatorname{Im}F\) is measurable. Moreover, \(|F(x)|\le g(x)\) on \(A^c\), and \(F=0\) on \(A\), so \(\int_X|F|\,d\mu\le\int_Xg\,d\mu<\infty\). Reversing the roles of \(X\) and \(Y\) proves that \(G\) is measurable and integrable.
Now \(u\) and \(v\) are real-valued integrable functions on the product space, since \(|u|,|v|\le|f|\). Apply real-valued Fubini separately to \(u\) and \(v\), using the zero assignments on \(A\). This gives \(\int_{X\times Y}u\,d(\mu\times\nu)=\int_X\operatorname{Re}F\,d\mu\) and \(\int_{X\times Y}v\,d(\mu\times\nu)=\int_X\operatorname{Im}F\,d\mu\). Combining these equalities proves \(\int_{X\times Y}f\,d(\mu\times\nu)=\int_XF\,d\mu\). Applying real Fubini in the other order to \(u\) and \(v\), with zero assignment on \(B\), proves equality with \(\int_YG\,d\nu\). \(\square\)
Worked Example: A Finite Complex Matrix
Take \(X=Y=\{1,2\}\), each with counting measure, and define \(f\) by the matrix \(\begin{pmatrix}1+i&-i\\2&-1+i\end{pmatrix}\). Both spaces are finite, and every function is measurable and integrable.
Integrating by rows gives the section integrals \((1+i)+(-i)=1\) and \(2+(-1+i)=1+i\). Their sum is \(1+(1+i)=2+i\). Integrating by columns gives \((1+i)+2=3+i\) and \((-i)+(-1+i)=-1\). Their sum is \((3+i)+(-1)=2+i\). The product integral is also the sum of the four entries: \((1+i)+(-i)+2+(-1+i)=2+i\).
The integral of the modulus is the sum of the four entrywise moduli: \(|1+i|+|-i|+|2|+|-1+i|=\sqrt{2}+1+2+\sqrt{2}=3+2\sqrt{2}\). Thus \(\left|\int f\right|=|2+i|=\sqrt{2^2+1^2}=\sqrt{5}\), and \(\sqrt{5}\le3+2\sqrt{2}\). Check: both orders of summation gave \(2+i\), and the modulus bound compares \(\sqrt{5}\) with the independently computed sum \(3+2\sqrt{2}\).
The Modulus Bound
Proof. Write \(z=\int_Xh\,d\mu\). If \(z=0\), the inequality follows from \(0\le\int_X|h|\,d\mu\). If \(z\ne0\), let \(\alpha=\overline{z}/|z|\); then \(|\alpha|=1\) and \(\alpha z=|z|\). For every \(x\), \(\operatorname{Re}(\alpha h(x))\le|\alpha h(x)|=|h(x)|\). The real-valued functions on both sides are integrable, so monotonicity of the real integral and linearity give \[ |z|=\operatorname{Re}(\alpha z) =\int_X\operatorname{Re}(\alpha h)\,d\mu \le\int_X|h|\,d\mu. \] This proves the claim in both cases. \(\square\)
Applying the proposition to the product-space function in Complex-Valued Fubini yields \(\left|\int_{X\times Y}f\,d(\mu\times\nu)\right|\le\int_{X\times Y}|f|\,d(\mu\times\nu)\). Applying it to the section functions also gives the useful estimates \[ |F(x)|\le\int_Y|f_x|\,d\nu\quad(x\notin A),\qquad |G(y)|\le\int_X|f^y|\,d\mu\quad(y\notin B). \] On the exceptional sets the left sides are zero by definition, so the inequalities hold there as well. Integrating them and using Tonelli gives \(\int_X|F|\,d\mu\le\int_{X\times Y}|f|\,d(\mu\times\nu)\) and \(\int_Y|G|\,d\nu\le\int_{X\times Y}|f|\,d(\mu\times\nu)\).
Worked Example: An Oscillatory Function with Strict Inequality
On \(X=Y=[0,1]\) with Lebesgue measure, set \(f(x,y)=e^{2\pi i(x+y)}\). This function is continuous, and \(|f(x,y)|=1\), so \(\int_{[0,1]^2}|f|\,d(m\times m)=1<\infty\). For every fixed \(x\), \[ \int_0^1 f(x,y)\,dy=e^{2\pi ix}\int_0^1e^{2\pi iy}\,dy =e^{2\pi ix}\left[\frac{e^{2\pi iy}}{2\pi i}\right]_{y=0}^{y=1} =e^{2\pi ix}\frac{1-1}{2\pi i}=0. \] Here \(e^{2\pi i}=1\). Thus the iterated integral is zero. Integrating first in \(x\) gives the same result, since \(\int_0^1 e^{2\pi ix}\,dx=(1-1)/(2\pi i)=0\).
By Fubini the product integral is zero, while the integral of the modulus is \(\int_{[0,1]^2}1\,d(m\times m)=1\cdot1=1\). Hence \(0=\left|\int f\right|<\int|f|=1\). Check: direct integration in either variable gives zero, and the rectangle has area one, confirming the modulus integral.
Worked Example: Piecewise-Constant Complex Sections
On \([0,1]^2\) with product Lebesgue measure, define \(f(x,y)=1+i\) when \(0\le x<1/2\), and \(f(x,y)=1-i\) when \(1/2\le x\le1\). The strict inequality in the first case and the weak inequality in the second make the pieces disjoint and cover the whole square. The dividing line has product measure zero, but the assigned values there are specified as well. The function is measurable and bounded, hence integrable.
For each fixed \(y\), the section integral is \[ \int_0^1 f(x,y)\,dx=(1+i)\frac12+(1-i)\frac12 =\frac{1+i+1-i}{2}=1. \] Integrating this section integral over \(y\in[0,1]\) gives \(1\cdot1=1\). The direct product integral gives the same result: the two regions each have area \((1/2)\cdot1=1/2\), so \[ \int_{[0,1]^2}f\,d(m\times m) =(1+i)\frac12+(1-i)\frac12=1. \]
On both regions the modulus is \(\sqrt{1^2+1^2}=\sqrt{2}\). Therefore \(\int_{[0,1]^2}|f|\,d(m\times m)=\sqrt{2}(1/2+1/2)=\sqrt{2}\), while \(\left|\int f\right|=|1|=1\). The estimate is \(1\le\sqrt{2}\). Check: the real parts add to \(1\), the imaginary parts cancel, and the two region areas add to \(1\).
Exceptional Fibers and a Common Pitfall
The zero assignment in the theorem is a deliberate measurability choice, not a claim that values on a null set are always irrelevant to measurability. The exceptional set \(A=\{x:\int_Y|f_x|\,d\nu=\infty\}\) is measurable and null. Setting \(F=0\) on \(A\) preserves measurability because \(A\) is measurable. On an incomplete measure space, replacing those zero values by arbitrary values may produce a nonmeasurable function: the preimage of a Borel set could become a nonmeasurable subset of \(A\). If the spaces are complete, arbitrary assignments on measurable null sets are measurable, but completeness is not assumed in the theorem.
A second common mistake is to replace the integral estimate by equality. The oscillatory example shows why that fails: its values have modulus one everywhere, but their phases cancel in the integral. The valid conclusion is the inequality \(\left|\int h\,d\mu\right|\le\int|h|\,d\mu\), not equality in general. Absolute integrability is essential to the stated complex Fubini theorem: it ensures finite section integrals almost everywhere and makes both iterated integrals well-defined and finite. The inequality itself applies once \(h\) is integrable; it does not by itself justify exchanging the order of integration.
Check Your Understanding
For each question, specify the measure space and verify the hypotheses before applying a result.
- For a complex-valued integrable function \(h\), explain why both \(\operatorname{Re}h\) and \(\operatorname{Im}h\) are integrable.
- In the complex Fubini theorem, why is the set of \(x\) for which \(\int_Y|f_x|\,d\nu=\infty\) a measurable null set?
- For \(f(x,y)=e^{2\pi i(x+y)}\) on \([0,1]^2\), compute both iterated integrals and the integral of \(|f|\).
- Explain why setting section integrals equal to zero on an exceptional measurable null set is safe even when the measure space is not complete.
- Give the phase factor used in the proof of \(\left|\int h\,d\mu\right|\le\int|h|\,d\mu\) when \(\int h\,d\mu\ne0\), and verify its modulus.