Product Integrals of Separable Functions

Why Separable Integrands Are Special

A function of two variables may depend on both coordinates in a complicated way. A particularly useful case occurs when its dependence separates: the function is the product of one function of $x$ and another function of $y$. Once the relevant measurability and integrability conditions hold, its integral over a product space factors into two ordinary integrals.

This is more than a convenient calculation rule. The formula is a direct application of Tonelli’s theorem or Fubini’s theorem, and the hypotheses matter. Nonnegative functions allow extended-valued integrals, while signed and complex functions require absolute integrability for the general factorization result.

Setting: Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be sigma-finite measure spaces, and let $\mu\otimes\nu$ be their product measure. Let $f:X\to[0,\infty)$ and $g:Y\to[0,\infty)$ be measurable, finite-valued functions. The associated function on the product space is $H(x,y)=f(x)g(y)$.

The Nonnegative Factorization Formula

Because $f$ and $g$ are measurable, their coordinate lifts $(x,y)\mapsto f(x)$ and $(x,y)\mapsto g(y)$ are measurable on the product space. Their product $H$ is therefore measurable as well. Tonelli’s theorem applies even if the integral of $H$ is infinite.

Theorem: For nonnegative measurable $f$ and $g$ as above, $$ \int_{X\times Y} f(x)g(y)\,d(\mu\otimes\nu)(x,y) = \left(\int_X f\,d\mu\right) \left(\int_Y g\,d\nu\right). $$ The integrals on the right may be infinite. In this extended-valued identity, a zero factor multiplied by an infinite factor is interpreted as zero.

Proof. Tonelli’s theorem states that the integral of a nonnegative measurable function on a product space equals either iterated integral, with values in $[0,\infty]$. For each fixed $x$, the inner integral is

\int_Y f(x)g(y)\,d\nu(y) = \begin{cases} f(x)\displaystyle\int_Y g\,d\nu, & f(x)>0,\\[6pt] 0, & f(x)=0. \end{cases}

Write $G=\int_Y g\,d\nu$. If $G<\infty$, the displayed inner integral is $f(x)G$ for every $x$, including where $f(x)=0$. Integrating over $X$ gives $G\int_X f\,d\mu$, with the result interpreted as zero if $G=0$ and $\int_X f\,d\mu=\infty$.

If $G=\infty$, the inner integral is infinite wherever $f(x)>0$ and zero wherever $f(x)=0$. If $\int_X f\,d\mu=0$, then $f=0$ almost everywhere, so this inner integral is zero almost everywhere and its integral is zero. If $\int_X f\,d\mu>0$, then $\{x:f(x)>0\}$ has positive measure: otherwise $f=0$ almost everywhere and its integral would be zero. The inner integral is therefore infinite on a set of positive measure, so its integral is infinite. In every case the iterated integral has the value stated in the theorem. Tonelli’s theorem identifies it with the product-space integral. $\square$

The convention concerning zero and infinity reflects the integral itself, not an ordinary arithmetic operation: a nonnegative function with integral zero vanishes almost everywhere. In particular, if one factor vanishes almost everywhere, the product function vanishes almost everywhere on the product space under the sigma-finiteness assumption.

Worked Examples

Worked Example: A Finite Product Space

Take $X=\{1,2,3\}$ and $Y=\{a,b\}$, each with counting measure. Define $f(1)=1$, $f(2)=2$, $f(3)=4$, and $g(a)=3$, $g(b)=5$. All sums here are finite, so the product-integral formula applies directly.

\int_X f\,d\mu=1+2+4=7, \qquad \int_Y g\,d\nu=3+5=8.

Summing the product over all six points of $X\times Y$ gives

\begin{aligned} \int_{X\times Y}f(x)g(y)\,d(\mu\otimes\nu) &=(1\cdot3+1\cdot5)+(2\cdot3+2\cdot5)+(4\cdot3+4\cdot5)\\ &=8+16+32\\ &=56. \end{aligned}

The separate integrals give $7\cdot8=56$, agreeing with the direct sum.

Worked Example: Exponential Functions on Infinite-Measure Spaces

Let $X=Y=[0,\infty)$, with Lebesgue measure, and set $f(x)=e^{-2x}$ and $g(y)=e^{-3y}$. The underlying spaces have infinite measure, but the functions have finite integrals:

\int_0^\infty e^{-2x}\,dx=\frac12, \qquad \int_0^\infty e^{-3y}\,dy=\frac13.

Tonelli’s theorem therefore gives the integral over the product space:

\int_{[0,\infty)^2}e^{-2x-3y}\,d(x,y) = \left(\frac12\right)\left(\frac13\right) = \frac16.

Thus an infinite measure for each coordinate space does not prevent the product integral from being finite. The relevant quantities are the integrals of the factors, not the total measures of the spaces.

Worked Example: Infinite Factor Integrals

Again take $X=Y=[0,\infty)$ with Lebesgue measure, but now let $f(x)=g(x)=1/(1+x)$. Each factor has infinite integral, since

\int_0^R\frac{dt}{1+t}=\log(1+R)\longrightarrow\infty \qquad\text{as }R\longrightarrow\infty.

The nonnegative factorization theorem says the product integral is infinite. This can also be checked on bounded squares: for every $R>0$,

\int_{[0,R]^2}\frac{1}{(1+x)(1+y)}\,d(x,y) = \left(\log(1+R)\right)^2.

These values grow without bound as $R$ increases. Since the full domain contains each square and the integrand is nonnegative, the integral over the full product space is infinite.

Signed and Complex Factors

For real-valued or complex-valued functions, the nonnegative theorem alone does not justify separating the integral: positive and negative parts, or complex cancellations, can obscure whether an integral exists. A useful sufficient condition is that each factor be integrable in absolute value. Here $L^1(\mu)$ denotes the measurable functions $f$ for which $\int_X |f|\,d\mu<\infty$, and likewise for $L^1(\nu)$.

Theorem: Suppose $f\in L^1(\mu)$ and $g\in L^1(\nu)$, with either real or complex values. Then $H(x,y)=f(x)g(y)$ is integrable on the product space and $$ \int_{X\times Y}f(x)g(y)\,d(\mu\otimes\nu)(x,y) = \left(\int_X f\,d\mu\right) \left(\int_Y g\,d\nu\right). $$

Proof. The absolute value factors as $|H(x,y)|=|f(x)||g(y)|$. Applying the nonnegative factorization formula gives

\int_{X\times Y}|H|\,d(\mu\otimes\nu) = \left(\int_X|f|\,d\mu\right) \left(\int_Y|g|\,d\nu\right) <\infty.

Thus $H$ is integrable. Fubini’s theorem now applies. For each $x$, the section integral is $f(x)\int_Y g\,d\nu$, since $f(x)$ is a fixed scalar. Integrating this expression over $X$ yields the asserted product. The same scalar argument holds for complex-valued functions, and the absolute-integrability calculation above ensures all the integrals are defined. $\square$

Worked Example: A Signed Factor with Cancellation

On $X=Y=\mathbb R$ with Lebesgue measure, let $f(x)=xe^{-|x|}$ and $g(y)=e^{-|y|}$. Both are integrable in absolute value. For the first factor, the integrand is odd, and its absolute integral is finite:

\int_{\mathbb R}|x|e^{-|x|}\,dx = 2\int_0^\infty xe^{-x}\,dx = 2.

Symmetry therefore gives $\int_{\mathbb R}f(x)\,dx=0$. For the second factor,

\int_{\mathbb R}g(y)\,dy = 2\int_0^\infty e^{-y}\,dy = 2.

The signed factorization theorem shows that the product integral is $0\cdot2=0$. This value comes from cancellation in the $x$-coordinate; it does not mean the product function vanishes everywhere. Its absolute integral is $2\cdot2=4$, so the product function is integrable and the cancellation is legitimate.

What the Hypotheses Do—and Do Not—Say

The theorem does not say that every iterated integral can be separated. For signed or complex integrands, absolute integrability is a safe, standard condition that permits Fubini’s theorem. Without it, an integral may fail to exist or a calculation may depend on an unjustified change in order. The assumption that the factors are individually integrable is also substantial: it ensures the product is absolutely integrable on the product space.

For nonnegative factors, Tonelli’s theorem needs no finite-integral assumption. It produces an identity of extended integrals, including infinite values, with the zero-times-infinity convention explained above. For signed or complex factors, use the integrable-function theorem rather than informally multiplying potentially undefined integrals.

Separable products are useful because they reduce a two-variable integral to two one-variable integrals, often making calculations and integrability checks straightforward. The factorization is a statement about product measure and its integration theorems; it does not itself assert anything about probability distributions or independence.

Check Your Understanding

Use the hypotheses, not just the algebraic form of the integrand, when answering.

What theorem permits the factorization for nonnegative measurable factors even when an integral is infinite?
Why does the factorization formula use a special convention when one factor integral is zero and the other is infinite?
If $f\in L^1(\mu)$ and $g\in L^1(\nu)$, what calculation verifies that $f(x)g(y)$ is integrable on the product space?
On $[0,\infty)^2$, what is the integral of $e^{-2x}e^{-3y}$?
In the signed example, why is it valid for the product integral to equal zero even though the product function is not identically zero?